User talk:Gill110951/Archive 1

Below are archived threads from my talk page (till 1 December 2010)

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Caroline Thompson

Hi Gill! You recently added a link to the article Caroline Thompson. The linked website belonged to another Caroline Thompson, who is listed in Wikipedia:Deceased Wikipedians. I'm a bit puzzled by this edit. Why would an article need an external link to the website of another person with the same name? A ecis^Brievenbus 21:52, 1 February 2007 (UTC)[reply]

I don't think I understand your answer. You mean to say that you have added the link to clarify to readers that the Caroline Thompson of the article is not the Caroline Thompson of Wikipedia:Deceased Wikipedians? I think the link only creates confusion, instead of preventing/solving it. A ecis^Brievenbus 21:34, 8 February 2007 (UTC) PS. Kunnen we niet beter gewoon in het Nederlands praten? ;)[reply]

Ja we kunnen best in het Hollands verder... Kijk: vele mensen zoals ik willen meer weten over Caroline Thompson (Scientist) en zoeker haar op in Wikipedia. Daar vinden ze iemand anders. Het is toch aardig om ze te vertellen waar ze moeten zijn? En zeker als er een wikipedia bladzijde komt, een deze dagen, over "mijn" Caroline Thompson. Die wil ik een deze dagen wel opzetten ... Als die er is, hebben we een "disambiguation page" nodig. Dus: ik vind niet dat een verwijzing naar een andere Caroline Thompson, op een bescheiden plek in het artikel, enige verwarring saait. Het is toch niet zo gek dat er nog meer mensen zijn die zo heten en dat als je een andere opzoekt dat je vriendelijk doorverwezen wordt? Gill110951

Sorry dat ik nu pas weer reageer. Ik ben het niet helemaal met je/uw redenering eens. Het punt is namelijk dat de Wikipedia:External links guideline zegt dat de link moet linken "to a page that is the subject of the article or an official page of the article subject." Verder staat er bij de Links normally to be avoided: "Sites that are only indirectly related to the article's subject: it should be a simple exercise to show how the link is directly and reciprocally related to the article's subject. This means that there is both a relation from the website to the subject of the article, and a relation from the subject of the article to the website." De Caroline Thompson van het artikel is niet de Caroline Thompson van de link, maar een toevallige naamgenoot. A ecis^Brievenbus 00:39, 25 February 2007 (UTC)[reply]

What should be linked: "Sites with other meaningful, relevant content that is not suitable for inclusion in an article". Ik zie niet dat er staat dat ELKE link MOET linken naar "a page that is the subject of the article or an official page of the article subject"Gill110951 02:40, 1 March 2007 (UTC)[reply]

Geachte Dr. Gill, Ik denk dat iemand een keer forensic statistics moet schrijven. Het is vreemd dat er wel een kategorie bestaat, maar geen hoofartikel. Zie category:forensic statistics. Misschien wilt u dat doen als u tijd hebt, want ik denk dat u, in tegenstelling tot ik, er voldoende van afweet en wellicht ook toegang hebt tot goede bronnen. Bij voorbaat dank. Andries (talk) 15:34, 3 February 2008 (UTC)[reply]

Reflist

Hi, I wanted to let you know that if you're using the ref tags to create citations you need to place the Reflist Template in the article. If you don't it may appear that your article is unsourced.--Torchwood Who? (talk) 05:56, 30 March 2008 (UTC)[reply]

Miscarriages of justice

Ha! Ik zie dat je op precies dezelfde pagina's actief bent als ik... Misschien kunnen we een edit-war beginnen? :-D --Hippalus (talk) 13:31, 5 April 2008 (UTC)[reply]

Waarom maak je geen wiki voor Ton Derksen? Er wordt al veel naar hem verwezen. Dat versterkt het net van links rondom Kevin, Lucia, etc. --Hippalus (talk) 13:47, 5 April 2008 (UTC)[reply]

WikiProject Physics participation

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On 2008-06-25, the WikiProject Physics participant list was rewritten from scratch as a way to remove all inactive participants, and to facilitate the coordination of WikiProject Physics efforts. The list now contains more information, is easier to browse, is visually more appealing, and will be maintained up to date.

If you still are an active participant of WikiProject Physics, please add yourself to the current list of WikiProject Physics participants. Headbomb {ταλκ – WP Physics: PotW} 16:08, 25 June 2008 (UTC)[reply]

Re: answer on de Berk

A more appropriate place for your note is the article talk page, not my user talk page. Thanks, mo ^talk 01:03, 13 November 2008 (UTC)[reply]

de Berk

Thanks for your answers. Often on Wikipedia, a controversial article will draw a variety of people with, to put it kindly, very strong opinions on the matter. They also draw conspiracy theorists (Jonestown comes to mind - there is an anonymous editor who regularly inserts that the entire Jonestown event was a US CIA assssination), "nutjobs", people with particular extreme biases and those that "climb on their soapbox" and deliver a new and contrary viewpoint about the topic. In some cases, "soapies" are extremists (Charles Manson shouldn't be in prison, he didn't kill anyone) whose basis of reference isn't productive or even in some cases, unsupportable. Wikipedia draws them all. Most often, they are operating from no basis of knowledge.

Unfortunately, this has led to a healthy dose of skepticism in most regular editors when someone brings in material of a new or controversial nature or is contrary to what leading sources say. What complicates the de Berk case is the unfamiliarity editors have with the justice system in your country and the language difference. Which leads us to now. I want to apologize if anything that was said or assumed offended you, it wasn't the intention. We were just working from a skeptic's viewpoint based on past experience with controversial material. Hopefully, all the issues in the de Berk case will be resolved. Wildhartlivie (talk) 20:18, 16 November 2008 (UTC)[reply]

Monty Hall problem

I found your comments on the Monty Hall problem article very interesting. I been arguing that the paper by Morgan et al, on which much of the structure of the article is based, is seriously flawed and that it generates a result that does not reflect any likely real scenario. I, with help from Nijdam, have been trying to do better myself on this page [[1]]and wonder if you would mind taking a look. Martin Hogbin (talk) 09:53, 27 March 2009 (UTC)[reply]

Thanks for your comments. Like Nijdam and Rick, I am familiar with the Paper by Morgan et al and I fully understand its conclusions, which are what you state in your conditional answer to the problem. Where I disagree with Rick and Nijdam is that I believe that, although not actually wrong, the Morgan paper is seriously flawed in many ways resulting in its being next to useless as a basis for an explanation of the Monty Hall problem. I am very happy to discuss this if you have time. Martin Hogbin (talk) 09:55, 28 March 2009 (UTC)[reply]

radical alterations to the intro to quantum mechanics article

Hi,

A new editor has unilaterally made many drastic changes to the article Introduction_to_quantum_mechanics to which you have made contributions. I do not think that the changes are desirable. I do not want to start an edit war. Could you please have a look at it? Thanks. P0M (talk)

I saw your message. Since I wrote the message above a couple of other users have reverted the wholesale dumping of the own article, the new editor put his own stuff back twice. There was more intervention and the guy took his stuff off and made a fork out of it.

I have been working on the article for about 3 years, but I never got beyond Heisenberg because I hit a snag that I have only just unraveled. If you are still interested, take a look at the article again in a couple of weeks. I hope that by that time I will have cut out a lot of stuff that has turned out to reflect little side trips. When I figured out what Heisenberg had actually done it indicated that some things that looked trivial in the beginning are actually very important. The rest can safely be cut without misleading the inquiring reader. P0M (talk) 14:48, 23 June 2009 (UTC)[reply]

Mediation of Monty Hall problem

A request for formal mediation of the dispute concerning Monty Hall problem has been filed with the Mediation Committee (MedCom). You have been named as a party in this request. Please review the request at Wikipedia:Requests for mediation/Monty Hall problem and then indicate in the "Party agreement" section whether you would agree to participate in the mediation or not.

Mediation is a process where a group of editors in disagreement over matters of article content are guided through discussing the issues of the dispute (and towards developing a resolution) by an uninvolved editor experienced with handling disputes (the mediator). The process is voluntary and is designed for parties who disagree in good faith and who share a common desire to resolve their differences. Further information on the MedCom is at Wikipedia:Mediation Committee; the policy the Committee will work by whilst handling your dispute is at Wikipedia:Mediation Committee/Policy; further information on Wikipedia's policy on resolving disagreements is at Wikipedia:Resolving disputes.

If you would be willing to participate in the mediation of this dispute but wish for its scope to be adjusted then you may propose on the case talk page amendments or additions to the list of issues to be mediated. Any queries or concerns that you have may be directed to an active mediator of the Committee or by e-mailing the MedCom's private mailing list (click here for details).

Please indicate on the case page your agreement to participate in the mediation within seven days of the request's submission.

Thank you, Rick Block (talk) 02:51, 14 January 2010 (UTC)[reply]

Request for mediation accepted

A Request for Mediation to which you were are a party has been accepted.
You can find more information on the case subpage, Wikipedia:Requests for mediation/Monty Hall problem.

For the Mediation Committee, Seddon ^talk and Xavexgoem (talk) 00:45, 22 January 2010 (UTC)[reply]

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Original research

Hi. I think your contributions to the Monty Hall problem are really well-written, mathematically correct, and interesting. But I want to make sure you are aware that everything must be sourced. This edit for example seems to be original research and lacks references. Are we on the same page about this? Wikipedia, unlike arXiv, does not publish original thought. Andrevan@ 04:58, 5 February 2010 (UTC)[reply]

Thanks! I edit bits of wikipedia often meaning to come back and add references later. I absolutely agree with you on the basic principles here. I see myself as a self-invited guest on the wonderful wikipedia show, and of course I try to behave properly in my host's public forum. I'm also human, forgetful, change my mind from time to time, love arguing, have strong personal points of view, get annoyed by what I perceive as stupidity, ignorance etc etc...

The edit you refer to is a nice point. It is not original research. I learnt it on wikipedia from Boris Tsirelson. It is common knowledge under mathematicians. It's plain simple logic. One cannot give a reference. One can say "a professional mathematician would argue as follows" and refer to wikipedia talk.

I put in the text there hoping that the other editors would a) understand it, b) improve the text, c) help search out for a reliable source.

It is meant to be universal common human logic! Common sense! How can you give authoritative sources to common sense? I hoped that the common sense would be recognised and adopted by the other editors. If it wouldn't be adopted, they'ld edit it away again. Fine by me. It's a dialectic process, looking for the truth together. Gill110951 (talk) 06:21, 7 February 2010 (UTC)[reply]

I came here to make roughly the same point and see that Andrevan has already broached the topic. I hope you don't mind if I add my own thoughts about this. First, let me make it perfectly clear that I think Wikipedia is extremely fortunate that experts such as yourself choose to do any editing here. Wikipedia's good fortune extends to the efforts of not just experts but to all of Wikipedia's volunteers including editors, administrators, bureaucrats, mediators, arbcom members, and even developers (!). However, there's a particular danger experts editing Wikipedia need to be aware of relating to the fundamental content policy of Wikipedia:Verifiability which is that all content must be verifiable according to a reliable published source. This is the same danger encountered in editing an article about yourself - see WP:AUTO - and amounts to the difference between adding or changing content based on what you know as opposed to what you can reference (to a reliable source). I believe this particular issue is one of the basic problems creating difficulty with the article on the MHP. Lots of people (not just experts!) think they know with absolute certainty The Truth (a humorous essay, but not without a point) about the Monty Hall problem - and, once again, to be absolutely clear I'm not in any way implying I disagree with the correctness of anything you've said about the MHP (or anything else).

I've suggested to Nijdam the analogy that Wikipedia articles should be considered to be what in the academic world would be more like comprehensive literature surveys (what you might put in the first chapter of a thesis) than an article you might write for publication. Articles here should say what others have said about the topic, and why, and then stop - with no value judgment implied.

As an expert, one of the main things you can contribute here is your knowledge of the sources (as opposed to your knowledge of the topic itself). Articles here are not meant to be "correct" or "incorrect" - they're meant to be a neutral accounting of what the sources say, fairly representing the prevalence of opposing views if there are opposing views. In math, I don't think the notion of equally valid opposing views comes up very often (I mean, one or the other must be correct, right?) so I suspect deferring to what the sources say is not a natural impulse for a mathematician. This comes up more in other sciences, and comes up constantly in non-scientific fields - but from Wikipedia's viewpoint the solution is always the same, i.e. say what the sources say. One point of this is to avoid any necessity for choosing sides where there are conflicts. Both sides can (should) be able to agree what the sources say.

If you already knew all of this already, my apologizes for belaboring the point. I know you're not a newcomer, but the first bullet point on my user page under "what I think" has for a very long time been

Wikipedia's newcomers are undoubtedly ignorant of many local rules and conventions, but some of them are experts and we desperately need them. Don't piss them off. Most of them mean well.

-- Rick Block (talk) 22:00, 7 February 2010 (UTC)[reply]

Thanks Rick, absolutely agreed. Except that you say "In math, I don't think the notion of equally valid opposing views comes up very often (I mean, one or the other must be correct, right?)". Of course, a theorem is true or false, a proof is correct or incorrect. But we are talking about mathematizations, about mathematical models. Mathematical descriptions of real world situations. That is a matter of taste, of debate. A matter of adequacy (more or less), not of right or wrong. There is not one and only one true way to answer Marilyn vos Savant's question. Of course, if like the bishop of Rome you claim to infallible knowledge, you might decree that so-and-so's mathematical formulation was the only true Monty Hall Problem. Well, and then I would become a protestant. It might lead to a schism, mutually excommunicating one another, and to a fork of wikipedia! By the way, creative mathematicians prefer to spend their time thinking creatively rather than looking for reliable sources; and they prefer to write reliable sources if they only find shaky ones. Also, one hopes that referees in peer reviewed journals will tell you if someone has already done it before, one cannot know everything. So: sorry, I am not going to be the greatest help here in finding you authoritative literature references, though I do try. Have you tried Googling "game theory" + "Monty Hall" yet? Gill110951 (talk) 22:15, 7 February 2010 (UTC)[reply]

The game theory aspect of the MHP has not been well represented in the article (it's been more or less absent, except for the briefest of mentions in the "Other host behaviors" section). If there is a significant body of literature on this it's an area that we should probably expand - although I highly doubt this is at all related to the mediation that's underway. It's somewhat dated by now, but Barbeau's literature overview is pretty comprehensive. It's curiously missing the Nalebuff reference (that IS in the article) which I think is the root of many of the game theory references. I've read someplace [2] that economists for a long time thought Nalebuff had originated the problem (!). -- Rick Block (talk) 04:52, 8 February 2010 (UTC)[reply]

I too read an economics and game theory paper that said Nalebuff originated the problem. His article however seems to be more about calculation of a conditional distribution than about the game theoretic content of Monty Hall. However: he did bring Monty Hall to the attention of the game theorists.Gill110951 (talk) 14:32, 10 February 2010 (UTC)[reply]

What is the Barbeau reference, @Rick Block ? Gill110951 (talk) 08:02, 12 February 2010 (UTC)[reply]

Either of the Barbeau references from the article, e.g. Monty Hall problem#refBarbeau1993. The 1993 one is slightly more detailed (but earlier). The 2000 one is a book. -- Rick Block (talk) 14:30, 12 February 2010 (UTC)[reply]

A question about game theory

In reading and participating in the MHP discussion I have learned more about probability then I ever intended, or thought possible. I must say it has been an interesting ride. I also noticed that Rick brought up Game Theory in the posts above. That reminded me that I wanted to ask someone knowledgeable about the bit under 'Variants' that says "The host is rewarded whenever the contestant incorrectly switches or incorrectly stays = Switching wins 1/2 the time at the Nash equilibrium". This seems to imply that if the producers want to make sure the contestant never gets a 2/3 chance to win the car all they have to do is give Monty a cookie if the contestant chooses wrong. Which is counter intuitive to say the least. Can you explain this in more detail? Thanks, Colincbn (talk) 12:44, 11 February 2010 (UTC)[reply]

@ Colincbn, that's really interesting, I hadn't noticed that one. I will have to go back to the sources to get the details exactly right, and then do the calculations to check the answer, and hope that doing all this give me a good intuition. If the rewards to the host don't match the losses to the player and vice-versa we don't have a zero-sum game. The minimax theorem is no longer true, but there is a weaker kind of equilibrium called Nash equilibrium. Non zero-sum games have counter-intuitive properties, since they are games where the contestants could actually both win more by collaborating. But since they don't know if the other will collaborate they both choose to do something else, which results in them both losing... To say it differently, if you know the other guy is a collaborative type you win more by not collaborating. But if the other guy is suspicious he will realise this...

Your remark about the cookie changes the game into a three-party game, and moreover a non zero-party game. Monty Hall wants a cookie. The producers want to hold on to the show's cadillac. The player wants to ride it home. (I'm assuming the producers have an infinite store of cookies so the value of the cookie to them is zero. I take it you want the value of a cookie to Monty Hall to equal the value of a cadillac to the producers to be equal to the value of a cadillac to the player. Now we can sit down and look up the Nash equilibrium on wikipedia and do the math).

PS for me it has been a roller coaster ride too. I have changed my mind dramatically several times, learnt a lot, had a lot of fun, hopefully made more friends than enemies... Gill110951 (talk) 08:09, 12 February 2010 (UTC)[reply]

One point that I think you should make more strongly in your papers is the similarity between the distribution of initial car placement and that of the hosts legal door choice. Morgan's main point is built on treating these two distributions inconsistently and, in places in your paper, you seem to do the same. At the bottom of page 5 you say the player can never do worse by switching. This is true only if the car is not initially placed uniformly at random. Martin Hogbin (talk) 10:36, 28 February 2010 (UTC)[reply]

I think @Martin Hogbin you are missing something, namely that I suppose that the player initially chooses his door uniformly at random. This is an important part of the game-theoretic solution. However the car is initially placed, and however the quizmaster chooses a door to open, it follows merely from the player actively making this randomization, that his (unconditional) chance of getting the car (by switching) is 2/3. That's the meaning of a saddle-point in this two-party game. The player's minimax strategy guarantees his winning the car with probability at least 2/3. The quiz-show-team's minimax strategy guarantees they lose the car with probability at most 2/3. In both cases this guarantee is given, whatever the strategy of the other party. Gill110951 (talk) 10:02, 2 March 2010 (UTC)[reply]

Yes, I think missed the plot completely, I misread what you wrote.

The point that many people do not get, but is not terribly relevant to your paper, is that if the car is arbitrarily placed and the player chooses a door uniformly at random, and the host has a fixed set of door opening parameters, the probability of winning by switching is always 2/3 and does not depend on the host's door opening policy. Martin Hogbin (talk) 11:41, 2 March 2010 (UTC)[reply]

OK. Though I don't know what you mean by "if the car is arbitrarily placed". Do you mean "however the car is placed"? If so I agree, and that is exactly my point; as long as the player uses their freedom to randomize, switching is a great policy for them.Gill110951 (talk) 10:46, 4 March 2010 (UTC)[reply]

That was what I meant. Martin Hogbin (talk) 14:32, 6 March 2010 (UTC)[reply]

Gill, do you know the authors of the pigeon paper? I would like to contact them with a possible reason that humans do worse than pigeons. Martin Hogbin (talk) 14:32, 6 March 2010 (UTC)[reply]

Have a look here, for example: [3] Herbranson WT, Schroeder J., Department of Psychology, Whitman College, Walla Walla, WA 99362, USA. herbrawt@whitman.edu Regards, -- Gerhardvalentin (talk) 14:52, 6 March 2010 (UTC)[reply]

Gerhard, I think you maybe misunderstood the meaning of my question (maybe Gill did too). I was hoping he might be able to help me contact the authors. Martin Hogbin (talk) 23:24, 6 March 2010 (UTC)[reply]

I misunderstood, indeed. I don't know them. I'm interested in your question and their answer!Gill110951 (talk) 07:22, 8 March 2010 (UTC)[reply]

My suggestion would be that humans have a higher expectation of success. As we all know, if the set up is uniform at random, the best you can do is succeed 2/3 of the time by always switching. Pigeons, and I guess most other animals, probabily have lower expectations of success that humans, winning most of the time is generally a good result in nature.

A distinguishing feature of human behaviour is to expect to achieve a very high success rate in our endeavours. Most people like to think of this as 'understanding how things work', although, as a physicist, I have learned that this a somewhat arrogant view and in fact we are just trying to do rather better that pigeons in predicting the future.

Humans dislike random events and we are programmed to try to discover patterns in nature. When presented with the MHP experiment, I would guess that most people believe that they can somehow 'work out the pattern' and get to win every time. Every year, millions of people go to places like Las Vegas and try to beat games based on random events where it is known that on average you will lose. Hope, and faith that there is an answer, is what keeps them going back. I am sure pigeons would quickly learn not to return.

By the way, you seem to have given up with the MHP article along with most other dissidents. Without the support of you all, I guess I will too. Martin Hogbin (talk) 12:39, 12 March 2010 (UTC)[reply]

I have said what I have to say about Monty Hall. It is a pity the mediator isn't a bit smarter. This is not something which can be settled by using medieval judicial procedures. It needs intelligence and taste. But I will continue to respond to rubbish, don't worry! Gill110951 (talk) 13:45, 19 March 2010 (UTC)[reply]

Opinion on MHP sought.

Gill, Your opinion on a current argument about the MHP would be welcome. Martin Hogbin (talk) 14:41, 1 June 2010 (UTC)[reply]

Hi Martin, I just made some revisions of the Monty Hall page. Cleaned up the mathematical solution to the conditional problem, added the game theoretic solution to the game theoretic version, cleaned the solution of the unconditional problem. I hope in a way which demonstrates a totally neutral POV. Still have to add lots of references. Gill110951 (talk) 06:58, 19 June 2010 (UTC)[reply]

MHP

Please refrain for such sweeping changes without consent. The old formal treatment is pretty much a standard treatment in textbooks and has been part of article the in slightly different version since it became featured. You are free to add additional material or approaches without consent but you cannot simply delete or modify old established content in such a manner without seeking consensus first.

Please also note that this is general article on MHP for a large variety of audiences not just about what you currently consider interesting about the problem or what your personal view of the best formal approach might be. Regards--Kmhkmh (talk) 13:19, 24 June 2010 (UTC)[reply]

That was a sweepling simplification which should have been made years ago. Gill110951 (talk) 13:21, 24 June 2010 (UTC)[reply]

I seek concensus by putting forward my edit and then I expect to get discussion about it. Not reversion. Gill110951 (talk) 13:22, 24 June 2010 (UTC)[reply]

I don't see that way. A common textbook treatment is definitely useful for the article (and I suspect that at least some of the other editors see it that way as well). That is no argument against having the odds approach in the article (on the contrary that's welcome addition), but I see no reason to use as replacement of the old treatment. Again I'd like to point out that this article is not about your personally favoured approach or interests regarding MHP.

However should all other editors agree with replacing the old formal solution by odds approach rather than having both, then I would have no objections. I'd still consider it a somewhat bad choice, since I consider the old formal treatment as useful, however it is not about my personal preferences either, i.e. should all other vote for its removal, so be it.--Kmhkmh (talk) 13:35, 24 June 2010 (UTC)[reply]

I disagree entirely. The common textbook formulation is out of date, not useful, not illuminating. Wikipedia shouldn't use outdated and useless mathematical formalisms when there are much superior alternatives. Try reading my text and understanding it. Gill110951 (talk) 13:42, 24 June 2010 (UTC)[reply]

PS or if you really think the old junk should be back, put it back in there. I think that nobody reads it and nobody bothers to look at it. That's why it's been there unchanged all these years.

EXACTLY because this article is for a wide and non specialist audience, the mathematical proof should be made as accessible as possible, and that is by using the "odds" version of Bayes. Which by the way is how many authorities do it in standard texts. So don't delete without studying first. And be constructive, not destructive. We want concensus, don't we? We want a beautiful article don't we? I am not pushing any POV except that as far as mathematics is put into such articles it should be first-rate and maximally accessible.

AND I'll tell you another thing. The wikipedia page is not there to promote any particular person's point of view. We (wikipedia editors) have reached the concensus, I thought, that there are various different formalizations of the Monty Hall problem, which have been discussed and promoted by various authorities. There is nothing "unmathematical" or "unformal" about the simple solution to what we are now calling the unconditional problem. On the other hand, there is a lot of criticism as to whether the conditional problem belongs anywhere except perhaps as an exercise in Probability 101, but still, it get's its own place under the sun. The relations between the different problem versions and different solutions are fascinating and illuminating to everyone who is interested in this article and they should be brought into the open. My edits are aimed at simplifying mathematics without making any concesions to mathematical correctness, and at reaching concensus by emphasizing links and connections between the various approaches.

I'm not sure what to answer to that. What you are claiming is plain wrong and that's obvious to see for anybody consulting the discussion page and archives. Frankly I find it utterly ridiculuous to argue with another mathematician on such a level.--Kmhkmh (talk) 14:14, 24 June 2010 (UTC)[reply]

I hope you have the same aims as I do

The book formulation is neither out of date nor is it junk. And again you seem to confuse your personal views or interests with the rest of the world.

If you claim the old formal solution was never challenged/discussed/read then that only shows that you never bothered to read old discussions or check history list or worse than you intentionally ignored them.

If you declare your personal views unilaterally as the best, then this is of course POV pushing.

And frankly from my perspective may you should take a step and look at your overall editing behaviour on this article (from pushing your own preprint papers to making large changes without seeking consent and even insisting on it, when the lack of consent was brougt to your attention). In particular from a professional mathematician, I'd expect better and that he'd be able to distinguish his personal views/interests/preferences from the bigger picture.

I'm not going to revert this again to avoid an edit war and leave the correction to others.

--Kmhkmh (talk) 14:09, 24 June 2010 (UTC)[reply]

I for one will not be 'correcting' it. Martin Hogbin (talk) 14:35, 24 June 2010 (UTC)[reply]

I don't think that comes as a surprise nor was I referring to you but possibly other than me nijdam, rick, glopk or coffetotheorem as far as current editors are concerned.--Kmhkmh (talk) 15:21, 24 June 2010 (UTC)[reply]

@Kmhkmh, if you honestly think that I really am doing this to push my own recent publications you ... don't believe in my good faith. I learn on wikipedia about Monty Hall, about probability, about doing applied maths, about communicating maths to nonmathematicians. I thank you all in my publications which are based on what I learnt here. But if you prefer, in future I'll write "except @Kmhkmh". Still wondering who you really are, BTW. It would be nice to get to know one another on non-anonymous terms but face to face, as it were. You know very well who I am. What have you got to hide? I don't bite.

I switch sides from time to time when I come to really appreciate the other's point of view. I state my current opinion clearly so that everyone can see what it is and I'm always prepared to modify it later. I often did. I think a real mathematician is always learning and especially in the tricky step from roughly, verbally, stated real world problem to formal math problem with unique answer (or to the discovery that there isn't a good one) one uses good taste, experience, wisdom, makes compromises... hence a wise person will often change their minds as they get deeper into a problem.

I like to state an extreme version of my current point of view in order to get discussion and refinement. It's called the socratic method. It's how we come to the truth, by engaging in a discussion, in which hyperbole and understatement and not hiding emotions can also play a role. Gill110951 (talk) 14:49, 15 August 2010 (UTC)[reply]

Mediation resumes

The mediation of the MHP case has re-started. If you wish to participate, would you be willing to check in on the case talk page? Note that the mediators have asked that participants agree to certain groundrules. Sunray (talk) 06:52, 11 August 2010 (UTC)[reply]

Comments to mediation comments by me

Nijdam's straw man of Game Theory

@Nijdam put up the following issue in an exercise to present opponent's views in as sympathetic way as possible

Nijdam: MHP has to be treated with game theory

Most people consider the MHP a psychological game, and hence it may best be treated with simple game theory.

I respond as follows.

@Nijdam, isn't this quite a straw man you're putting up here? I'm not aware of anyone in the world that thinks this, and I cannot imagine how you could possibly think they exist. Suggestion for rephrasing (but of course, it's your perceived issue of a perceived opponent):

Nijdam: Game Theory gives insight to MHP

Some people (perhaps many, at first exposure to MHP) start instinctively, it seems, to think in terms of psychology. Is Monty Hall trying to trick me? If I do that he might... Does he know that I know that he knows...? Correspondingly, one might say, it has also been treated mathematically with elementary game theory in the economics / decision theory / game theory world. Correspondingly, vos Savant did not ask 'What is the conditional probability?' but 'Should you switch?'

Vos Savant asked for a decision, please note, not a probability!

Now go on to try to mention the useful insights. Let me give you a little help

From game theory we learn that the wise player would randomize his initial choice secretly at home in advance, in order to get the best of all possible worlds; to be totally free of any mathematical "assumptions", how academically natural or conventional they might seem to be.

You're supposed to present a perceived view of a perceived opponent with as much empathy as you can muster.

It's about being on a game show. But we're not nuts. You're not allowed to ask advice when you stand there on the stage! You'ld better ask advice in advance.

Maybe vos Savant should have written "You're going to be on a game show. If you answer the silly questions correctly, you'll be faced with a choice of three closed doors..."

I'm not trying to put words into your mouth here, @Nijdam me old mate, just trying to help things along. Looking forward to beer with you soon! Gill110951 (talk) 12:50, 15 August 2010 (UTC)[reply]

Monty Hall is a Statistics 101 Problem [I don't think!]

You learn in Statistics 101 not only how to solve a formal problem but also how to decode a verbal problem description into a formal problem. If the wording is ambiguous you have to figure out what are the assumptions which you have to make in order to solve the problem with the tools that the teacher has provided you. Since the MHP is an example/exercise in the early chapter on Bayes' theorem in discrete probability, and since you cannot do this kind of probability without writing down a sample space explicitly and making sure that all elementary probabilities are defined, you have to fill in a lot of "gaps" in Marilyn vos Savant's words in order to compute the right answer "2/3" to the right formal question "what is the conditional probability that...".

In class, and in textbook examples, the teacher/writer always uses a uniform distribution when no distribution is mentioned, and when she said "random" she means "uniform at random". This means that the correct answer is "2/3" and it is completely clear what is the correct mathematical question to which this is the correct answer.

More advanced material giving variations to the just described standard MHP as it is called in reliable sources, or addressing the messy complications which arise when you don't make the right assumptions, or using fancy mathematical tricks like "symmetry", "invariance" or "odds formulation of Bayes theorem", let alone dragging in von Neumann's minimax theorem which is never treated in elementary statistics texts and which only a few professional mathematicians and economists know, hardly belongs in Wikipedia

Whether the MHP is to be solved using subjectivist/personal interpretation of probability or by an objective/frequentist interpretation is a purely philosophical issue totally irrelevant to what you should do when you are on a game show.</

The best/correct way to get from the correct question to the correct answer is by bare hands computation from first principles, ie using the definition of conditional probability. This route is infallible and requires no further knowledge or thought. You just have to be careful in the calculations and, of course, careful in writing down the right words to justify each step.

It turns out you didn't need to know Bayes' theorem at all, even though the problem is in a chapter on that topic; you just needed to know the definition of conditional probability and the law of total probability. But it is useful to do the whole thing over again using Bayes' formula for a conditional probability in a discrete probability space in all its gory detail, even though this requires a lot of skill (and time) writing imposing math formulas on wikipedia pages.

To @Kmhkmh. I am not just trying to formulate how I see @Nijdams's point of view but also to merge it with the points of view - as I see them - of quite a lot of editors. Of course it is a legitimate point of view. I try to express it as sympathetically as I am able. Of course I have a very different point of view. I am not trying to write it down at all, now. Maybe you care to try. Someone else should try that, no-one tried yet.

I think we're missing a meta-issue

The fact is that vos Savant's short and semantically ambiguous verbal reformulation of her correspondent Craig Whitaker's pretty garbled question lifted the MHP from the puzzle corner in elementary statistics teaching and blasted it into the public domain and into the domains of psychology, economics, game theory, ethology. Vos Savant's little note (co-)generated a vast and continually evolving social-cultural eco-system which is not "owned" by any particular narrow professional grouping.

I think that the only thing that everyone agrees on is that Vos Savant's original words - supposedly a quote of Whitakers' - are a focal point to the MHP and should be a point of departure for the article. Yes! That's it! The question is not: What is the answer? (The answer is well known: "Switch"). It's: How to make sense of the question, as step towards understanding why/when/if the answer really is the right answer. We have to admit (just by consulting reliable sources) that there is not One True Way (though every editor may have a favourite).

The ambiguity of vos Savant's words means that even within mathematical sciences, and even within probability and statistics, there are different formalizations and routes to sound answers, with good common sense behind them, reliable sources to back them up, and giving different valuable and interesting insights.

IMHO the article we all care so much about must reflect the fact that MHP has escaped from the elementary statistics classroom into the wild, where it has been happily mutating and proliferating wildly for years, and hopefully will continue to do so for years into the future. Great! That's why our article is a Much-Consulted Wikipedia Article and must become a Great Wikipedia Article again.

I want to make a plea to celebrate and reflect the richness of the phenomenon which we are writing about in order to produce a balanced article which addresses the diverse needs of the diverse people who will come to it. The article should be about the meta-MHP, about the MH Problem - Problem; and about the MHP phenomenon and MHP eco-system. About how people have attempted to make sense of the question and to justify their answer and what others have done with that. On the way, we find out why unaided individual pigeons mostly find the right answer, but unaided humans mostly can't. If we accept this then there is resolution for a great deal of issues (they become non-issues), room for a several "formal" re-formulations, and room for several solution methods. Gill110951 (talk) 11:54, 15 August 2010 (UTC)[reply]

I agree with all that you say provided that the first part of the article deals properly with the 'simple puzzle' aspect of the problem. It should show readers why the answer is 2/3 and not 1/2, and convince them why this is so. It should also explain why the host's knowledge of where the goats are matters. This is what makes the problem so famous and what all the questions to vos Savant and WP talk pages were all about. One thing not needed at the early stage is anything that makes the problem and solution harder to understand. It is clear to me that it was Selvin's intent when he first formulated the problem that it should be a simple, unintuitive, probability problem. I sense a certain academic arrogance in some people: 'This basic problem is much too easy for me, I need to make it more complicated before it becomes interesting'. Martin Hogbin (talk) 13:32, 15 August 2010 (UTC)[reply]

Absolutely, Martin! And in a moment I'll write some stuff here about Hellpimp's brilliant recognition of the elementary game theoretic nature of the problem and why the answer is indeed 2/3. Gill110951 (talk) 14:43, 15 August 2010 (UTC)[reply]

The problem is that some people will say that Hellpimp's solution does not address the question 'as asked'. Of course, we all know that it perfectly well addresses the question that Whitaker meant to ask (and maybe actually did ask before vos Savant added to it, and arguably did ask anyway). If you were going on a game show, why would you only ask about possible case? Martin Hogbin (talk) 15:44, 15 August 2010 (UTC)[reply]

That is because the question as asked needs to be mathematized and if people ask stupid questions a good mathematical scientist will not just answer the stupid question but will also try to help the questioner might rethink the original question in view of some further information. "The question is: what is the question". Anyway, please let's take it step by step. 1) Do you agree that Hellpimp would makes a valuable and possibly unique contribution to almost any late night discussion at the bar on the MHP? With a genuinely key insight: it pays to start by choosing your own door uniformly at random!!!! 2) Might we not be able to convince a few other co-editors that this is interesting and true and different from the usual stories which the forces of reaction want to keep us stuck with? 3) Since we have the mathematical economics game theory literature to show that this informal solution corresponds to a reliable source formal solution, and since it is so patently common sense true, doesn't it get a chance to see the light of the day somewhere on the wikipedia page, one day, after mediation has cut some of the bigger and more obnoxious ego's down to size? I am not out to dictate the MHP page. I do feel a strong need to correct blatant errors and do some justice to the richness of the topic, and I do get bloody pissed off by what comes across to me as a combination of arrogance and stupidity and abuse of power. Now, everyone has blind spots, me included. Also, some people come over arrogant and bossy when they feel threatened. Too bad. But we have got some sensible mediators now so let them let arrogant bossy behaviour cause loss of face by being told off by the mediators. By the way, one and the same person can come across as / behave as arrogant and bossy on one day, and be the best of constructive collaborators on another. Myself included, no doubt about that. Truly seeing people always of necessity have blind spots, but they are not always the same blind spot, because they move their head around and walk about as well as just staring in fixed position at a fixed view. Let's not take it all so seriously. It's not about your or my ego or Glkanter's or Rick Block's. It's about having fun and learning and teaching on Wikipedia. Gill110951 (talk) 18:18, 15 August 2010 (UTC)[reply]

Hellpimp had the answer all along - 2/3, game theory

Hellpimp on the MHP talkpage, way back

Hellpimp wrote way back, November 2008, according to Martin Hogbin, the following.

Maybe this isn't any simpler to understand, but this is what I thought of as I slowly convinced myself the answer really is 2/3. Hellpimp (talk) 06:41, 7 November 2008 (UTC)

The answer to what??Nijdam (talk) 09:17, 16 August 2010 (UTC)[reply]

As I understand it, if a contestant chooses a door at random,

We do not know how the contestant has chosen.

the remaining doors will both have goats 1/3 of the time and a goat/car mix 2/3 of the time.

He seems to say: P(C=x|X=x) = 1/3.

1/3 of the time the remaining doors both have goats, so by switching you would lose on this 1/3 of occasions.

At this stage you are not yet offered to switch, so what does he mean to say?Nijdam (talk) 09:17, 16 August 2010 (UTC)[reply]

But on the other 2/3 of occasions the host will pick the door with a goat to avoid the door with the car. Since this is the case 2/3 of the time, by switching to the other door not chosen by the host you will succeed 2/3 of the time.

This guy got it right. He says (my capitalization, sorry, I'm excited):

IF A CONTESTANT CHOOSES A DOOR AT RANDOM!

You HAVE to choose YOUR INITIAL door at random YOURSELF, in order to get the guarantee of 2/3!!!!

This guy is a born winner (ie instinctive game theorist who knows you have to randomize to optimize)

Gill110951 (talk) 15:00, 15 August 2010 (UTC)[reply]

Hellpimp, continued, to Pierre

Let me discuss this further.

The question "Suppose you're on a game show..." is really the wrong question. If you're on a game show, you'll have known in advance you were going to be on it. You might even have realised that it would be easy to answer the dumb questions so the only thing worth worrying about is "which door to open". But if you are really smart you also know that you also get to choose your own initial door, first.

If you're on a game show, there is no way you can ask for advice when you stand there on the stage in front of closed door 1, with door 3 open. It's a point in time when you can't ask for advice, and since you could have guessed you might have got to that position, you had an opportunity to learn some wisdom about game shows in advance.

Whatever you do, don't blindly initially choose Door 1! Choose a door blindly, uniformly at random!

Now you don't have to make *any* assumptions whatever about "what is the chance that the car is behind doors 1, 2 or 3" initially. Because you are smart you can act in a way which doesn't depend on your making unverifiable, "out of the air" mathematically convenient assumptions. Note, Marilyn correctly asked "what do you do?" not "what is the conditional probability that...".

So let's reformulate the question as "Suppose you're going to be on a game show ... and then the usual story follows on the lines of: And suppose, say, you choose door 1, Monty opens 3...

The smart person is not trapped by the words *say* door 1, *say* door 3. The smart person takes a step back and takes a wider view.

Hellpimp got it right. He is absolutely right. Is his answer complete? Well, with a bit of game theory - or with a bit of careful thought - you can easily convince yourself that there is no way you can guaranteed do better than 2/3. But that is a pretty irrelevant refinement since nobody would ever imagine that anyway, and if it could be done, someone would have told us how long ago. So I'd say his answer is essentially complete and only a nitpicker (or a mathematician who wants to show that he does know his stuff) will take the trouble to answer the question, is Hellpimp's strategy not only brilliant but also optimal?

The whole conditional probability story is a red herring, from the point of view of vos Savant's words, as understood by a smart person, who knows that on a game show you have more choice than only switching or not switching. You also on that game show have a choice which door to choose first of all.

Of course it is mathematically wonderful how the unconditional probability, conditional probability, and game theory problems are related. You can use the minimax theorem to prove that the conditional probabilities must all be at least 1/2, under the appropriate conditions, without doing any probability calculations at all! This is because if you know the minimax theorem (eg look it up on wikipedia) and can guessed the minimax strategies (not difficult, when you see the symmetries and know about the power of randomization), you can prove that you have indeed found the saddle point. Now the the stuff about the conditional probabilities can be deduced since otherwise the saddle point couldn't have been a saddle point.

It's all so beautiful. "How could I not have known this before" - Pierre, in War and Peace.

Gill110951 (talk) 15:16, 15 August 2010 (UTC)[reply]

Yes we have a POV but it's not about Rick's (1) conditionalists vs (2) unconditionalists

I think it's about (1) or (2) OR both/neither. An I go for the third option on the grounds of NPOV. As a side effect I also get compliance with The Truth

Here's why:

Rick Block wrote:

I have been suggesting for some time that we have a POV issue and that we're actually arguing about

1) whether the article (as of the last FARC or even currently) is NPOV or whether it endorses the POV of the "conditionalists" (those sources who argue that the simple solutions typically presented are not quite responsive to the question).

Just for the record: Right now, I think recent versions endorse the conditionalists though the strength of the endorsement has been reduced. I think we need to go further. Gill110951 (talk) 09:00, 16 August 2010 (UTC)

2) whether the structure of the article should be to focus first and foremost on presenting the simple solutions in an attempt to convince the reader that the chance of winning by switching given the standard interpretation is indeed 2/3 and not 1/2, or whether doing this creates a structural POV effectively saying these solutions are the correct and proper way to understand the problem (i.e. endorses this POV).

Just for the record: Right now, I think the structure of the article should be to focus first and foremost on *good* simple solutions (for many, many sound reasons), and in particular those simple solutions which *are* correct and proper ways to understand the problem or for instance correspond to informal versions of correct and proper ways. I believe that there is not a unique correct and proper way to understand the problem. I don't think this constitutes violation of WP by picking and choosing on a basis of a POV. I think it corresponds to good taste and balance in selecting from the wealth of material which is "out there". Thus I think one certainly *may* use The Truth to guide the collaborative and distributed selection and editing process in pursuit of and in compliance with Wikipedia aims and policies. If an editor seems to claim (or seems to report that a reliable source claims) all three of (A) and (A => B) and (not B) simultaneously, with no comment at all, one might wonder if he/she is up to his/her job. So other editors should bring this little matter of The Truth up in discussion, in good collaboration with all concerned Wikipedia editors, naturally. Wikipedia is never finished. Because time never stops still. We are following a moving target. Gill110951 (talk) 09:00, 16 August 2010 (UTC)

I propose the both/neither alternative. There are good and proper interpretations of the MHP which the conditionalists will love. There are good and proper interpretations of the MHP which the unconditionalists will love. Both those parties also seem to miss that there are good and proper interpretations of the MHP which fall in neither category, especially those that I one could label economical or opportunist or taking a higher view: it's the point that you don't just get the choice switch/stay but also the initial choice of door in the first place. Choose it uniformly at random and you are safe, you needn't make any "unwarranted" assumptions whatever.

OK I see this through the POV of a professional pure and applied mathematician. On the one hand, biased. On the other hand, trained and competent to spot holes in arguments, to spot structural relations between apparently different results. I have made an in depth study of the reliable sources.

Hence all we have to do is to adopt an NPOV. Now indeed there exists verifiably a controversy between the fundamentalists. And of course these guys make a lot of noise and get a lot of attention. But there is actually also a sound middle of the road in which everyone gets there due and by integrating the use of basic clear logical thinking and all the Wikipedia Policies there are, one can come to a great article. But not if you first insist on the wrong question being answered in favour of one or the other fundamentalist school. We want a NPOV, right? We also want to get the article back to being a justly great article again. It got a bit unbalanced and new guys elbowed in with new ideas. And the world turned round and Morgan et al retracted, Tsirelson exhibited some beautiful mathematical short cuts, some other people finally figured out what all the economists had been thinking about all these years...

vos Savant's words are "out there". They still stimulate original thought. Taking those words as describing informally a lay person's idea of a problem which needs to be mathematized to be solved, the professional mathematical scientist will also watch out that the chosen wording doesn't lead them into a narrow blind alley when the girl who posed the question would obviously have been more than delighted if her words had been reformulated a tiny bit and thereby making obvious what is the question you really want to ask. vos Savant asks: what should you do. I say: you will of course have thought a bit ahead. You have of course already noticed that you have TWO decision moments. You're focussing on the second but actually the first is equally important, let me tell you why ... Gill110951 (talk) 10:06, 16 August 2010 (UTC)[reply]

Forest fires

Gill - There's an internet forum phenomenon called a forest fire where a thread about a topic spreads becoming multiple threads, which makes the "current" thread difficult to find and hard to follow. We're basically in a meeting being held by the mediators. Side discussions on other pages are highly distracting. I know if feels slow, but can you please limit conversations pertinent to the mediation to the mediation page itself? Thanks. -- Rick Block (talk) 15:36, 16 August 2010 (UTC)[reply]

Thanks Rick, and please call me Richard. I felt that your long piece about the issues contained a great deal of side issues and, in as far as it was about the mediation issues, it presumed, to me, a whole lot of prior assumptions which I know for sure many editors don't buy. So I tried in the spirit of the mediation to discuss this with you "aside". I think it could be a useful discussion. I think we have pretty complementary expertises so if we could get the ego thing out of the way maybe we'ld all get along a bit better. But it doesn't matter, the mediators seems really capable, I like their strategy. Gill110951 (talk) 05:35, 17 August 2010 (UTC)[reply]

Richard - I am perhaps coming across here not very friendly but I really would prefer to have discussions on the mediation page. In other circumstances I might actually enjoy a 1-1 discussion with you about this, but since we're in mediation anyway it seems like sticking to the prescribed format is advisable. One advantage of this is anything we discuss is visible to everyone else. BTW - thanks for clarifying how you prefer to be addressed. I've often wondered about this. -- Rick Block (talk) 15:10, 17 August 2010 (UTC)[reply]

Dear Rick, thanks. Very good. We keep discussions on the mediation page but/and we do our best to keep within the mediation framework. I know I often feel a dire need to blow off steam in this kind of situation but I guess it can also very often be wise to go take a walk and come back the next day. I am having enlightening discussions, I think, with a few other of the gang (of the mediated) on their or my talk pages. That's going to happen anyway. If it throws up something substantial it'll come back to the mediation or even main talk page later. When I think I see that someone is missing a point in probability or logic, for instance, I like to take it up with them. This often clarifies things for one or both of us if not both. It's the way science progresses, and also meta-scientific activities too. You can be sure WP:MHP is going to stimulate a few more great publications on MHP. Gill110951 (talk) 04:14, 18 August 2010 (UTC)[reply]

Glopk replies.

Gill, I appreciate that you spent a day (as you wrote in my talk page) on the mediation issue, as well as your enthusiasm for discussing the problem. Unfortunately, for me Wikipedia is strictly a time-off endeavour, after demands of work, family and study have been satisfied. I simply do not have the bandwidth to follow 20 separate threads about the MHP (even if I had the inclination). Therefore, in the interest of bringing the current mediation to a successful outcome, with a fruitful endstate for all parties involved (if that's possible), I would greatly appreciate if you joined me in limiting the discussion about the topics under mediation to the mediation page, and in allowing the mediators to conduct the conversation. Attention to being concise and to the point, as we have all agreed to do upon joining the mediation, would also be greatly appreciated. glopk (talk) 00:32, 17 August 2010 (UTC)[reply]

For me too, Glopk! I agree. I had been satisfying demands of work family and study for about two months since last I spent some time on wikipedia MHP. I spent a day on the page, honestly trying to follow the mediation guidelines in letter and spirit, since there was a lot to read and a lot of mediator's questions to answer. The mediator also suggested to hold talks about matters of fact, truth and whatever, in other locations. So I indicate to all concerned my willingness and openness to do so. OK? Gill110951 (talk) 05:31, 17 August 2010 (UTC)[reply]

Why MHP-WP is so difficult

Experts versus Laymen?

The MHP-WP is the Problem of writing a Wikipedia article on MHP. The collective task of many voluntary editors is to summarize the immense literature on the MHP, where we must restrict ourselves to literature which is W-reliable, i.e. reliable in the sense of Wikipedia. There are several W-reliable books on MHP. The topic is enormous. Those books take differing, particular, points of view which are not shared by other W-reliable sources. Wikipedia policies say that we must reflect this variety of views by adopting a neutral proportionality principle: if nine times as many reliable sources say X than the number who say Y, then Y is a minority point of view which gets 10% of the attention. But suppose an author first writes X but later repudiates and switches to Y? Suppose the majority of non-experts in a certain field presently believe X (and write it in W-reliable sources) while a majority of experts believe Y; and suppose there are 9 times as many W-reliable sources written by laypersons as by experts? Do books count more a couple of pages in a journal?

Further problems arise when a certain topic has been argued by different experts, and by different non-experts, to belong to different expert-domains.

Every wikipedia editor naturally has a personal favourite solution of the MHP-P (by which I mean the problem to be solved, the route by which to solve it, and the answer which comes out).

As a mathematician, I see how mathematics can be used to organise this huge amount of material in a coherent way, and to defuse controversy. Mathematics is a great tool for synthesis. For instance the debate between the conditionalists and unconditionalists about whether or not the conditional solution adds anything to the unconditional, is no more than the question: could one do better (unconditionally) than 2/3? Well, I think no one would believe you could, and certainly no one has yet shown how to do it. And the reason no one did that yet is because it can be mathematically proved, in many many ways, that 2/3 is indeed the best you can do. Showing that all the conditional probabilities are at least 1/2 is one way. Guessing and checking (all easy to do!) the game-theoretic saddle-point is another way.

However, most WP editors are not professional mathematicians.

Sigh. I guess that just shows it's my job to write W-reliable sources, not to contribute to wikipedia articles.

Gill110951 (talk) 15:09, 18 August 2010 (UTC)[reply]

Sorry Richard, as a mathematician you should know better (sigh). The debate between the "conditonalists" and "unconditionalists" is not about whether the conditional solution adds anything to the unconditional solution, but about the logic that the unconditional solution is no solution to the general accepted form of the MHP in which the player is offered to switch after the host has opened a goat door. Nijdam (talk) 21:06, 18 August 2010 (UTC)[reply]

Come off it Wietze, as a mathematician I do understand the difference. On the other hand, do you as a mathematician understand that the mathematical question: "can you do better than 2/3 unconditionally" is mathematically equivalent to the question "are all conditional probabilities at least 1/2"?

Frankly, no I can't. If you mean to say that when the car is placed randomly (what used to mean uniformly), the conditional probs are at least 1/2, okay. But what do you mean by equivalent? If we do not make assumptions about the distributions, the conditionals being at least 1/2 is not equivalent to the overall being 2/3.Nijdam (talk) 15:19, 19 August 2010 (UTC)[reply]

Sorry, let me spell it out. Sometimes I'm too fast, often I'm too slow. You naturally agree that

Prob(switching wins) = sum (over 6 pairs of possible door chosen, door opened)

Prob( switching wins | door chosen, door opened ) times Prob( door chosen, door opened) .

We know the left hand side equals 2/3 under the assumption that the initial choice had 1/3 chance to hit the car. The only way we could have a strategy with a larger hit-chance than than 2/3 would be by not switching for some configuration which we know satisfies Prob( switching wins | door chosen, door opened) < 1/2. We know that if all cars are initially equally likely, that all conditional probabilities are at least 1/2. Proving 2/3 (unconditional) can't be beaten is equivalent to proving that all the conditional probabilities are known to be at least 1/2. So yes, the conditionals being provably at least 1/2 (under assumptions known to be true) is mathematically equivalent to it being impossible to beat unconditional 2/3. QED.

I don't see your point. As I remarked you have to make assumptions about the distributions, and so you did. Nijdam (talk) 23:45, 20 September 2010 (UTC)[reply]

One can make more or less assumptions and get more or less conclusions. I am telling you about the equivalence of two of the conclusions. They are mathematically equivalent. Let me try again. There are six configurations (door chosen, door opened). Hence there are 2^6=64 non-randomized strategies for the player and a 64 dimensional continuum of randomized strategies: for each of the 64 configurations one can stay or switch or randomize between them. Under no assumptions at all, the law of total probability tells us that the strategy "always switch" is the best strategy of all (with respect to unconditional hit-chance), if and only if no conditional probability disfavours switching. If we only know the players initial choice is right with probability 1/3, we know that "always switch" beats "always stay" (unconditional 2/3 versus 1/3) but we don't know anything else. If however the host is constrained to hide the cars uniformly at random, one can show with Bayes that no conditional probability disfavours switching, hence "always switch" is optimal - beats all possible strategies. Alternatively one can try to prove in that situation in another way that "always switch" is optimal (2/3 can't be beaten) and if you succeed you've proven that no conditional probability disfavours switching, without Bayes. The conditional solution completes the unconditional solution by showing that "always switch" is globally optimal, unconditional 2/3 can't be beaten. At the cost of a further condition on the host's behaviour. Gill110951 (talk) 05:43, 21 September 2010 (UTC)[reply]

But the idea that it's only *after* another door is opened that you get to think about the choices we had/have is nuts. Right now you and I are *talking* about a game show, we're not *on* the game show. We're talking about knowledge about the game show which in real life we could only have before we stood on the stage under the lights in front of the three doors. We *must* have known in advance of the show, that if we got to this stage, the quizmaster will certainly open a different door with a goat, and certainly ask us to switch. We didn't suddenly "know" this by instantaneous divine inspiration *after* actually choosing our door. We must have been guaranteed this information before. Just because we saw it happen on the last 100 shows doesn't make it certain it'll happen again, either. You can only *know* it is going to happen because, before agreeing to participate, you and Monty Hall signed a contract in which Monty Hall committed himself to what he would do after you chose your door irrespective of what door you are going to choose, if you did make it to the last round.

Again, what are you heading at? Being the player in the game, or the mathematician reflecting on the MHP, we have to consider what have happened or what may happen. in the last case we reason: if I choose door 1 and the host opens door 3, then ..., or if ... etc. Nijdam (talk) 23:50, 20 September 2010 (UTC)[reply]

It's inconceivable that if you know what you're supposed to know, you only knew it at the moment you had to make your decision. You must have known it before, hence you were able to plan a strategy before. Wise strategy: choose door at random and then switch. Now you don't have to assume *anything* about the quizmaster beyond what you have been told.

Economists are a lot smarter than probability theorists. Are more in touch with the real world. The only way the problem description makes sense as a description of a real world problem is that you know in advance certain things that the quizmaster certainly will do. Therefore you have time to consider BOTH your decions moments in advance. On the other hand the quiz-show team is not committed in advance to any particular probability distritions. You have no idea. All you know is what they are forced to do (put a car behind one door, open a different door revealing a goat after you have chosen a door. The more I think about it, the less vos Savant's problem makes sense as a probability problem, and the more it makes sense as a game theory problem. Selvin talked about a game show. He was a statistician so he used some elementary probability. But if he had known game theory he would have used game theory. Silly man. (This is what I call solution driven science). PLEASE imagine a game show without preconceived notions of what kind of mathematics should be used. Gill110951 (talk) 22:20, 18 August 2010 (UTC)[reply]

... or maybe rational subjective versus irrational frequentist?

Here's a an exchange I found useful:

Subject: This Is What I'm Hearing...

Initiated by: Glkanter

1. Suppose you're on a game show...

2. "If the contestant chooses doors uniformly at random, the likelihood of initially selecting the car is 1/3, a goat 2/3."

Therefore:

3. "If the contestant does NOT choose uniformly at random, the likelihoods may be different values."

4. I reject the notion that the contestant's method of making that door choice affects the probabilities in the Monty Hall Problem Paradox. Glkanter (talk) 14:34, 18 August 2010 (UTC)[reply]

I hope you are not hearing that, because it wouldn't be logical, and you can be sure I only use logically correct arguments. But you are sure you will hit a goat 2/3 of the time, however you choose a door. It's a mathematical theorem that this means that the car is being hidden by the organizers of the quiz show uniformly at random. Gill110951 (talk) 15:29, 18 August 2010 (UTC)[reply]

Gill, I'm not formally trained. There's nothing more that I can tell you. Somewhere recently (maybe on one of Martin's talk pages) you wrote something like 'as long as the contestant chooses uniformly at random, he has a 2/3 chance of selecting a door with a goat behind it'. That is the statement I do not agree with. If those were not your words, then I apologize, and we can end this discussion.

Otherwise, please tell me which of the above statements is flawed. Thanks. Glkanter (talk) 16:07, 18 August 2010 (UTC)[reply]

I am formally trained, which can be both an advantage and a handicap. Let me rephrase my sentence. If you choose your door uniformly at random, you can be certain the chance is 2/3 that your choice hides goat. If you don't choose your door uniformly at random, you can't be certain the chance is 2/3 it hides a goat. (I don't say it isn't 2/3, I say you can't know if it is 2/3 or not). Interesting discussion! Probability is a tricky concept! If you say that the probability is 2/3 because you don't know anyting at all, then you are a subjectivist. Does probability reside in our heads, it is about the information we have in our heads about a situation, or does it reside in the situation, objectively. People who use probability have been fighting about this for 250 years, without making any progress. Gill110951 (talk) 22:46, 18 August 2010 (UTC)[reply]

Yeah, I'm not trying to convince you that I'm right, but I will go on happily in my ignorance, making the best decisions I know how.

(A)

Richard: How did you arrive at your selection of door #1?

Glkanter: I used a random number generator.

Richard: Then the probability is 2/3 that you picked a goat.

or

(B)

Richard: How did you arrive at your selection of door #1?

Glkanter: It's my lucky number.

Richard: Well then, you can't be certain the chance is 2/3 it hides a goat. I'm not saying it isn't 2/3, rather that you can't know if it is 2/3 or not.

I guess that is meaningful to you and those that care to argue about it. Sorry, that does not include me. Glkanter (talk) 23:42, 18 August 2010 (UTC)[reply]

It's meaningful to me because it brings me a step closer to understanding your meaning and hence the total rationality and correctness of your argument in its own terms. You're saying, I think, that for you a distinction between (A) random number generator and (B) lucky number is completely irrelevant. That means that in your world, as you walked onto the stage and was asked to choose a door, each door was equally likely to hide the car, and of course your choice method was irrelevant. For the same reason, Monty Hall's choice method when he has one is irrelevant. For the same reason, the door numbers (or colours, or positions) themselves are irrelevant. Since they're strictly irrelevant there is no implication whatsoever in naming them for convenience or concreteness as Selvin, vos Savant and so many before and after have done. Probability for you is about your rational beliefs. It's epistemological, not ontological; "subjectivist" not "frequentist". Like Laplace and a great many other very great and very rational thinkers, past, present, and future. I think that Martin Hogbin also holds instinctively to the position that probability describes rational belief. When one looks at the roots of the word one finds the same view. When one tries to look in the mind of the typical intelligent man in the street, you'll find the same view. According to many scientists in the field now, the "objectivist" position has been a brief aberation for a large part of the 20'th century (roughly: the 20's to the 90's). However, as such, it certainly has played a major role in making great advances in science and technology. Right moving to a new synthesis. Personally I am on the side of the synthesis, the debate had been polarized hence fruitless for too long already. The first step is for each side to realise the self-consistency of the other's view, and to realise that there are situations which call for either view. Gill110951 (talk) 06:54, 19 August 2010 (UTC)[reply]

Conclusion: it is totally correct to believe that the short intuitive solution is not only correct but also complete. ie it is the last word on the MHP ... *if* we take a certain very wikipedia-reliably sourced metaphysical position in epistemology (the science that studies what it means to know something). Reliably sourced both at the expert and at the layperson's level.

I'm pretty sure that you can reliably source the claim that ancient Greek, Hindu and Chinese writers and thinkers have discussed these problems. Gill110951 (talk) 07:23, 19 August 2010 (UTC)[reply]

Richard, I'm glad you found our exchanges of value. Some random (?) thoughts:

I guess the best way I could describe the doors without implying uniqueness would be 'left, center, and right'.

*Somebody*, but not the contestant, does know the method used for car placement. Likewise, *somebody* knows where the car is for each play of the game. The goats and car don't place themselves behind those doors. But this puzzle begins, 'Suppose you're on a game show..."

Carlton's decision tree (formally?) proves the extension of this indifference (?) to the host decision by virtue of not assigning any values to how he chooses between 2 goats.

Therefore, all solutions, including Bayesian (Bayes Theory?) which use 50/50 are of only partial utility.

And Nijdam's ponderous insistence that the "1/3 prior <> 1/3 posterior" is erroneous.

For the MHP, Wikipedia's only concern is reliable sources, and, I guess, the prevalence of each POV. Not 'right or wrong'.

Are you agreeable to taking out the phrase, 'If you choose your door uniformly at random' from your statement:

If you choose your door uniformly at random, you can be certain the chance is 2/3 that your choice hides goat.

Which leaves: You can be certain the chance is 2/3 that your choice hides goat.

And eliminate this one altogether:

"If you don't choose your door uniformly at random, you can't be certain the chance is 2/3 it hides a goat. (I don't say it isn't 2/3, I say you can't know if it is 2/3 or not)."

Or do I not understand which school you belong to? Glkanter (talk) 08:33, 19 August 2010 (UTC)[reply]

I don't belong to either school. As a mathematician and scientist I can take the position of being a rational subjectivist, and I can take the position of being a being a frequentist; I can tell you what you are assuming, what you can get, and what comes out at the end.

The subjectivist starts being indifferent about various things and so chooses a door any which way, etc. etc. He would initially bet equal odds on each of the three doors and he would bet equal odds on the quizmaster's choice if the quizmaster did have a choice. He would therefore rationally bet at odds 2:1 that after switching he had the car, whatever door he actually chose first and whatever door was actually opened.

The clever guy chooses his door uniformly at random.

If the frequentist can know for sure that in the long run, each door will hide the car equally often, then switching will get him the car 2/3 of the times. If you assume it just within those cases where he initially chooses door 1, then within those cases he'll get the car 2/3 of the time too. If you further restrict attention to the cases where Monty opened 2, and the cases where Monty opened 3, he'll get the car at least 1/2 the time by switching. Moreover in one of the cases he'll get it at least 2/3 of the time.

Does that clarify things? I certainly can't say that one and only one of these stories is "the right" story for MHP. And within wikipedia that is an irrelevant question. Out of wikipedia I like to take the meta-view and I delight in the variety of approaches and solutions which all say something different. Gill110951 (talk) 16:47, 19 August 2010 (UTC)[reply]

What is wrong with the conditional approach to the MHP

Nijdam asked me:

Question 1. What criticism is there on the conditional approach?

Question 2. Is the MHP a real world problem?

Question 3. What more "givens" are needed for the (conditional) solution?

Let me answer Question 3 first of all. Let's look at three sets of assumptions and corresponding conclusions which can be drawn from them. Remember, I always start with Whitaker/vos Savant's verbal description of the problem together with the uncontroversial "explanation" that the quizmaster always opens a door revealing a goat and he can do this because he knows where the car is hidden. But I do not a priori assume anything else.

Assumption A. In order to conclude that switching gives you the car with (unconditional) probability 2/3, it is necessary and sufficient to assume that the door you have initially chosen has probability 1/3 of hiding the car.

Assumption B. In order to conclude that switching gives you the car with conditional probability at least 1/2, in any of the six conditions of door initially chosen & door opened by host, it is necessary and sufficient to assume that all doors have initially equal probability 1/3 of hiding the car

Assumption C. In order to conclude that switching gives you the car with conditional probability 2/3 in any of the six conditions of door initially chosen & door opened by host, it is necessary and sufficient to assume that all doors have initially equal probability 1/3 of hiding the car and that the quizmaster opens a door uniformly at random when he has a choice.

The more you want to get out, the more you have to put in!

Now let me discuss the meaning of these conditions in the real world, even though it's one of the questions put on the table: Question 2, is MHP a real world problem? I come back to that later (my brief answer is: it is a wonderful toy problem, a model problem, for learning about decision under the face of uncertainty in the real world, as well as for learning about Bayes' theorem in an introductory statistics course. Statistics and probability are certainly useful tools for decision making, but not the only ones.)

Depending on what you mean by probability, you will find it more or less easy to "buy" stronger assumptions. That sounds good, but it isn't necessarily good: if what you mean by probability is weaker, you can more easily sign up for heavier assumptions, but what you get out is weaker too, since it is a probability statement to be interpreted according to your "weaker" notion of probability.

A smart player chooses her door initially uniformly at random and thereafter switches. (She can do this since she knows in advance she might make it to the last round, and she's smart). She *engineers* the truth of Assumption A and therefore gets its conclusion. And by the way, by the von Neumann's minimax theorem this is the best she can do in a very meaningful sense, in the situation that the host is not bound to use any particular probability law, or indeed any at all, to govern his choices. Note that she gets Assumption A by her own action, but has no way of knowing whether or not Assumption B is true, let alone C.

Some people like to solve the problem with conditional probability. If they are frequentist oriented, they will add as an assumption that it is given that each door is initially equally likely to hide the car. Possibly they will also add that in the case when the host has a choice, he is also equally likely to open either door. This corresponds to Assumptions B and C. Note that extra assumptions are being added to the vos Savant story, and it is not clear where they come from, except perhaps from a desire to solve the problem with the particular tool of conditional probability. This is called "solution driven science". It's hard for me to imagine a game show where you know for sure that the car was hidden by a fair randomization, or that the host used a fair coin toss to choose the door to open when having a choice. Is it written in a signed agreement which was available to you in advance? Or do you just have some statistics from viewing the show in the past? This doesn't prove that what you think you saw in the past will happen again today.

Some people are subjective Bayesians in their use of probability. Given the vos Savant story only they have no reason whatever to assign any particuar meaning to the specific door numbers. The story could just as well have been told with any other door numbers instead of "say, door 1", and "say, door 3". Thus their prior distribution over the location of the car and the choice of the quizmaster is uniform, because of the absence of any information distinguishing any of the doors from the others, apart purely from an arbitrary labelling. If they are clever they will, like Glkanter, not condition on the door numbers at all since they understand in advance that they are irrelevant, and they'll compute the unconditional probability 2/3, knowing that all doors are initially equally likely to hide the car, hence also the door initially chosen by the player. Of course the conditional probability is the same, they knew that in advance so it's not interesting to verify it by a computation. For the same reason they don't condition on the day of the week.

OK, anyway, the subjectivist does naturally go the full Monty and does buy Assumptions C. But what does this give him? It just tells him that he would be rational now to bet at odds 2 to 1 that the car is behind the other door, since it is the only rational bet, given that he was also prepared to bet at equal odds that each door initially hid the car, and that he's not interested in the door numbers (or would bet equal odds that the host opened either when he had a choice). It's amusing that the subjectivist doesn't bother to choose his door by randomization because he a priori knows that all doors are equally likely. 20th century science objectively made great advances by using randomization in statistical design of experiments, in statistical sampling, in optimization, decision theory and game theory, and many more places besides. This tells me that dogmatic subjectivists are missing out something. They are so charmed by their own personal inner consistency that they don't care much about the real world or about communicating with others who might have different expectations.

Teachers of conditional probability may prefer to specify the full symmetry assumptions C and get the conditional answer 2/3 by direct computation. Mathematicians with a love of symmetry and an abhorrence of unnecessary computations will observe that the unconditional probability is 2/3 hence the conditional probabilities are all 2/3 too, since they must all be equal by symmetry, and we know they average out to 2/3 by the law of total probability. Statisticians will advisedly take this opportunity to teach the odds form of Bayes' rule, which actually is a theorem giving you a quick way to get the right result, which has incredible power and application all over science. The usual treatment of computing the conditional probability from first principles while paying lip service to the Reverend Thomas is not an example of using Bayes' theorem, it is an example of avoiding using Bayes' theorem, by carefully and patiently working through tedious computations.

OK, so what are my criticism's of the full conditional solution? I have nothing against the solution as such, but I object to the way it is sold, namely by papal decree, or iman's fatwah, rather than by argument. People say "you must use a conditional probability" but don't explain why. Give me a good explanation why you *must* compute the conditional probability, and I would be happier. Now actually there are some good reasons, but they are not because it is dogmatically the case that you have to do that. Here is the real reason: under assumption B, you can do better than unconditional 2/3 if and only if if there is a specific situation where the conditional probability that switching doesn't give the car is less than 1/2. Well, I didn't ever meet anyone who thought it would actually be *better* to stay rather than switch, nor did I ever meet anyone who thought one can do better, unconditionally, than 2/3. And of course it can be proven, by computing a conditional probability (or even via the minimax theorem if you prefer that to Bayes! But it is a longer and tougher route, partly because the game theoretic approach tells us something when assumptions B and C are not given).

So in my opinion the conditional approach is important because it tells you that always switching is optimal in the context of Assumption B, not just that it is better than always staying. I want to find a short intuitive but simultaneously rigorous argument for this.

Finally, is the MHP a real world problem? It is a problem which is used in many fields for many purposes. In probability and statistics it is often used as a vehicle for getting the student interested in conditional probability, and familiar with calculations thereof. Nothing wrong with that. However I think that we teachers of statistics and probability could get much more out of it, since it can also be used as an exercise for teaching model building, for teaching mathematization. It also allows us to touch on game theory and the minimax theorem; statisticians and probabilists ought, in my opinion, to know about this fundamental and beautiful theory (in elementary settings) too. Teaching model building and mathematization is actually much more interesting, and much more challenging, than teaching probability calculus. It is also more of an art than a science, and that is why there are different good answers to the MHP.

I am especially critical of teachers of elementary statistics and probability who chuck in assumptions of uniform distributions independence and whatever all over the place just to get a nice answer. This brainwashes students into thinking that statistical science needs these assumptions and that these assumptions are somehow justified and scientific. And that leads to disasters cf. Lucia de Berk, OJ Simpson, and so on and so forth. Gill110951 (talk) 11:52, 20 September 2010 (UTC)[reply]

The essential MHP

I summarized what I have learnt about MHP from wikipedia discussions in [4]. An earlier writing of mine was accepted for publication but I am going to offer this completely rewritten text as "final revision". Comments are welcome either here or by personal email. Gill110951 (talk) 17:43, 7 October 2010 (UTC)[reply]

Learning

There seems to be some evolution in your thoughts about MHP. I received your email about the above mentioned article for Statistica Neerlandica, and I will respond to it shortly. In the mean time I want to try to know where you stand.

Is it your opinion, as well as of most (all) sources, that the player is offered the possibility to switch, only after the host has opened the door with the goat?
If the MHP is considered a probability puzzle, should the decision be based on some probability?
What probability should the decision be based on? (I do not ask how it may be calculated.)

Please just answer in short, no need to argue. Nijdam (talk) 21:27, 10 October 2010 (UTC)[reply]

Hi Wietze! And we should not forget that beer together. I am at NIAS in Wassenaar on sabbatical all this academic year. Are you ever in the West of NL?

I do occasionally frequent the West, but till now didn't find an opportunity to visit you. We'll see.

To answer your questions as briefly as possible:

I think everyone agrees that the player is only offered the switch after the host has opened the door with the goat, but I think that the player knew in advance that this was certainly going to happen.

Okay, although I think showing a goat should come as a surprise to the player, for the MHP it may be different.

His decision should be based on his knowledge. If the player knows probabilistic properties of the situation he should try to use them.
Vos Savant's words do not tell us anything about the probabilistic properties of the game. In order to advise the player, we have to do some mathematical model building. We should do this in a two-way dialogue with the player, to find out what the player really wants from us, to find out what he would be happy with. Let me offer two rational solutions.

(Rational Subjectivist). The player wants to base his decision on his rational subjective personal probability that switching will give the car, given *all* the information available to him. This is by definition of "relevant" the same as the probability that switching gives the car, given all relevant information. For such a player, the car was initially equally likely to be behind any of the three doors. (This doesn't mean that the host hides the car completely at random, it means that the player doesn't know anything about how the host does it; before the host opened a door, the player was completely neutral in his opinions about where the car was; the player chose Door 1 because "1" is his lucky number.) For the player, the host would next be equally likely to open Door 2 or Door 3 if the car happened to be behind Door 1. (This doesn't mean that the host doesn't have a bias in door-opening, it means that the player's opinions about that bias are completely symmetric under arbitrary relabelling of the doors). So: the subjectivist wants to know Prob(switching gives car| everything he knows), where "everything he knows" includes the specific history of door numbers in this case. By symmetry the door numbers are irrelevant, along with everything else, so he just wants to know Prob(switching gives car).

(Rational Objectivist). The player knows nothing about the way the host hides the car and nothing about the way the host chooses a door to open when he has a choice. Conceptually he can imagine the host's car-hiding process as a random process and the host's door-opening process as a random process but these are two unknown random processes and could well have any bias whatever. He solves the problem by introducing probabilities known to himself: he chooses his own door initially uniformly at random. He knows now that Prob(switching gives car) = 2/3. He doesn't know Prob(Door 2 has car|Player chose Door 1, Host opened Door 3), he doesn't even know if it is at least 1/2, since he doesn't know anything about bias.

Well, is this you want to present the average wikipedia reader? And although you argued here that the solution may come down to the calculation of the unconditional probability, this is not the same as the discussed "simple solution". Nijdam (talk) 16:19, 11 October 2010 (UTC)[reply]

What I think myself, and what I want to communicate to colleagues in mathematics and statistics, is not the same as what the average wikipedia reader wants to know. But not all wikipedia readers are "average". The article should be structured so that a lot of average readers can learn what they need to know from the initial sections. Think of this as the "popular" part of the article. Later sections can and should be more sophisticated, to reflect the various types of "professional" writings on MHP. Most of the active wikipedia editors of the MHP page are amateurs, not professionals. They will determine the main outline of the article. The professionals will stand on the sideline and offer advice and make sure that the basic text doesn't contain nonsense. It's interesting to see how the average layperson, even if highly educated, is not able to appreciate basic mathematical reasoning, i.e., reasoning about reasoning. I think this is the reason why the fights go on and on. Several editors think about MHP purely instinctively and seem unable to distinguish between the answer to a problem and the reasoning which gets you there. Several editors seem still completely in the dark as to what a conditional probability means. Several editors seem completely in the dark that there are different ways to think about what probability means. The "reliable sources" on MHP don't help since they nowhere give a clear layman's reasoning why the player would be wise to ask himself "what is the conditional probability". The reliable sources on MHP don't discuss the fact that whether or not you can do this at all can depend on what you mean by probability. (Jason Rosenhouse's book comes the closest to this, but I think he is a bit muddleheaded on the matter. He's not an experienced applied mathematician or statistician and I don't think he really understands the issues). These are the things which I try to bring out in my article, which is intended for professionals like you or me, not for average wikipedia readers or writers.

Regarding the simple solutions on the MHP page: I tried to get across to the other editors the fact that Monty Hall's own solution in Selvin's second letter *only* assumes that the initial door is right with probability 1/3, it doesn't assume that each door is equally likely to hide the car. I try to raise the question: how could we know this, what do we mean by it? I argue that if you use probability in a subjectivist sense you will easily agree that *for you* all doors are equally likely to hide the car and *for you* Monty is equally likely to open either door if he has a choice: all this is merely an expression of the fact that your prior knowledge about the situation is completely neutral with respect to the labelling of the doors. It follows that your final conclusion that the other door is twice as likely to hide the car as your initial choice is also merely an expression of your knowledge at the moment you have to act. The only empirical meaning it has is that *you* would be completely indifferent with respect to the one-time choice: "switch and take what is behind the other door" and "toss a fair die and take a car if the outcome is 1,2,3,4; take a goat if it is 5,6".

If you want to apply probability in a frequentist sense you have to have frequentist probabilities as input. Vos Savant doesn't give us any information about how the two host actions (hiding a car and choosing a door) are actually done. You could guess and then give an answer which is conditional on your guess being a realistic description of the real world situation. Or you could act by creating objective frequentist probability yourself.

I don't think that any other editor has taken this on board. I didn't succeed in what I tried to do. Never mind, it has been an interesting experiment, and one learns more from a defeat than from a victory. I want people to be carried by arguments, not by authority; if the arguments don't come across, then apparently the man in the street is not ready for them yet.

My personal opinion about which way MHP should be solved is not important. As a mathematician I can offer a spectrum, a menu, of strictly different solutions and I don't lay down the law which one is best. There is no "best" solution.

I think that MHP gives a fantastic opportunity to people who teach statistics and who teach mathematical modelling to get students (in the broad sense: I also "teach" MHP to journalists and lawyers) acquainted with the creative process of mathematical model building. An art not a science. Be wary of solution-driven science. Be suspicious of "default" assumptions which only have convention or authority behind them, but no substantive argument. Gill110951 (talk) 09:28, 12 October 2010 (UTC)[reply]

Okay Richard, of course all that is of interest may appear in the article. My main (only) concern is the introductory part. So let us concentrate on this. I took part in the discussion, because I'm worried about the lack of argumentation in the "simple solutions", being yet presented as complete full solutions to the "standard" MHP, say the Kraus & Wang version, and widely used and quoted. Several sources do so, but several others criticize this, and for good reasons in my opinion, and I hope in yours too. Nijdam (talk) 09:49, 13 October 2010 (UTC)[reply]

I agree that if you accept all the usual uniformity assumptions, then the simple solution is not complete. It tells you that always switching is better than always staying. It doesn't tell you that always switching is better than all other possible strategies. This follows from noting that by symmetry,

Prob(switching gives car|player chose 1, host opened 2)

=Prob(switching gives car|player chose 1, host opened 3)

so both these probabilities must equal their average: Prob(switching gives car|player chose 1) = 2/3. I think the simple solution ought to be much more simple too. If you only assume that Prob(initial choice hides car) = 1/3 then it follows that Prob(switching gives car) = 2/3. I think that the sources are mostly very poor quality. That's why I felt the need to write my own. I wonder what you think of my latest attempt. Gill110951 (talk) 14:07, 13 October 2010 (UTC)[reply]

Glopk's reply

Hello, please see here. glopk (talk) 00:29, 16 October 2010 (UTC)[reply]

MHP: the elusive concensus

MHP is a puzzle, a paradox. Usual intuition gives the wrong answer. One has to take a step back, think clearly and think out of the box in order to see that the player ought to switch. This means that at some point we have to suspend our direct intuitions and switch to careful conscious logical thinking. Later, we may develop an intuition which helps us give the right answer in similar situations in the future, but for newcomers to MHP that intuition is not there.

When we change mode and start using conscious rational thought instead of immediate intuition, we must distinguish in the original real world story between those elements which are important to keep in mind, and those which are merely ornamentation. I think that the disagreement between conditionalists and unconditionalists is actually not a disagreement between using conditional probability formalism or not, but is a disagreement whether the specific door-numbers are ornamentation or essential. After all, both are interested in the situation after a door has been opened, and clearly the probability that there is a car behind door 3 (or after "a door") after it is opened revealing a goat has indeed dramatically changed.

In a sense the disagreement is harmless, because under the usual supplementary assumptions about equal probabities all over the place, people who take the door-numbers on board, and work carefully through the more complicated problem do discover, after going to all that trouble, that whether or not you should switch does not depend on the specific door numbers in hand. However, what if we had discovered that the door numbers were relevant??? Admittedly you would have to change the problem a lot to make them relevant! See the paper by Jef Rosenthal. But please remember that a priori, the first time people hear about MHP, almost no-one thinks it is of any relevance that of the two doors closed after the host has opened Door 3 and revealed a goat: one was selected by the Player and the other (the other door left closed) was selected by the Host. To say it differently: newcomers to MHP see at this stage of the game two closed doors with numbers 1 and 2 on them and one opened door, Door 3, with a goat standing there; they don't "see" that the two closed doors got to be in that state of being still closed in a very asymmetric way. It's not the numbers and the open/closed which counts, it is the specific history which led up to that situation. It's important that Player chose 1 and Host left 2 closed, rather than Player chose 2 and Host left 1 closed. In fact: it's only important that Player chose A_DOOR, Host left AN_OTHER_DOOR closed.

It's precisely because the disagreement is "kind of harmless" that the debates carry on endlessly. The conditionalists don't have a good argument why you *have* to do it by conditional probability. The unconditionalists are straight-forward down-to-earth practical people who have no patience for subtle points of logic.

Couldn't the two parties agree that the big difference of opinion is not about conditional or unconditional, but about whether the door numbers are taken care of outside or inside the formal analysis? The advantage of keeping them outside, in the informal, pre-processing stage (or if you prefer, as post-processing, as afterthought), is that newcomers to MHP are much more easily taken on board, and that the essence of MHP comes across to anyone. You don't have to have a PhD to understand that its smart to switch. The point to taking them inside is that this does make the solution "more complete". Because, as I have said before, by taking explicit note in the logic/mathematics of the specific door numbers, one does rule out the possibility that some mixed strategy of switching or staying depending on the door numbers in the case at hand, is better still.

With the K & W premises in the Wikipedia MHP article, there is no 'mixed strategy' available to the contestant. Selecting the car is a random event that occurs 1/3 of the time, and does not announce itself. The only strategies available to the contestant are 'switch' as a rule, or "don't switch" as a rule. The contestant would have no way of recognizing when 'not switching' would win the car. Therefore, 'taking explicit note in the logic/mathematics of the specific door numbers' is not a requirement to determine that 'always switching' is the contestant's optimal strategy. That is derived quite clearly from the simple solutions. Glkanter (talk) 10:34, 11 November 2010 (UTC)[reply]

The big first step in enjoying MHP is to understand that switching beats staying. For the connoisseurs it is important to realise that these two are not the only options and that we can with just a bit more work say something better still. Clearly the editors on the MHP page are divided in the accent which needs to be put on the stuff for the connoisseurs. There is at least one editor who thinks the connoisseurs ought to be totally ignored, there is at least one editor who thinks that MHP is only for connoisseurs. Personally I find both of those two points of view not conducive to constructive editing of an important wikipedia page.

I realise that all this is a personal opinion. However it is a personal opinion which I have come to out of the challenge to find a synthesis of the different positions held by different editors here, which respects them all, does them all justice, and results in something richer.

Moreover it is strongly influenced by my professional-life-long experience consulting in statistics, with clients from law and medicine, psychology and linguistics, computer science and quantum physics. This has taught me that solving real world problems with mathematics and logic is an art. One must make choices which are a matter of taste. One has to balance degree of relevance to the actual problem, and amenability to formal solution. A solution to a problem in medicine or law which is absolutely correct from a scientific point of view but which one can't explain to a medical doctor or a lawyer is deeply unsatisfactory. We always have to compromise, and finding a beautiful compromise is a creative process. "The solution" depends not just on the problem but also on who asked the problem.

Wikipedia is an encyclopaedia, and people come from many backgrounds to learn about MHP. We want most everyone to be able to find "their" solution. That means that we do have to present different solutions in an unprejudiced way. I think that means that all editors working on the page have to work hard to try to understand all the main points of view which are out there.

Intuitionist and conditionalist quarrel

I would say the quarrel is more about the difference between

COND: Prob(car is behind Door 1 | host opens Door 3) - which I think nearly all people view the problem to be and is the problem for which inappropriate conditioning leads to the 50/50 answer

and

INT: Prob(car is behind Door 1 | host opens another door) - which I think virtually no one (at least initially) views the problem to be, but trivially reduces to Prob(car is behind Door 1) because it is given that the host will always open "another door"

Although it is quite true that these are conceptually different but (under the standard conditions) numerically the same, understanding the difference is not easy. Note that if the "wrong" answer is based on

Prob(car is behind Door 1 | car is not behind Door 3)

shouldn't the "right" answer be based on

Prob(car is behind Door 1 | host opens Door 3)

rather than

Prob(car is behind Door 1 | host opens a door [Door 2 or Door 3])

Doesn't the puzzle lead you to the wrong answer by encouraging you to think about the conditional probability? You can certainly conceptually reframe the problem - but by doing this I think you are (more or less) ignoring the crux of the problem, which is that you're standing in front of two closed doors that don't have equal probabilities. -- Rick Block (talk) 19:18, 7 November 2010 (UTC)[reply]

Yes I agree that the conditionalists and intuitionalists are fighting about whether the problem should be solved by considering COND or INT. Understanding the difference is indeed not easy! I think that Glkanter does not really understand the difference at all. Martin sort of understands there is a difference but he does not think it particularly interesting. The fact that the numerical answers are the same, and that one can get from INT to COND by symmetry, further confuses the issue.

Then your second point. I don't talk about what is the right or wrong problem. I only tried to explain what Carlton is saying. He says that a wrong intuition about conditioning leads to the wrong answer (1/2). The right way to do conditioning leads to the right answer, and moreover, he thinks that we ought to be interested in the conditional probability. So does Wietze Nijdam. Martin and Garry don't agree. I am not impressed by the reasons which the reliable sources give, that you "must" look at the conditional probability. On wikipedia we have to weigh the opinions of a small number of mathematicians who can't explain their dogmatic stance against hundreds of "ordinary people" who are not interested in the subtelties of conditioning or not conditioning on information which turns out to be irrelevant and/or could have been argued in advance to be irreleant. Then there is the further complication that many simple solutions don't assume the full force of the standard conditions and are therefore strictly better in the sense that they are applicable in wider circumstances. I for instance, do not think that the standard conditions are warranted, except as an expression of ignorance, and the argument that I must switch because I'm ignorant does not appeal to me. Richard Gill (talk) 21:35, 7 November 2010 (UTC)[reply]

Rick and Richard, I find your discussion about Garry's and my understanding of the problem vacuous and patronising. I fully understand the difference between the different probabilities and am very willing to discuss it with either of you. There are many important points that you have both missed in this stylised attempt to discuss the subject rationally. I look forward to having a proper discussion with either of you, without personal insults, wikilawyering, or bluster. Martin Hogbin (talk) 22:52, 7 November 2010 (UTC)[reply]

Martin - I said nothing whatsoever about your or Glkanter's understanding here. Richard definitely did - and I agree he shouldn't have. -- Rick Block (talk) 06:40, 8 November 2010 (UTC)[reply]

I don't know, Martin. Under normal circumstances, I would agree. But we're dealing with a very special, unique situation here. These people have the demonstrated ability to divine other writers' intentions, despite the actual words those writers have selected. They've been doing it for months now with reliable sources, and they've shown, like yesterday, that they can do it with the likes of you or I. You need to open your eyes, amigo. GLKANTER 05:06, 8 November 2010 (UTC) 4 tildes added by GLKANTER a 2nd time to eliminate any ambiguities. Glkanter (talk) 12:25, 9 November 2010 (UTC)[reply]

This was a remark by Glkanter? I'm sorry if it insults Glkanter and Martin (and everyone) that I try to figure out how you think about probability and how you think about logical reasoning. Tell me if I got you wrong and then we can have an interesting discussion. As Einstein (I think) said: "ah, now we have a contradiction, now we can make progress!" Then there is the anonymous remark that I also demonstrate an ability to understand Carlton's mind too. I do think I have the qualifications to maybe understand him better ... after all I work in the same field as he does, I know the culture, the hidden assumptions. But you can ignore me if you like.

I do my best to understand how other people think about the things we're talking about here all the time. You can c all it a professional deformation, if you like. I'm alert to hidden assumptions, cultural differences, slips in logical reasoning, and so on. I usually succeed in understanding the people I work with and it usually pays off (consultation, teaching, research, scientific collaboration - 36 years experience which has taken me to the world top in my profession). If you think I get you wrong tell me, because that's a learning opportunity for at least one of us, usually for both of us. I think I understand writers like Carlton who, after all, are working professionally in the same business as me. I know the culture and I know the hidden assumptions. It's not magic. I have no quarrels here. I am not pushing any point of view. I don't have a hidden agenda. I *am* trying to learn how people think about probability and about MHP, because I'm very very concerned why probability reasoning keeps leading to disaster in court and in medical research, to nonsense in the media, and so on. Richard Gill (talk) 12:16, 9 November 2010 (UTC)[reply]

If you were such a great reader of people, you would realize my name on Wikipedia is Glkanter, not something else. I've told you on Wikipedia and in e-mails that you are mis-stating and failing to comprehend my MHP arguments and/or point of view. I've asked you to stop sending me e-mails, but they continue. So yes, despite the fine CV you present, you have demonstrated repeatedly that you do not know what I am thinking, and you are the first person I have ever encountered in my life who has been so bold as to state that they do. Glkanter (talk) 12:25, 9 November 2010 (UTC)[reply]

It's true I often get people extremely angry with me. Sometimes it's not kind to be kind. In Dutch we say "a gentle doctor makes stinking wounds". And no, I am not perfect, despite my fine CV. My apologies. Richard Gill (talk) 12:47, 9 November 2010 (UTC)[reply]

@Richard - If you agree the typical 50/50 answer is based (incorrectly) on

Prob(car is behind Door 1 | car is not behind Door 3)

then doesn't

Prob(car is behind Door 1 | host opens a door)

address a different problem (i.e. is the situation that Door 3 is already open, or not)? There are various phrasings of this other problem out there (Morgan et al. and Gillman both propose a wording), and maybe this problem is as counterintuitive as the usual phrasing, but if in the situation as given Door 3 is already open the sample space has already been reduced, excluding any possibilities involving Door 2 being opened. Again, you can certainly reframe the problem (for example, to something you find more appealing) - but if you do so it seems like you should mention that this is what you're doing. AFAIK, there aren't any sources that explicitly do this. -- Rick Block (talk) 06:40, 8 November 2010 (UTC)[reply]

I agree Rick that in probability theory

(INTUITIONIST) Prob( car is behind Door 1 | host opens a door)

refers to something different, means something different, (even if it has the same numerical value under the standard conditions) from

(CONDITIONALIST) Prob( car is behind Door 1 | host opens Door 3).

Do they address different problems? Well, clearly Carlton thinks so. He thinks that the second probability is the one which you need to know in order to advise Whitaker, not the first one. My personal opinion is not relevant here, but I will remark that it is quite amazing that there seems to be no reliable source which explains WHY you *MUST* condition on *everything* you see. Morgan et al., Nijdam, and Glopk all seem to know that you MUST, but none of them can tell me WHY. Rosenthal argues by example but his examples don't convince me at all. It seems to be a dogma. You just have to believe teacher.

Actually, I can think of several good reasons. The reason I would give to any particular person would have to be tuned to the meaning of probability which that user has in mind. I would give a different answer to a subjectivist as to a frequentist. But I am a scientist and a mathematician, not a Pope. So I don't tell you that you *must* do this or that. Science does not supply moral imperatives.

I would merely try to explain to you why it might be to your advantage to do this or that. (I suppose in a way, the Pope does too - he says that you will burn in hell if you don't do it his way).

This is why it is important for me to figure out how the various co-editors think about probability, though they don't like it when I suggest that I am beginning to get some idea. It's just in order to be able to communicate. We all write English words but actually we all speak subtly different languages. Hence the impasse. Richard Gill (talk) 12:41, 9 November 2010 (UTC)[reply]

So let us discuss the subject then

I agree with you Gill that to find the exact point on which we disagree would be useful. Let me ask who disagrees with the following. Please no rude remarks, bluster, or wikilawyering:

A condition in a probability problem is any event that it is considered might affect the probability of interest.

Rick, Richard, Glkanter who agrees with me? Martin Hogbin (talk) 19:12, 9 November 2010 (UTC)[reply]

My opinion, fwiw is: that is a true statement. Glkanter (talk) 19:41, 9 November 2010 (UTC)[reply]

I agree, but I disagree. – Martin, you are so right. There obviously is no need to condition on "the day of the week" in this concern, but you are free to do so, if you just should like to. Yes, anyone is free to take anything he likes as a condition. Even the fact that the host doesn't open his preferred door, but opens his avoided door. Because in that case switching will win for sure, otherwise only 1/2. Anyone is free to choose "his" conditions as he likes. Why not the "number" of the door (if there are any)?
Let me try to say what Falk says: It does not depend on the "number" of the door the host opens, but nevertheless it could depend on "which one" of the two other doors he has opened. See Ruma Falk. Regards, Gerhardvalentin (talk) 01:24, 10 November 2010 (UTC)[reply]

You seem to be agreeing with me Gerhard. What I am saying is that what is to be taken as a condition in a probability problem depends on the way that you decide to formulate the problem mathematically. When formulating the problem you need to decide which events are independent of which other events and, in particular, which events might affect the probability of interest. There is simply is no formal mathematical way to do this.

In your example, it is quite possible that the day of the week might affect the probability of winning by switching, if the host is not taken to choose a goat-door uniformly at random. On the other hand, if the host is taken to choose a goat door uniformly at random, then we might well take the view, at the start of formulation the problem, that which one of the two doors opened by the host cannot possibly affect the probability of interest and need not therefore be taken as a condition in solving the problem.

I am still waiting to hear some logical argument against what I have said but I fear that all I will get will be bluff and bluster. Martin Hogbin (talk) 09:39, 10 November 2010 (UTC)[reply]

I held off on editorializing, I thought this was a binary 'yes/no' question. If the 'condition' is 100%, the my answer changes. btw the question you asked was "Who agrees with me?". So, once again, I'm a little confused. Glkanter (talk) 09:47, 10 November 2010 (UTC)[reply]

I am not sure what you mean. My assertion is that the only way to decide whether a given event should be taken as a condition in a probability problem (assuming this is not specified in the question) is to ask whether the occurrence of that event might affect the probability of interest. Do you agree with that? Martin Hogbin (talk) 14:57, 10 November 2010 (UTC)[reply]

I agree iff that excludes a 100% 'condition', such as [paraphrasing] 'the host will then reveal a goat behind one of the other doors'. Glkanter (talk) 15:07, 10 November 2010 (UTC)[reply]

I take that as an 'agree'. A 'condition' which occurs with certainty is not really a condition since we need not consider the case that it does not occur. Martin Hogbin (talk)

Rick, Richard, any thoughts? Martin Hogbin (talk) 20:29, 10 November 2010 (UTC)[reply]

Quite nonsense such a question! What do you aim at? Do you mean to give a new definition of the term "condition" yourself? Do you try to rephrase the probabilistic definition? What is it? At least your definition does not coincide with the usual term. We discussed this over and over. Nijdam (talk) 10:39, 11 November 2010 (UTC)[reply]

Nice to hear from you, rather than just call my question nonsense perhaps you explain to me what is wrong with it. You have indeed given me a couple of suggestions as to what makes an event a condition in a problem but none of them stands up to scrutiny.

The most reasonable definition you have proposed is that any event that might reduce the sample space is a condition. The problem with this definition is that it depends on how you choose to set up your sample space in the first place, in particular it depends on which events you consider independent from which other events. For example, you might consider that the probability that the car was originally placed behind door 1 is independent of whether the host says the word 'pick'. In such a case you construct your sample space to not include this event in any way. Clearly the host saying 'pick' will not then reduce your sample space and, under your definition, it is not a condition of the problem. If, however, you consider that the host might be trying to give the player a hint (or mislead then) by the wording he uses, for example by saying 'pick' when the car is behind door 1 and 'choose' when it is not, then your sample space must include the event that the host says 'pick'. The occurrence of this event then reduces the sample space and becomes a condition of the problem. In other words the definition of a condition as an event that might reduce the sample space reduces to my definition. Martin Hogbin (talk) 11:30, 11 November 2010 (UTC)[reply]

Okay Martin, glad to notice you didn't forget what I told you. In fact the term condition is not defined, but the term conditional probability is. Any event may serve as a condition. We discussed that for a conditional probability to real differ from the (unconditional) probability, the event that forms the condition must reduce the sample space. This not just depends on the construction of the sample space, as the sample space has to be rich enough to describe the experiment. Nijdam (talk) 15:19, 11 November 2010 (UTC)[reply]

But whether an event reduces the sample space depends on the sample space that you choose. In solving the problem you have to decide which events might possibly be relevant to the probability of interest. For example, how do you know that the host saying the word 'pick' should not be included in your sample space, or the day of the week, as Gerhard suggests?

On the other hand I might reasonably contend that the sample space for the MHP consists of only two events, the event that the player originally chooses a car (having probability 1/3), and the event that the player originally chooses a goat (having probability 2/3). I assert that no other events are significant, certainly not door numbers or any other method of identifying doors since the standard formulation of the problem is completely symmetrical with respect to door number.

With the particular sample space that you prefer, the problem is indeed conditional, but it need not be. Martin Hogbin (talk) 22:33, 11 November 2010 (UTC)[reply]

The sample space generated by the event Car consists of 4 events: Car, Goat, Certain and Impossible. With this space you cannot describe the MHP. How would you for instance indicate the host opening a door? A (the) sample space for the MHP consists of 27 elementary events, indicating the combinations of Car, Choice and Opened door. I have described this many times. There is no mentioning in the MHP of other relevant information.Nijdam (talk) 23:07, 11 November 2010 (UTC)[reply]

Your last sentence says it all, 'There is no mentioning in the MHP of other relevant information'. In order to decide on your set of elementary elements you need to decide which information is relevant. There are two reasons that you may decide that an event mentioned in the problem statement is not relevant to setting up of the sample space. One reason is that the event happens with certainty. For example, we do not include the possibility that the player does not choose a door. The other reason for not considering an event to be relevant is that we do not consider the occurrence of the event could possibly make any difference to the outcome.

Martin, we do include the certain and the impossible event. Your second reason is no reason for not including the event. What the outcame is, we only know after we have defined our sample space.

Your event 'the host opening a door' is generally taken to happen with certainty; no one ever considers the case that the host does not open a door at all. It is therefore not relevant and need not be incorporated into the sample set. I think we all agree that it would be a different problem if the host did not open a door.

The empty event is the event the host does not open a door. It has probability 0.

More relevant would be the events 'the door opened by the host is number 3' and 'the door originally chosen by the player is number 1'. If these events are considered to be described by the problem as happening with certainty then we do not need to incorporate them into our sample set. If we do we are, in the opinion of most people, answering a different problem.

??????

I have not expressed myself well here. I do not think that we disagree about this, I was just saying that the rules do not state that the player must originally pick door 1 and the host must open door 3. We can apply these conditions though, if we wish. Martin Hogbin (talk) 10:00, 13 November 2010 (UTC)[reply]

No, but the player has to pick some door, call it x, if you like. And the host opens door h, h<>x, h<>"car". This is the same problem, but is more difficult to picture and less comprehensive for the layman.Nijdam (talk) 12:18, 13 November 2010 (UTC)[reply]

We both agree that the player might have originally chosen any of the three doors and that the host then opens any door hiding a goat and not originally chosen by the player. In the standard problem formulation (K&W) we agree that the door number chosen by the player and the door number opened by the host turns out not to make any difference to the outcome. These events therefore need not be incorporated into in our sample space. The only point we disagree on is whether the exclusion of the door numbers from the sample set is justified at the start of the problem as being due to an obvious symmetry with respect to door number or whether it can only be regarded as a conclusion from study of a wider sample set.

The door numbers do matter, we discussed this, although the answer may be the same for all combinations.

If the answer is independent of the door numbers,in what way do they matter? Martin Hogbin (talk) 10:05, 13 November 2010 (UTC)[reply]

The point is, you only know the answer is the same, after the calculation. Nijdam (talk) 12:18, 13 November 2010 (UTC)[reply]

No you may only know the answer is the same after the calculation but I know that it must be the same because of the symmetry with respect to doors. There is nothing in the K&W problem statement to suggest that there is any significance to door numbers. It is therefore intuitively obvious to me, and others, that P(C=1|X=1,H=3) = P(C=1|X=1,H=2) = P(C=1). Given that the host chooses randomly, how could it be otherwise? How can one door be preferred? Martin Hogbin (talk) 16:13, 13 November 2010 (UTC)[reply]

This has been explained to you several times before. You may intuitively know this, or proof it using symmetry, yet it is about the conditional probabilities, as you write down yourself. These are the probabilities of interest.Nijdam (talk) 15:10, 14 November 2010 (UTC)[reply]

And I have responded several times before that, as it is obvious that the (agreed conditional) probability P(C=1|X=1,H=3) that we wish to calculate is equal to the unconditional probability P(C=1), we need only calculate the unconditional probability to get the answer that we require. It is not uncommon in mathematics to first show that the (difficult) thing you wish to calculate must be equal to some other (easier to calculate) thing and then calculate the easy thing to get the answer that you require. In the case of the MHP the intuitive and obvious symmetry makes the first step superfluous. As you know, I have no objection to solving the problem the hard (and more general) way later. Martin Hogbin (talk) 16:21, 14 November 2010 (UTC)[reply]

(editing)

(outindented)

Doesn't it strike you as odd, that you say on the one hand that the conditional probability is the one we wish to calculate, and on the other hand defend that the problem doesn't need conditional probabilities?Nijdam (talk) 16:52, 15 November 2010 (UTC)[reply]

We wish (according to some at least) to calculate a conditional probability but we can calculate this quite easily by noting that its value must be equal to a far more easily calculated unconditional probability. What is wrong with this? Martin Hogbin (talk) 18:38, 20 November 2010 (UTC)[reply]

See my comment below.Nijdam (talk) 00:38, 21 November 2010 (UTC)[reply]

There are some points that we do agree on however. To describe the problem reasonably fully we should start with a sample space with 27 elements rather than 9 as in our, alleged, complete solutions in the article.

If we take the door chosen as given, there are only 9 relevant events.

There is no rational reason to take the door originally chosen as given but the door opened by the host as not so. No version of the problem suggests that this is the case. Martin Hogbin (talk) 10:05, 13 November 2010 (UTC)[reply]

It is better policy, especially in cases of doubt, to start with a wider sample set even if it may be considered unnecessary duplication. Martin Hogbin (talk) 10:53, 12 November 2010 (UTC)[reply]

Do I understand that you also consider the 27 events? Nijdam (talk) 19:05, 12 November 2010 (UTC)[reply]

The richer the sample set, as you put it, the more comprehensive the answer. My sample set of 2 is based on the strict (and correct as it happens) assumption that only the door originally chosen player makes any difference to the outcome for the standard version of the problem.

A sample set of 9 is based on the bizarre assumption that the door originally chosen by the player does not matter but the door opened by the host does matter. There is no rational justification for this for any version of the problem.

A sample set of 27 is based on the assumption that the door originally chosen by the player and the door opened by the host might both matter. This turns out not to be true but it might be wise, if you are in doubt, to address the problem this way.

A sample set of 54 allows for the possibility that the host's saying the work 'pick' might affect the outcome.

An infinite, and therefore somewhat impractical, sample set covers every option.

You have to choose your assumptions before you start the maths. Martin Hogbin (talk) 23:44, 12 November 2010 (UTC)[reply]

This is an interesting question and an interesting debate! Sorry, I am on the road at the moment. I hope to give some reactions this weekend. Richard Gill (talk) 23:39, 12 November 2010 (UTC)[reply]

This is the point that we discussed earlier. The problem has, at some stage, to be converted from a collection of words into a formal mathematical description.

Martin's question was, whether I agreed with the statement: "A condition in a probability problem is any event that it is considered might affect the probability of interest". Though I wouldn't put it in exactly those words, I do agree with him. If you have to make a decision and you want to optimize your *unconditional* probability of making the right decision, then by conditioning on *all* your information you will make an optimal decision. If the conditional probability does not change on leaving out some of your information in the conditioning event, then clearly you'll still be making an optimal decision by only conditioning on what is left.

In our case, under the usual conditions, Prob(door 2 hides car | player chooses 1 and host opens 3) = 2/3. This can be rewritten as Prob(switching gives car | player chooses 1 and host opens 3) = 2/3, because under the specified condition, "switching gives car" and "door 2 hides car" are the same event. It is a fact that Prob(switching gives car | player chooses 1 and host opens 3) = 2/3 = Prob(switching gives car). Therefore computing the marginal probabilities that switching gives car versus staying gives car does give you the optimal decision.

The unconditional probability tells us that "switching, whatever" beats "staying, whatever". Any good argument why the conditional probability (given everything) does not change from the unconditional, tells us that "switching, whatever" beats anything.

If you can find any decent argument why "unconditional 2/3" can't be improved you are done. For instance, the remark that the conditional probability that switching gives the car can't depend on the specific door numbers chosen and opened, is enough.

Once people have been convinced that always switching gives the car with probability 2/3 it will be easy to convince them that one cannot do better. It would not even occur to anyone that you could do better. There clearly can't be a way to win the car with certainty, and it seems totally implausble that there could be a way to win the car with unconditional probability between 1 and 2/3.

Richard, are you now agreeing with me that the unconditional solution alone proves that 2/3 & 1/3 is the optimal solution, and that the conditional solution adds no value in this regard? Glkanter (talk) 12:32, 14 November 2010 (UTC)[reply]

No, the conditional solution improves on the unconditional solution. The unconditional solution tells us that that always switching has overall success probability 2/3, which beats always staying with its overall success probability of only 1/3. The conditional solution tells us that there is no way to improve on the 2/3 overall chance of success attained by always switching. It tells us that any strategy of sometimes switching, sometimes not, depending on the specific door numbers of the specific occasion, and possibly throwing in some extra chance as well, has overall success probability somewhere between 1/3 and 2/3. Richard Gill (talk) 08:06, 15 November 2010 (UTC)[reply]

Maybe you could point out what is wrong with my logic, here, as per the K & W premises? Thanks. Glkanter (talk) 08:55, 15 November 2010 (UTC)[reply]

The discussion about the size of the sample space can be postponed or ignored. We can just as well say what we want to model as a (possibly) random variable. There is a door C hiding the car, a door X chosen by the player, a door H opened by the host revealing a goat, and the remaining door Y to which the player is offered a switch. Everyone agrees that C and X are independent, and H is certainly different from C and X (this makes Y well defined). Everyone (except me) assumes that C is uniformly distributed and that H, conditional on C and X, is uniformly distributed on the remaining doors. Many people take X as nonrandom (and equal to 1), some take it uniform random.

Richard Gill (talk) 08:20, 14 November 2010 (UTC)[reply]

Sorry, I have the idea that Martin's question actually refers to "independence" rather then to "condition". I do not know what you Richard want to demonstrate here. A condition to a probability problem is theoretically any event. A condition may influence the value of some probability, and leave the value of another unchanged. As you well know in the last case the relevant events are called independent. So, what are we discussing here?Nijdam (talk) 15:26, 14 November 2010 (UTC)[reply]

Maybe I have used the wrong terminology, but earlier you said that a condition was any event that reduced the sample space. Are you now saying that a condition is any event at all? Martin Hogbin (talk) 16:49, 14 November 2010 (UTC)[reply]

One can compute conditional probabilities given *any* event of positive probability whatsoever, and an event is just a subset of the sample space. But I think, Martin, you are asking *when* should you condition on something, and if so *what*, when someone asks you a probability. If you want to maximize your *unconditional* chance of success you should condition on everything you can which influences the chance of success. If you want to show that your answer is indeed optimal, you have to show that what you didn't condition on, doesn't matter. Richard Gill (talk) 07:56, 15 November 2010 (UTC)[reply]

Richard, I do not take C to be uniformly distributed, except in the 'standard' (K&W) formulation of the problem, where it is specified to be so. For the moment, at least, I am considering only the K&W' formulation, in which I take C, X, and H to be all uniform at random (even though K&W do not actually specify 'uniform').

No problem. Anyway, I hope we all agree that we *can* talk about the problem in terms of random variables C (car), X (player's first choice) , and H (host) where C and X are independent and H is different from C and X. We can also talk about Y, the door to which the player can switch, it is defined in terms of X and H. The independence of C and X represents the fact that the player makes his choice without knowing the location of the car. Where the car is hidden, doesn't influence the player's choice. This framework accommodates those who take X always equal to 1 because that's the player's favourite number, and those who take X uniformly distributed, and anything else in between, as well. A sample space of size 12 will do fine, it is large enough to contain the six events {C=a,X=b,H=c} where a,b,c is one of the six permutations of 1,2,3, as well as the six events {C=a,X=a,H=b} where a,b are one of the six different pairs of numbers 1,2,3. Now we can start thinking about how to assign probabilities to these events subject to the independence requirement Prob{C=a,X=b}=Prob{C=a}Prob{X=b} for any a, b (same or different) from 1,2,3. And we can figure out what conclusions can be drawn from what assumptions. After agreeing on a framework which is wide enough to incorporate everyone's asssumptions and everyone's approach, we can objectively discuss what can be said about Prob(Y=C) and what can be said about Prob(Y=C|X=1,H=3). That is what I do in my Statistica Neerlandica paper [5] (the paper will come out in vol. 65: January, 2011). See propositions 1, 2, 3, 4 on pages 6 and 9. Richard Gill (talk) 08:38, 15 November 2010 (UTC)[reply]

How would you reword my statement, '"A condition in a probability problem is any event that it is considered might affect the probability of interest"? Martin Hogbin (talk) 09:52, 14 November 2010 (UTC)[reply]

To optimize a decision in a probability problem, it is wise to condition on everything which might affect the probability of success. Richard Gill (talk) 08:42, 15 November 2010 (UTC)[reply]

My urn problem

Nijdam, you never gave me a satisfactory answer to this question.

A game consists of an urn containing 10 balls numbered 0 to 9, otherwise identical. To win you must pick the 9-ball.

The first player picks a ball that proves not to be the 9. This ball is not returned to the urn. The second player picks a ball that also proves not to be the 9. This ball is not returned to the urn. What is the probability that the third player will win? What is the correct sample set to use in this calculation? Martin Hogbin (talk) 10:25, 13 November 2010 (UTC)[reply]

A correct sample space has 720 = 10*9*8 ball-numbers in the first three draws as elementary outcomes. Then P(X3=9|X1=1,X2=2) = P(X3=9,X1=1,X2=2)/P(X1=1,X2=2) = 1/8. The numerical answer will be the same for all other possible combinations of X1 and X2. As only the number 9 ball is of interest, we may prove the situation to be equivalent to 9 black and 1 white ball. Then the answer is given by P(X3=white|X1=X2=black) = ... We even may prove, by reasoning that after 2 balls not being the number 9 ball have been removed, an urn problem with 7 black and 1 white ball remains. It is then however not possible to answer other questions, e.g.. the probability of drawing number 1. Nijdam (talk) 12:11, 13 November 2010 (UTC)[reply]

Would you say that your last answer, 'after 2 balls not being the number 9 ball have been removed, an urn problem with 7 black and 1 white ball remains' is deficient in any way?

What would be you answer if only the 9-ball was numbered? Martin Hogbin (talk) 15:58, 13 November 2010 (UTC)[reply]

Martin initially asks two questions. The zero'th question to answer is whether in his question 1, the questioner is asking for the unconditional probability that the third player will win, or the conditional probability that the third player will win given that the first two players didn't pick winning balls. Given that Martin asked the question and taking account of the order of his sentences, I guess that a conditional probability is intended. I think that the second question needs to be rewritten as "what is a correct sample space to use in this calculation?".

One can give an argument for "1/8" first, and then figure out what are minimal / reasonable/ ... sample space(s) to carry the parts of the solution which are probabilistic in nature afterwards. One might like to imagine that all ten balls are removed from the urn one by one, and take as sample space all 362880 permutations of the numbers 0,1,...,9, assigning equal probability to each. Richard Gill (talk) 08:46, 14 November 2010 (UTC)[reply]

My main point was that, as you say, we always have to, 'figure out what are minimal / reasonable/ ... sample space(s)' to use. There is no mathematical formalism to do this, as Nijdam seems to suggest. To make my point clearer let me add two other slight extensions to the problem.

The question starts, 'A game consists of an urn into which 10 balls numbered 0 to 9 are placed...'. Do we now need to expand our sample space to include the order in which the balls were placed into the urn?
The question explicitly confirms the standard convention regarding urn problems, which is that a ball taken from an urn is deemed to always be selected uniformly at random from the balls in the urn, regardless of the the history.

Nijdam, would you expand the sample set for case 1? Would accept that the 1/8 solution is sufficient for case 2? Martin Hogbin (talk) 09:35, 14 November 2010 (UTC)[reply]

Indeed, I made the conventional assumption that balls are picked uniformly at random, and assumed that the questioner had this in mind too. The idea behind that is that the physical mixing of the balls in the vase ensures that this procedure is close to what anyone would want to call completely random. We do not have to appeal to the principle of ignorance. Note that repeatedly picking a ball from repeatedly well mixed vases will empirically confirm the uniform distribution. Provided of course the balls are the same size, weight, etc etc.; in fact, indistinguishable, as far as the mixing and selection process is concerned. Richard Gill (talk) 07:32, 15 November 2010 (UTC)[reply]

Richard, I presume you are answering my second question. I take it that you agree that if you define balls taken from an urn to be selected uniformly at random from all the balls in the urn then there can be no quibble with the 1/8 answer. Do you agree? What about you, Nijdam?

The objective of giving these two scenarios is to try to find out how we each choose our starting sample space. In the second case I can see no reason whatever to include the history of of how 8 balls, only one of which is the 9-ball, got to be in the urn. Our sample space needs only 8 elements. Does anyone disagree?

In the second scenario I am considering the opposite possibility, that we do not assume the normal convention for urns but we consider the likely case for a real physical urn from which a person draws balls naturally. In this case, the way in which balls were put into the urn could well matter. If the urn is not shaken, the balls put in last might be more likely to be taken first. Also it might matter which balls were removed by previous players. To deal with this problem properly we need to start with a very large sample space. Can we all agree this?

As you will see, I am trying to investigate the rules we use to set up our sample space based on a verbal question. Martin Hogbin (talk) 09:39, 15 November 2010 (UTC)[reply]

It seems to me that the question where one makes a transition from verbal to mathematical mode is a matter of taste. One can make a mathematical model for three successive and completely random drawings without replacement from a vase containing 10 labelled balls and within that model consider the conditional probability that the third draw yields the "9" given that the first two draws didn't; or one can argue verbally that after the first two "not-9" draws we are in the situation of one completely random draw from an urn of 8 labelled balls of which just one is labelled "9". The unconditional model will involve more complex mathematics; the conditional model is much simpler. Actually, more or less the same physical intuitions are being used to reduce the problem to the 8-ball-vase as are used to solve the 8-ball-vase problem.

Yes, of course it is a matter of taste how you model the problem and I fully accept that. Some people may, for whatever reason, want to model version 2 of my problem as three successive draws from an urn, but that is not my point. The real question that I am asking is whether the 1/8 solution is can be reasonably criticised. I would particularly value Nijdam's opinion on this. Martin Hogbin (talk) 15:20, 15 November 2010 (UTC)[reply]

I don't like the hang-up on sample spaces. In probability theory we talk about events and about random variables, about probabilities, conditional probabilities, and expectation values. The sample space is a kind of scaffolding used to put up the building; or a kind of foundation which we understand is there, but never go down and inspect. It plays a major role in chapter one of introductory books on probability and statistics but gradually recedes from view as we slowly become accustomed to probabilistic reasoning. The important things to decide are: what is the basic unit of repetition; what are the random events and random variables which we want to have in the probability model; what do we want to attain. Richard Gill (talk) 10:54, 15 November 2010 (UTC)[reply]

I am not the one hung up on sample spaces, I am happy to discuss the problem any way that you want but others, including Morgan and Nijdam, talk about the 'correct' sample space. I am trying to show that the 'correct' sample space is a matter of choice, much as you say.

I think my urn problem (version 2) is fairly specific about what the random variables are and what I would like to obtain is the probability that the third player will draw a 9. Martin Hogbin (talk) 15:20, 15 November 2010 (UTC)[reply]

You say "the" probability Martin, but all probabilities are relative to something. And the word "probability" has different meanings to different people. Under the standard assumptions, the probability that the third player will draw a 9 is 1/10. Under the standard assumptions, the probability that the third player will draw a 9 given the first two players didn't is 1/8.

But you know all this already! (What was version 2 exactly?) Richard Gill (talk) 15:44, 15 November 2010 (UTC)[reply]

This is the full question,

A game consists of an urn containing 10 balls numbered 0 to 9, otherwise identical. To win you must pick the 9-ball. The first player picks a ball that proves not to be the 9. This ball is not returned to the urn. The second player picks a ball that also proves not to be the 9. This ball is not returned to the urn. A ball taken from the urn is deemed to always be selected uniformly at random from the balls in the urn, regardless of the the history What is the probability that the third player will win?

No doubt it is possible to contrive bizarre answers with sufficient imagination but I can see no reasonable interpretation of the question or 'probability' that gives an answer other than 1/8. Martin Hogbin (talk) 20:08, 15 November 2010 (UTC)[reply]

(outindented) The answer is 1/10. The conditional probability, given the first two did not win, is 1/8. Nijdam (talk) 15:52, 16 November 2010 (UTC)[reply]

How about for the little green man that just arrived from outer space only after the 2 balls had been removed? Glkanter (talk) 16:05, 16 November 2010 (UTC)[reply]

Nijdam, are you seriously telling me that you would answer 1/10 to my question above? I can only imagine that you are alluding to my recent insistence on having the word 'given' in a recent MHP discussion.

Well there you are, you have to ask the right question. If the information you gave about the first two players are intended to have taken place and the third player knows this, your question about the third player is a CONDITIONAL probability. You may call it a probability given the known past, or the posterior probability, or the probability after the given events have taken place, my pleasure, they all refer to the CONDITIONAl probability. This should be clear by now. If you on the contrary are just asking for THE probability, I have to assume you are referring to the (unconditional) probability. Nijdam (talk) 22:16, 16 November 2010 (UTC)[reply]

I've left you for a moment with this, but now have to confront you with your own inconsistency. In your urn problem you picture a situation, in which two forgoing players have picked a no winning ball, and then as a solution for the next player to win, you think it natural to calculate the conditional probability. In the MHP, however, you insist in a complete comparable situation, to calculate the unconditional probability. Nijdam (talk) 10:08, 17 November 2010 (UTC)[reply]

Martin, do you notice the inconsistency in your arguments??Nijdam (talk) 10:45, 20 November 2010 (UTC)[reply]

There is no inconsistency in my arguments. In both cases (assuming door numbers are intended to be specified in the MHP) we wish to calculate a conditional probability that is, a posterior probability given certain events. The argument is about how we calculate this conditional probability. How would you do it for my urn example? Would you start with a huge sample space including every possible order that the balls were put into the urn and every possible pick by the previous two players and then condition it, or would you simply note that the probability of interest is independent of all these events (this is specified in my problem statement) and simply calculate the unconditional probability given 1 9-ball and 7 not-9-balls because you know that this must be equal in value to the probability that you wish to calculate. Martin Hogbin (talk) 18:48, 20 November 2010 (UTC)[reply]

Happy throng

[outdented]I'm glad to hear from you that you also wish to calculate a conditional probability. That's the only issue I'm interested in. I won't dwell in the past and look for supposed statements of yours otherwise. Stick to this and you are in good company, i.e mine, Rick's, Glopk's, Kmhkmh's and in the end also Richard's. This is our one and only issue! Really. And be sure, the simple solutions, as posed by the sources, do not wish to calculate a conditional probability. It is not a matter of how it is calculated, but only what is to be calculated. And indeed, with certain assumptions, there is symmetry, that may be used to ease the calculation. Even an intuitive argujment may be given, BUT has to be given (mentioned). And the simple solutions are not only incorrect because they fail on this argument, but mainly because they do not intend to calculate a conditional probability. I hope, after reading this, you are converted to the one and only right belief. Nijdam (talk) 00:27, 21 November 2010 (UTC)[reply]

Sorry Nijdam, I an not yet quite ready to join your happy throng of true believers. For a start you will note that I said, 'assuming door numbers are intended to be specified in the MHP'. Although this is fairly clearly the intention in the K&W formulation it is not so with Whitaker's question, especially as the door numbers were added by vos Savant. You have never given me a reason why Whitaker would have wanted advice only on a specific combination of doors.

Regarding the symmetry argument, I have suggested a compromise, which Rick has rather made fun of, which is to add some positive and not off-putting wording to the simple solution section stating that we have made use of a obvious and intuitive symmetry. As Morgan put it in response to our letter, 'had we adopted conditions implicit in the problem...'. To make clear what I am aiming at I am proposing an unintrusive justification for using the simple solution to calculate what some consider to be a conditional probability just to cover our mathematical integrity but not a warning or disclaimer. Would you like to suggest something?

I presume you would also wish to add some kind of rationale in the so-called 'Conditional probability solution' for ignoring the player's original choice of door and calculating P(C=1|H=3) (for a game in which the player can only pick door 1) when you really want to know P(C=1|X=1,H=3). ~~We could even use the same words as the reason is exactly the same, the obvious and intuitive symmetry with respect to door number~~. Martin Hogbin (talk) 11:03, 21 November 2010 (UTC)[reply]

There is no virtue in myopia. The rigidness and doctrinaire needed to overlook the equivalency of the ten less two balls and the eight balls scenarios in Martin's urn problem does not indicate superior knowledge of the subject matter. Glkanter (talk) 02:58, 17 November 2010 (UTC)[reply]

I agree. There comes a point where attempts at additional sophistry do nothing but discredit the author. As Boris has pointed out, if you want to be really rigorous in mathematics you can barely move off the spot, and and Richard has said, there still no universal agreement on the meaning of the word 'probability'. On the other hand my question was pretty clear to most people, although as I expected, with sufficient ingenuity, it is still possible to misinterpret it. Martin Hogbin (talk) 10:12, 17 November 2010 (UTC)[reply]

The question that I asked was perfectly clear in what was being asked. I really do not care whether you use the word conditional or not, all Bayesian probabilities are conditional. I have amended the question below to suit you, now perhaps you would care to answer it. Martin Hogbin (talk) 22:54, 16 November 2010 (UTC)[reply]

In the problem above we are told many facts, that there are balls in the urn instead of say, boxes, that the balls are numbered 0 to 9 instead of, say 1-10, and we have to decide which of these facts are relevant to calculating the probability of interest. The numbering of the balls seems very unlikely to affect the probabilities thus we do not take this as a significant condition of the problem. On the other hand it seems likely that the removal of two balls, neither of which is the 9-ball, will affect the probability of interest. Thus it would seem sensible on the above basis to take the removal of the balls as a condition of the problem but not the numbering of the balls. This was the very point that I was making earlier. A condition is only relevant and only needs to be considered if it might affect the probability of interest.

What is your answer to the question below question and what is the appropriate sample space to use?

10 balls, numbered 0 to 9. are placed into an urn one at a time. Given that two balls have then been removed from the urn, neither of which is the 9, what is the (conditional if you insist) probability that the next ball taken will be the 9? A ball taken from the urn is deemed to always be selected uniformly at random from the balls in the urn, regardless of the the history. Martin Hogbin (talk) 18:16, 16 November 2010 (UTC)[reply]

You were asking for Nijdam's answer but this is my talk page so I'll give you my answer. The probability that the next ball will be a 9 given the preceding two balls were not, is 1/8. Your question "what is *the* appropriate sample space to use" is a bad question. Many sample spaces (and assignments of probabilities) can be considered appropriate and all appropriate sample spaces would lead to the same answer. Richard Gill (talk) 18:39, 17 November 2010 (UTC)[reply]

Thanks for your answer, I am still waiting for Nijdam's. You know the point that I am getting at, which is that given that the question specifically states that, 'A ball taken from the urn is deemed to always be selected uniformly at random from the balls in the urn, regardless of the the history', a sample space of 8 elements is not incorrect. To put it another way, would you say that the sample space chosen must include the possible balls that were taken by the first two players and the order in which the balls were put into the urn? Martin Hogbin (talk) 19:11, 17 November 2010 (UTC)[reply]

Well, there isn't much to add to Richard's answer. Your problem, Martin, is, you use terminology, which is not well defined. You asked for "the probability". And since you've given some experiment, I have to interpret "the probability" as the probability linked to this experiment. All other probabilities are conditional ones. That's why I gave the amswer 1/10, and commented it. Most unexperiencied people automatically think of the conditional probabilty after given some events has happened.for the probaility. You BTW do not do so in the MHP.Nijdam (talk) 16:38, 18 November 2010 (UTC)[reply]

My question is perfectly clear let me repeat it to make sure that you are reading the right version:

10 balls, numbered 0 to 9. are placed into an urn one at a time. Given that two balls have then been removed from the urn, neither of which is the 9, what is the (conditional if you insist) probability that the next ball taken will be the 9? A ball taken from the urn is deemed to always be selected uniformly at random from the balls in the urn, regardless of the the history.

There is only one possible answer to this question which is clearly 1/8. Please explain to me how it can be anything else. Martin Hogbin (talk) 20:27, 18 November 2010 (UTC)[reply]

Nothing to explain, as long as you know this 1/8 is the value of the conditional probability. Got it? Nijdam (talk) 23:27, 24 November 2010 (UTC)[reply]

Yo seem to be avoiding the point. What sample space would you start with to get this conditional probability? Martin Hogbin (talk) 23:50, 24 November 2010 (UTC)[reply]

Response to Richard's Comments On The Mediation Page

One could state that many of K & W's premises are vital to the simple solution being valid. Why single out any one of them? Glkanter (talk) 19:02, 17 November 2010 (UTC)[reply]

Because it seems to be the one that we disagree on. I am trying to find our exactly where we disagree.Martin Hogbin (talk) 19:12, 17 November 2010 (UTC)[reply]

That's not exactly my question, I don't think. Why should it be a *requirement* to single out for greater emphasis *any* premise when giving a solution? Glkanter (talk) 19:49, 17 November 2010 (UTC)[reply]

Actually, the door numbers are not premises per K & W. Opening another door to reveal a goat is the premise. How can the door numbers be a *requirement* for a solution, then? Glkanter (talk) 19:54, 17 November 2010 (UTC)[reply]

It's not a question of singling out premises. It's a question of being careful. And by the way, the simple solution is valid without most of K&W's premises. But that's a different issue.

It's all so simple. The simple solution says that always switching beats always staying. In order to get that conclusion, you only need to assume that your initial choice is right with probability 1/3. The conditional solution tells you that always switching beats everything else. In order to conclude that,, you only need to assume that each door has probability 1/3 to hide the car (Morgan et al solution). If you want an easy argument why always switching beats everything (uncondtional+symmetry), you need to assume all K&W's conditions.

The more you want to conclude, the more you'll have to assume, and the more hard work you'll have to do. If you're happy with a Volkswagen, buy a Volkswagen. If you want an Alpha-Romeo, raise the cash and buy an Alpha-Romeo. Most sources are happy with Volkswagens. Some insist on an Alpha-Romeo. But I think that this is a matter of taste. The tastes of the sources differ, the tastes of the editors differ. Please let's keep matters of taste and opinion separate from matters of fact (logic, mathematics). Wikipedia has to report what sources say, we can't arbitrate on matters of opinion. Richard Gill (talk) 20:19, 17 November 2010 (UTC)[reply]

I agree regarding the premises. The decision tree I derived from Carlton's simple solution demonstrates that very few premises are required, that the simple solution *is* (100%) conditional, that 1/3 prior = 1/3 post, and that the result is 2/3 & 1/3. I wrote above how the simple solution, and the application of logical reasoning, demonstrate that there are no 'mixed' strategies available to the contestant, so that switching with 2/3 success has to be optimal, no conditional solution required to prove that. I post these things all the time (wish I didn't have to), and yet you and the others choose to keep arguing around them, endlessly. Bunch of talk, for no benefit. Glkanter (talk) 20:29, 17 November 2010 (UTC)[reply]

Martin asked on the mediation page "Is it the fact that the host opens a door to reveal a goat, which is important since it influences the probability of interest or is it the fact that the host chooses door 3 which does not influence the probability of interest for the standard case?" This is exactly backwards. The "condition" that the host "opens a door to reveal a goat" influences no probabilities whatsoever. It always happens, does not change the sample space, so does not change ANY probabilities. The probability the car is behind the player's door (given this) is still 1/3, and the probability the car is behind each of the other two doors is also still 1/3. You know one of them is open, but you don't know which one so you can't say the probability of one of them is 0 and the other is 2/3 - the probability of each of them is still 1/3. Although extremely counter-intuitive, this is what refusing to identify the door the host open does - and I would argue it is not how anyone actually thinks about the problem.

On the other hand, the condition that the "host chooses door 3" does influence all the probabilities. Knowing the host opens door 3 reduces the sample set, excluding all cases where the car is behind door 3 (allowing us to say the probability the car is behind door 3 is now 0 not 1/3) but also excludes some cases where the car is behind the player's chosen door as well. The probabilities of all the doors are now conditional probabilities - (in the standard case) 1/3:2/3:0 rather than the "obvious" 1/2:1/2:0. Saying "the probability of the door the host opens is 0 so the probability of the other door is 2/3" (and implying it applies to the specific case of player picks door 1 and host opens door 3) is talking about conditional probabilities and fundamentally depends on knowing which door the host opened. If you don't know which door was opened, then you simply can't say this (all you can say is the probability of both unpicked doors is 1/3). If you do know which door was opened, then (and only then) can you say its probability is 0 and talk about the probabilities of the player's door and the 3rd door - but these are all now conditional probabilities.

Richard - please confirm for Glkanter and Martin that everything I've said here is The Truth. Assuming they want to continue arguing about it (I don't), feel free to try to explain it to them. -- Rick Block (talk) 02:35, 18 November 2010 (UTC)[reply]

What Rick says is The Truth. Glkanter gave the same argument himself, I thought, in his decision tree. The probability the host will open a door and reveal a goat is 1 (conditional on the player choosing door 1). Hence conditioning also on the host revealing a goat by opening a door does not change any probabilities whatsoever. Therefore if the probability that the car is behind door 1 given the player chose door 1 is 1/3, then the probability the car is behind door 1 given the player chose door 1 AND the host opened a door revealing a goat is also 1/3. Hence the probability the car is behind the door the host didn't open given the player chose door 1 and the host opened a door revealing a goat, is 2/3, since the car has to be behind the door the player chose or the remaining door.

If you want to know what is the probability the car is behind Door 2 given the player chose Door 1 and the host opened Door 3 you will have to do more work.

With the K & W premises (the decision tree derived from Carlton's simple solution shows that from the contestant on a game show's uninformed SoK - each door has the same 1/3 chance of hiding the car, the host always revealing a goat is the only premise that needs to be stated), I know that the host is equally likely to open either door I haven't chosen (the unchosen doors are still 1/3 and 1/3 before he opens a door), and various simple solutions determine the overall probability is 2/3 & 1/3. If the overall (average) probability for the unchosen doors is 2/3, and they're equally likely, then either unchosen door is 2/3. That's a shortcut to the door 1 and door 3 problem, not misunderstanding or avoidance. Just like Monty Hall explained, and Selvin heartily endorsed. Glkanter (talk) 07:23, 19 November 2010 (UTC)[reply]

This is indeed cute and counter intuitive. The probability the car is behind Door 2 given the player chose Door 1 and the host opened a different door revealing a goat is 1/3. The probability the car is behind the door the host didn't open given the player chose Door 1 and the host opened a different door revealing a goat is 2/3. Richard Gill (talk) 22:22, 18 November 2010 (UTC)[reply]

Richard - say more explicitly what "you will have to do more work" means, i.e. that the argument that the probability given the host opens "a door" results in a 2/3 chance of winning by switching DOES NOT MEAN the probability given the host opens door 3 results in a 2/3 chance of winning by switching. This is the precise point of disagreement. I would venture a guess that Glkanter would argue about this until his dying breath. -- Rick Block (talk) 06:50, 19 November 2010 (UTC)[reply]

You are correct, Rick. That's pretty much what I said, and my understanding of the problem hasn't changed since I posted this on the MHP talk page on February 7, 2009. My point of view is supported by reliable sources, including Selvin and vos Savant, and I understand that there are other reliably sourced solutions and POVs as well. Glkanter (talk) 07:08, 19 November 2010 (UTC)[reply]

By "do more work" in order to figure out what is Prob(win by switching| player chose 1, host opens 3) I mean that a logical argument or a calculation is needed to justify the numerical answer which you might find by guessing or intuition. For instance intuition might tell you that this cannot possibly be different from Prob(win by switching| player chose 1) and in this case intuition is correct; but then most people's initial intuition that the probability is 50-50 was incorrect, so it is wise to be careful. Especially if we are using MHP to develop our powers of reasoning and problem-solving, and we are using it to show people that probabilistic reasoning can be tricky.

Glkanter's remark about reliable sources is not relevant to this issue. Every reliable source gives a logical argument for the correctness of their solution. But reliable sources clearly differ in what they consider to be the problem, since some are interested in A: Prob(win by switching| player chose 1) and others in B: Prob(win by switching| player chose 1, host opens 3). Some reliable sources are clearly confused themselves, Selvin for instance.

My only interest in any of these discussions is to support the removal of the NPOV violations and OR from the Wikipedia MHP article. I believe its valuable to demonstrate that my arguments are not OR, but come from reliable sources. Glkanter (talk) 08:43, 19 November 2010 (UTC)[reply]

But not much work is needed. A couple of short sentences is enough to get from A to B. Prob(win by switching| player chose 1, host opens 2 ) = Prob(win by switching| player chose 1, host opens 3 ) by symmetry. These two probabilities average out to Prob(win by switching| player chose 1) = 2/3. Therefore they must both equal 2/3. Richard Gill (talk) 08:31, 19 November 2010 (UTC)[reply]

I don't care about labels. And your explanation is hogwash. I say 'simple' or 'conditioned on the 100% a goat will be revealed' (the tree derived from Hall's/Carlton's simple solution), or 'combining doors', etc... The conditional decision tree based on door #s, like Chun's has no claim to 'exclusively right'. And this is all based on reliable sources. Glkanter (talk) 02:45, 18 November 2010 (UTC)[reply]

What does "right" mean? Under the usual assumptions, there are easy ways to correctly and logically deduce that the switcher wins 2/3 of the time. Under the usual assumptions, by employing symmetry, one can extend this to correctly and logically deduce that the switcher wins 2/3 of the time that the player chose door 1 and the host opened door 3. I should hope that any wikipedia editor who can get dressed in the morning could agree on these two statements and could agree that they are not the same.

What is a matter of opinion according to wikipedia criteria, is whether MHP is solved by showing that the switcher wins 2/3 of the time, or by showing that the switcher wins 2/3 of the time that the player chose Door 1 and the Host opened Door 3. The sources apparently disagree in what constitutes a solution to MHP, therefore we must agree that the sources disagree in what constitutes MHP. We editors cannot resolve this disagreement, we can only report it neutrally. Even though some of us think that the MHP is problem 1, others think that MHP is probem 2, and others think it is neither.

Moved to here

I have no objection to the word 'conditional', especially as it strictly applies to all Bayesian probabilities. By all means refer to the MHP as a conditional probability. The point is what is the relevant condition. Is it the fact that the host opens a door to reveal a goat, which is important since it influences the probability of interest or is it the fact that the host chooses door 3 which does not influence the probability of interest for the standard case?

Anyone, please feel free to move this discussion back to Richard's talk page, assuming he does not mind, or to my talk page, if you feel that it is not appropriate here. Please join in if you wish. Martin Hogbin (talk) 15:29, 17 November 2010 (UTC)[reply]

Martin, you put it in a splendid way. You don't need to condition on the fact that the host chooses door 3 in order to find the conditional probability is 2/3, but you do need to know that this fact doesn't change the probability! So you can solve *the standard* MHP the simple way and just remark afterwards that by symmetry the specific door numbers don't matter, or you can do it the long way and find it out as a result of a long computation. Some people will say that this is so obvious and intuitive it is a waste of time. But that is a poor argument, since ordinary people's use of obvious intuition leads to the 50-50 result. MHP teaches us that immediate intuition is dangerous when solving probability puzzles. So lovers of the simple solution ought to admit that the finicky and pedantic conditionalists have got a point.

[This seemed to have got lost in the move] I agree that the conditionalists have a point, as a general rule it is better to take into account all possible conditions in problems such as this and thus the conditional solution is probably a better approach, particularly for students of probability. I would argue that the best approach is to show all possible doors that the host might have chosen. To show only one door creates an inconsistency in the way player's choice of door and the host's choice of door are treated.

All I am saying is that the simple solution must not be branded as 'wrong', 'incomplete', or 'answers the wrong question', especially in a way that might put readers off understanding or reading it. It is correct even though most people who read it may not know exactly why.

Let me give an example from my school days. We had two French teachers. Occasionally a pupil would, by accident, use a word that was perfectly correct French but one which they had not been taught yet. One teacher would mark it wrong but put a note in the margin explaining that it was technically correct, the other would mark it right but put a note in the margin explaining that the pupil had made a lucky mistake. We all had a lot more respect for the second teacher.

So it is with the MHP, I have no objection to, after the simple solution, explaining what assumptions are behind it and how more general cases should be treated, but to have a 'heath warning' in the simple solution section is both technically incorrect and off putting. Some editors here seem stuck on the concept, 'simple solution wrong - Morgan solution correct'. There is much more to it than that and to present the issues underlying the MHP in that overly simplistic way does a service to no one. Martin Hogbin (talk) 15:37, 18 November 2010 (UTC)

Martin, Morgan's original paper, not being aware of Selvin's 50/50 goat door premise (or that on a game show the host doesn't tell the contestant where the car is), had the conditional decision tree as a false solution. Their 'correct solution' said it was >= 1/2 in all cases, so switch. Later they said 'its 2/3, period'. So Morgan never said 'the conditional decision tree is right, and the simple solutions are wrong'. I'm not sure which (any?) reliable sources have said that, but there are certainly editors here who are advocates of that POV. Glkanter (talk) 08:28, 19 November 2010 (UTC)[reply]

[This also got lost during the move] I've made my living finding shortcuts to solving problems. I don't agree '...as a general rule it is better to take into account all possible conditions in problems such as this and thus the conditional solution is probably a better approach...' I find the *meaningful conditions* then solve the problem. Both approaches work. Each has its place. One has more utility in academia, the other in practical application. Glkanter (talk) 15:51, 18 November 2010 (UTC)[reply]

I've made my living in acadaemia finding short cuts to solving problems. That's what creativity is about. It's also important to be precise and to develop language and tools which allow a higher degree of precision. Mathematics made enormous strides with enormous pay-off to science and technology by investigating situations like MHP where intuitions clash and people can't come to agreement using the existing tools. Richard Gill (talk) 07:51, 19 November 2010 (UTC)[reply]

Quote: "Martin, ... You don't need to condition on the fact that the host chooses door 3 in order to find the conditional probability is 2/3". Which conditional probability do you mean than?? Nijdam (talk) 16:57, 18 November 2010 (UTC)[reply]

Sometimes I have the feeling that what I and Rick consider the simple solution, and what we explained several times, and what is found in the sources, is not what other people consider to be the simple solution. Worse, the notion of what the simple solution is, seems to change now and than. Nijdam (talk) 17:00, 18 November 2010 (UTC)[reply]

Yes, I do mean that you do not have to condition on the fact that the host chooses door 3 to get the conditional probability P(C=1|X=1,H=3) in the standard case because you know it must be exactly equal to P(C=1). Sometimes in mathematics there are shortcuts. Martin Hogbin (talk) 19:46, 18 November 2010 (UTC)[reply]

Martin is right. In order to conclude that P(C=1|X=1,H=3)=1/3 you do not need to compute it explicitly from first principles. You can argue that P(C=1|X=1)=1/3 and then argue by symmetry that P(C=1|X=1,H=2) must equal P(C=1|X=1,H=3) and hence both of the latter two probabilities must equal P(C=1|X=1)=1/3. Thus: the simple unconditional argument tells you that always switching is smarter than always not switching, and symmetry tells you that it is not only smart but that it is also the smartest thing to do. Richard Gill (talk) 22:53, 18 November 2010 (UTC)[reply]

I'm really getting tired of this discussion. Tell me, why do you want to conclude that P(C=1|X=1,H=3)=1/3?? Nijdam (talk) 10:40, 20 November 2010 (UTC)[reply]

And I would argue that most people spot the natural symmetry of the problem with respect to door number and do something like is done in the the argument above intuitively. Those who do not see the symmetry (or who want to check that their intuition is correct) must do it the hard way. Martin Hogbin (talk) 00:02, 19 November 2010 (UTC)[reply]

I think you are right Martin. This insight needs to get published in the scientific literature (especially on mathematics education and problem solving) since it seems to be another original contribution to MHP studies. Maybe time for another letter to The American Statistician? Richard Gill (talk) 07:57, 19 November 2010 (UTC)[reply]

Why so excited about this symmetry argument? And where in the simple solution is symmetry mentioned? Any idea what the simple solution is?Nijdam (talk) 10:40, 20 November 2010 (UTC)[reply]

I don't agree that Martin has developed a new 'insight'. Rather he has stated, once again, what the typical English speaking Wikipedia reader knows about about game shows. Absent any information, uniformity across all alternatives is the best/only model available. Regardless, all door symmetries are clearly stated in the K & W premises. [User:Glkanter|Glkanter]] (talk) 08:15, 19 November 2010 (UTC)[reply]

Glkanter, you said that you yourself wrote "how the simple solution, and the application of logical reasoning, demonstrate that there are no 'mixed' strategies available to the contestant, so that switching with 2/3 success has to be optimal". I'm happy to credit you as well as Martin with this insight, which seems to be shared now by all of us. I agree that the typical English speaking Wikipedia reader ought to be able to gain this insight too. Now, please give me a reliable source where it is written out clearly that "the simple solution and the application of logical reasoning show together that always switching is optimal". Richard Gill (talk) 08:45, 19 November 2010 (UTC)[reply]

That's not necessary for the Wikipedia MHP article. All that's needed is that its better to switch, and 2/3 is greater than 1/3. You have been claiming that the conditional decision tree is necessary to determine that always switching is optimal. The simple solution plus logic isn't for the article, it was to end your argument that the conditional decision tree answered a question that the simple solutions in the Selvin or K & W MHPs can't answer. Now you've agreed that the simple solutions are enough to determine that 2/3 is optimal. Glkanter (talk) 08:57, 19 November 2010 (UTC)[reply]

The conditional decision tree is one way to obtain the conditional probability given the specific door numbers at hand, and hence the result that switching is optimal. I have never claimed that it is the only way. I've agreed, in fact I have always been pointing out, that the simple solution plus simple logic together does the same. The simple solution alone only shows that always switching beats always always staying.

What's necessary for the wikipedia article? What should be in there is determined by the so-called reliable sources, not by The Truth. There are reliable sources which say that the simple solution is definitely wrong so I'm afraid the article will have to pay some attention to their contribution. But it is good that we are approaching agreement, I think, concerning The Truth. I have always said that sensible editors ought to be able to agree on the The Truth and thereafter use it as a guide to writing a good article.

If there are no reliable sources which say that the simple solution plus simple logic together prove optimality and render all complicated solutions redundant then it is a pity for the article that we can't reproduce that point of view. We could try to produce a new reliable source for future generations of wikipedia editors. Richard Gill (talk) 09:19, 19 November 2010 (UTC)[reply]

I think your focus on 'the optimal strategy' is a tangent of interest, perhaps, only to you. My struggle has been to eliminate criticisms of the simple solutions from appearing with the simple solutions, or from building up the conditional decision tree as somehow superior. Of course the criticisms will be addressed later in the article. Its pretty clear Nijdam's requirements for the simple solution section are intended to undercut the simple solutions, hence they are an NPOV violation, and give Undo Weight to a minority POV. Morgan's paper calls the conditional decision tree solution false, too. Glkanter (talk) 09:27, 19 November 2010 (UTC)[reply]

Well, I think it is also of interest to Rick, Nijdam, glopk, Kmhkmh; and Martin sees the point of fussing about conditional probabilities. And regarding your struggle: I have exactly the same aims and I try to promote them by emphasizing that an easy logical argument provides a bridge from the simple solutions to the conditional solution and indeed makes long formal calculations completely superfluous. I also tried to support this struggle by emphasizing that the simple solutions make less assumptions hence are more widely applicable. Richard Gill (talk) 09:35, 19 November 2010 (UTC)[reply]

You seem to miss the whole point in our controversy., which is that that the simple solution lacks any necessary arguments, be it logically, mathematically, intuitively or whatever. Nijdam (talk) 10:40, 20 November 2010 (UTC)[reply]

No, I accept that point, but I ask you to accept my point that the symmetry argument that the conditional probability P(C=1|X=1,H=3) required must be equal to the unconditional probability P(C=1) is intuitively obvious to many people and happens, in this case, to be correct.

Really Martin, you still don't get it! The point is not that there is a symmetry argument - I mentioned it 3 lines above - but the LACK of it in the simple solution. Nijdam (talk) 12:43, 20 November 2010 (UTC)[reply]

I believe that the argument is sufficiently obvious and intuitive to most people that it can be omitted however, see my suggestion below and on the mediation page. Martin Hogbin (talk) 19:02, 20 November 2010 (UTC)[reply]

I personally would not be against some kind of statement to this effect in the simple solution section so long as it was phrased in a positive way so as not to put of understanding the simple solutions. I cannot speak for others however. Martin Hogbin (talk) 12:38, 20 November 2010 (UTC)[reply]

Nijdam: the simple solution is easy to formalise, I don't know what you mean by "lacks all any necessary arguments". Let X = door chosen by player, C = door hiding car, H = door opened by host, Y = door to which player is offered a switch. H is different from X and C. Y is different from X and H. Simple solution: Prob(X=C)=1/3 implies Prob(Y=C)=2/3. Switching wins the car with probability 2/3 if we assume the initial choice X has 1/3 chance to be correct. Now Prob(Y=C) = sum over x, h Prob(Y=C|X=x,H=h) x Prob(X=x,H=h) . If initially all doors are equally likely to hide the car and the host is equally likely to open either door when he has a choice, the problem is invariant under renumbering of the doors. Thus all the conditional probabilities Prob(Y=C|X=x,H=h) as x, h vary must be equal. They average out at 2/3 so they must all equal 2/3. Conclusion: switching gives the car with conditional probability 2/3, given the initial choice of the player and the door opened by host. Corollary: not only does "always switching" beat "always staying" in terms of overall success chance (2/3 versus 1/3), it also beats any other strategy (2/3 versus something in between 2/3 and 1/3). Richard Gill (talk) 14:21, 21 November 2010 (UTC)[reply]

Really Richard, I'm rather amazed you still defend the simple solution. I think I made it rather clear its flaw not only lies in not giving the necessary arguments to show the relation between the unconditional and the conditional probability, it simply does not mention the starting point of the solution has to be the conditional probability. I'm happy with A simple solution that starts by mentioning in some form the need to base the decision on the posterior (conditional, or given the presented situation) probability and reasons, by symmetry or simpler by intuition, that this probability will be (numerically) the same as the original (prior, unconditional) one.Nijdam (talk) 17:02, 21 November 2010 (UTC)[reply]

Richard: It also beats any other strategy (2/3 versus something between at most 1/2 and 1/3). Correct? Gerhardvalentin (talk) 15:34, 21 November 2010 (UTC)[reply]

I'm focused on the editors who insist on including criticisms of the simple solutions in the solution sections. As for 'superfluous', that opinion doesn't belong in the solution sections, either. Whatever is reliably sourced belongs in the solution section. Nor does the discussion of 'less assumptions' and 'more widely applicable' belong in the solution section. Glkanter (talk) 10:02, 19 November 2010 (UTC)[reply]

What about some positive explanation of what is intuitive obvious to many for the benefit of those poor should for whom it is not? Martin Hogbin (talk) 12:38, 20 November 2010 (UTC)[reply]

Each editor will give his own opinion regarding 'optimal solutions'. I don't think this issue serves the Wikipedia reader trying to understand why its 2/3 & 1/3 instead of 50/50, which is what I believe the solution sections should address. In fact, I think it would cause confusion to the reader if your 'optimal solution' discussion, which many readers will infer intuitively, was in the solution sections. We're only putting in reliably sourced solutions, why do we have to make a case that any of them are worse, as good as, or better than any others in the solution sections? Glkanter (talk) 09:49, 19 November 2010 (UTC)[reply]

Who are your sources for all this optimal solution stuff? Is it addressed often in the reliable sources? Glkanter (talk) 09:53, 19 November 2010 (UTC)[reply]

My source is Gill (2011) (peer reviewed and accepted for publication in January 2011). It is the result of an attempt to explain to "simplists" like yourself why the "conditionalists" insist on figuring out the conditional probability. The reason being that the simple solution only tells you "always switch" beats "always stay". The conditional solution tells you "always switch" beats everything. I found it striking that the conditionalists only state as some kind of dogma that you base your decision on the conditional probability, but they never tell you a good reason for doing so. Richard Gill (talk) 14:49, 21 November 2010 (UTC)[reply]

Richard, I think you've forgotten this:

"Glkanter, you said that you yourself wrote "how the simple solution, and the application of logical reasoning, demonstrate that there are no 'mixed' strategies available to the contestant, so that switching with 2/3 success has to be optimal". I'm happy to credit you as well as Martin with this insight, which seems to be shared now by all of us. I agree that the typical English speaking Wikipedia reader ought to be able to gain this insight too. Now, please give me a reliable source where it is written out clearly that "the simple solution and the application of logical reasoning show together that always switching is optimal". Richard Gill (talk) 08:45, 19 November 2010 (UTC)" Glkanter (talk) 16:11, 21 November 2010 (UTC)[reply]

Gill (2011) [6] is such a reliable source. I sent you a copy some months ago. Richard Gill (talk) 20:50, 21 November 2010 (UTC)[reply]

Want To Talk About Morgan?

"In general, we cannot answer the question "What is the probability of winning if I switch, given that I have been shown a goat behind door 3?" unless we either know the host's strategy or are Bayesians with a specified prior. Nevertheless, in the vos Savant scenario we can state that it is always better to switch. The fact that Pr(W | D3) >= 1/2, regardless of the host's strategy, is the key to the solution."

Let's Make a Deal: The Player's Dilemma

Author(s): J. P. Morgan, N. R. Chaganty, R. C. Dahiya, M. J. Doviak

Source: The American Statistician, Vol. 45, No. 4 (Nov., 1991), pp. 284-287

Published by: American Statistical Association

"From this and her previous solutions, one is tempted to conclude that vos Savant does not understand that the conditional problem (of interest to the player) and the unconditional problem (of interest to the host) are not the same, and that 2/3 is the answer to the relevant conditional problem only if p = q = 1/2. Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one. But then, it may be that only an academician, and no one connected with a game show, would ever consider p # q."

Let's Make a Deal: The Player's Dilemma]: Rejoinder

Author(s): J. P. Morgan, N. R. Chaganty, R. C. Dahiya, M. J. Doviak

Source: The American Statistician, Vol. 45, No. 4 (Nov., 1991), p. 289

Published by: American Statistical Association

"To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period."

Morgan, J. P., Chaganty, N. R., Dahiya, R. C., and Doviak, M. J. (1991),

“Let’s Make a Deal: The Player’s Dilemma,” The American Statistician,

45 (4), 284–287: Comment by Hogbin and Nijdam and Response

The American Statistician, May 2010, Vol. 64, No. 2 193

Sounds to me like they've recognized the host chooses uniformly when faced with 2 goats. That would take the conditional decision tree and the Bayes' theorem solution in the article off their 'false' list. It also means, as per their rejoinder above, and as I just discussed with Richard, that when p = q = 1/2, "It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one." That would take the 4 simple solutions off their 'false' list, leaving only the erroneous 50/50 solution.

So, there's your 'reliable' source, Richard, that says when the host chooses uniformly between 2 goats the simple solutions solve the door 1 and door 3 conditional problem. Glkanter (talk) 13:36, 19 November 2010 (UTC)[reply]

I have to agree that Morgan now seem to give the impression that the conditional probability solution is a somewhat long winded way to get that which might have been got far more easily by simpler means. Martin Hogbin (talk) 20:09, 19 November 2010 (UTC)[reply]

I've always admired their work, as you know. Glkanter (talk) 20:34, 19 November 2010 (UTC)[reply]

Yes, Morgan et al. gave the symmetry argument to show that the conditional probability is the same as the unconditional probability in the "everything uniform" case. It does not take the simple solutions off their "false" list. It means that an extra sentence is needed to expand those simple solutions into solutions of the problem "what is the conditional probability?" The simple solutions need to be completed, according to Morgan et al., with a sentence which explains why the conditional probability is the same as the unconditional probability. Richard Gill (talk) 14:28, 21 November 2010 (UTC)[reply]

So the whole argument that the simple solutions are flawed boils down to the sources that use the simple solutions don't include a sentence about the obvious symmetry that makes the two solutions equivalent? That's an opinion about writing/teaching styles, not an indictment of the obviously/intuitively valid mathematical shortcut used by those sources. Glkanter (talk) 16:18, 21 November 2010 (UTC)[reply]

Yes that's right Glkanter, that is what I have been saying for months. I don't think however it's an opinion about writing/teaching styles. I don't think that all those sources which gave the simple solution realised that they could have made a remark about symmetry so that their simple solution also gave the answer to the question "what is the conditional probability"? You are a fan of Carlton, but he didn't notice that it could be done that way. And I don't think that many of their readers would have thought it up for themselves. But anyway, since it is such a small point anyway, we needn't make a big issue out of it on the wikipedia page. That's a good sign (concensus could be nigh). Richard Gill (talk) 20:46, 21 November 2010 (UTC)[reply]

No Richard, the starting point of the so called simple solutions is wrong. There is no such thing as a simple solution. People confuse - and I hope you don't - an easy way of calculating the conditional probability, with the simple solutions, that do not mention the conditional probability at all. Nijdam (talk) 21:14, 21 November 2010 (UTC)[reply]

Answers to K&W

In the light of the above comments it is interesting to see how various people might answer the K&W formulation. I will refer to then as 'ordinary person' and 'expert'

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.

At this stage we can ask the unconditional question. What is the probability that the player will win if she switches?

You may ask any question, but not all can be answered. Why ask for the unconditional probability? The question asked is: would you like to switch? And the player considers her situation, and uses all the info she has. She reasons: I have chosen door No x, the host has opened door No h, I calculate the conditional probability given x and h. Etc. Nijdam (talk) 23:10, 24 November 2010 (UTC)[reply]

Ordinary person: 1/2

Expert: 2/3

Ordinary person (after varying times of study) Oh yes, I see now it is 2/3

Now add in the condition:

Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

And ask for the probability that the player will win if they change their choice.

Ordinary person: 2/3 of course. Duh!

Expert: Now let me see, we need to consider the full sample space of........then we need to condition on the player's and host's door choice......so that is......[long calculation]... 2/3. That is quite remarkable, by some quirk I have got the same answer as you. How did you get your answer ordinary person?

Ordinary person: The door numbers obviously do not make any difference, we already have the answer.

Expert: Hmm...I will have to look into that...sample space...Bayes' rule........... No, I think you were just lucky. Martin Hogbin (talk) 19:57, 19 November 2010 (UTC)[reply]

Outstanding! Too bad nobody else will acknowledge this. Glkanter (talk) 20:36, 19 November 2010 (UTC)[reply]

Martin, I suspect what you wrote above isn't all quoted from sufficiently reliable sources. – Nevertheless what you wrote is exactly what anyone can see in reading the sources with open eyes, and in comparing what they say. Gerhardvalentin (talk) 22:26, 19 November 2010 (UTC)[reply]

Although this was put in a lighthearted way, the more serious point that I was making was that intuition can often lead us astray, but in some cases it can prove to be correct. In the first (unconditional) case, the natural and intuitive (to most people) assumption of a symmetry between the two doors that the player has to choose between proves to be wrong and leads to the wrong answer of 1/2. For this reason experts may train themselves to mistrust their intuition and always do a full calculation just to be sure.

In the case of the host door choice, however, the natural assumption that it makes no difference which door the host chooses proves to be justified and gives the right answer. Some experts may, through their training, not trust their intuition and insist on doing the full conditional probability calculation even though it is not necessary. Martin Hogbin (talk) 19:12, 20 November 2010 (UTC)[reply]

The true expert won't do the long calculation. The true expert will note that by the law of total probability the unconditional probability is equal to the average of the conditional probabilities. By symmetry the conditional probabilities are all the same, therefore equal to their average, 2/3. Richard Gill (talk) 14:32, 21 November 2010 (UTC)[reply]

By the way, there is at least one reliable source where this is all written down: the two recent papers by yours truly. Richard Gill (talk) 14:40, 21 November 2010 (UTC)[reply]

Let's Summarize Briefly

Richard (and anybody else who would like to respond), do you agree with the following statements, all as per the Selvin and K & W premises?:

1. The 'optimal strategy' of 'always switching' can be derived from the simple solutions.

2. The conditional probability of door 1 and door 3 can be derived from the simple solutions as doors 2 and 3 are equally likely to be opened by the host.

3. Morgan (as per the rejoinder in the Morgan section above) is a reliable source for #2.

4. There is nothing the contestant can do to alter his initial 2/3 likelihood of selecting a goat.

Thanks. Glkanter (talk) 08:34, 20 November 2010 (UTC)[reply]

Richard's Response

1. If the player's initial choice is correct with probability 1/3, then by always switching he gets the car with probability 2/3. Always switching beats always staying in terms of overall (unconditional) success chance. This does not yet prove that always switching is optimal. Could there be some mixed strategy of sometimes switching, sometimes staying, depending on chance and/or the specific doors chosen and opened, with a better overall success rate than 2/3?

This isn't obvious for the player in a given situation, but need to be argued. Nijdam (talk) 23:21, 24 November 2010 (UTC)[reply]

2. The symmetry of the K&W assumptions shows that the conditional probability of winning given initial door chosen and door opened does not depend on these two specific numbers and hence must equal the unconditional probability which by 1. is 2/3. It follows that the stategy of always switching is optimal (it cannot be improved by sometimes staying). This follows by consideration of the law of total probability: the overall success probability equals the sum over the six possibilities of the door initially chosen and the door opened of the conditional probability of winning in each of these six situations, times the probability that each of these situations will occur. Since in each situation we can't do better than 2/3 we can't do better on average than 2/3, either.

3. Morgan et al's response to Seymann already mentions that in the fully symmetric case one can get the conditional probability from the unconditional by symmetry.

4. The point is that under symmetry, the contestant's initial 2/3 likelihood of selecting a goat is not changed by seeing which door the host opens. Since the two conditional likelihoods must be the same they must both equal their average, 2/3.

Richard Gill (talk) 10:05, 21 November 2010 (UTC)[reply]

What If?

When the contestant is initially faced with three closed doors, each door has a 1/3 likelihood of hiding the car and a 2/3 likelihood of hiding 1 of the 2 goats.

After the contestant has selected a door, these values are unchanged, all 3 doors each still has a 1/3 likelihood of a car and a 2/3 likelihood of a goat.

What if the premise about the host choosing equally when faced with 2 goats was removed?

In the simple solutions, there is no affect. Since each of the unchosen doors has a 2/3 likelihood of being a goat, the host will open them an equal number of times, or 50/50. This is the same outcome as what the uniform host premise results in.

However, the conditional decision tree solution and the Bayes' theorem solution in the Wikipedia article can no longer be used, as they require the uniform (50/50) host premise to return the 2/3 & 1/3 result.

Its been reliable sourced by Morgan (see their rejoinder in the Morgan paragraph above) that the simple solutions, when the unchosen doors are equally likely to be opened by the host, answer the conditional door 1 and door 3 question. It turns out the simple solutions provide exactly as much precision to the door 1 and door 3 problem as the conditional solutions, even without the 50/50 host premise, while the conditional solutions are of no value without that superfluous premise. Glkanter (talk) 12:09, 20 November 2010 (UTC)[reply]

Not true. If the host will open door 3 with probability q when he has the choice between doors 2 and 3 because the car is behind the door, 1, chosen by the player, and if the car is initially equally likely behind every door, then the a posteriori odds on Door 1 : Door 2 are q : 1. Thus the conditional probability also favours switching, whatever the value of q. In other words, the strategy of aways switching is optimal (beats any other strategy, simple or mixed, in terms of overall success chance) if only we know that the three doors are initially equally likely. The simple solution only tells us "always switching" beats "always staying" in terms of overall success chance. It doesn't tell us "always switching" beats anything. Richard Gill (talk) 14:08, 21 November 2010 (UTC)[reply]

Discussion with Martin

If the question asks for a conditional probability given specified door numbers then the simple solutions are plain wrong if there is a known asymmetry in the host's door choice, or the producer's initial car placement choice. Martin Hogbin (talk) 13:02, 20 November 2010 (UTC)[reply]

As an example, to make things clear, the simple solution fails for this question:

There is a game show in which there are three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed uniformly at random behind the doors before the show. The player chooses a door uniformly at random and, after the player has chosen a door, the door remains closed for the time being. The game show host, Monty Hall, then has to open any one of the two remaining doors and ask the player to decide whether they want to stay with their original choice or to switch to the remaining door. Given only the above information and that the player has chosen door 1 and that the host has opened door 3, and that the host's probability of choosing door 3 when the car is behind door 1 is q. What is the probability of the player winning the car if they switch to door 2?

Is that the MHP though? Not in my opinion. Martin Hogbin (talk) 13:11, 20 November 2010 (UTC)[reply]

I have no idea what this is about. Glkanter (talk) 13:27, 20 November 2010 (UTC)[reply]

You say above, 'It turns out the simple solutions provide exactly as much precision to the door 1 and door 3 problem as the conditional solutions, even without the 50/50 host premise'. I am saying that that is not true in some cases as is demonstrated above. The simple solutions can be plain wrong without the 50/50 host choice. Martin Hogbin (talk) 14:01, 20 November 2010 (UTC)[reply]

Not from the contestant's POV. *Maybe* its 0 & 1 the other way, as its not stated as a premise to be of *any* values. Glkanter (talk) 15:37, 20 November 2010 (UTC)[reply]

I agree not from the contestant's POV, but my question specifically states, Given only the above information and... thus the basis on which the question is to be answered is made clear. The question above is the one to which Morgan's solution is the only correct answer, unfortunately it is not the MHP in my opinion. Martin Hogbin (talk) 15:43, 20 November 2010 (UTC)[reply]

Martin, I feel as if you've hijacked this section. Because I didn't state 'as per K & W rules'? What's up with that? I never discuss anything but the MHP as per Selvin's and K & W's premises as the starting point. To be honest, I wish you'd delete that stuff that I still don't understand the point of. Glkanter (talk) 16:12, 20 November 2010 (UTC)[reply]

I will shut up if you like but if you overstate your case you will easily be proven wrong. You have claimed that, 'It turns out the simple solutions provide exactly as much precision to the door 1 and door 3 problem as the conditional solutions, even without the 50/50 host premise'. I do not think you will get much agreement from others with that statement. Other people will make different assumptions from you and therefore get different answers. Martin Hogbin (talk) 16:39, 20 November 2010 (UTC)[reply]

Very nice solution! And please, I did not use the words 'shut up'. I realize this will be controversial, but I think its accurate and defendable within the very narrow constraints of the Selvin/K & W MHP. Because of the likely resistance to this argument, and since it has no real meaning as far as the article goes, I put it here on Richard's page, not on the talk page or mediation page. I never fear saying something that I believe is true (and perhaps novel), provided it will not hurt someone for no reason. Glkanter (talk) 17:07, 20 November 2010 (UTC)[reply]

The point is that you make certain assumptions, such as the the question must be answered from the Bayesian perspective from players SoK and that the principle of indifference should be applied. I personally agree that these are the most natural assumptions but others may take a different view, possibly without realising it, thus you will be doomed to continue arguing for ever. I was trying to show that you could be wrong, but only within a very narrow band of assumptions that were stated in my question. Martin Hogbin (talk) 18:18, 20 November 2010 (UTC)[reply]

I make no assumptions beyond what Selvin and K & W provide. I DO NOT invoke the principal of indifference anywhere it does not already exist per those sources. In fact, my argument removes the 50/50 host premise, showing that the simple solutions do not rely on it (another HUGE F-ING Morgan error), while the conditional solutions do. Glkanter (talk) 19:23, 20 November 2010 (UTC)[reply]

Selvin and K&W both include the 50/50 assumption, in which case the simple solutions are but if you abandon the 50/50 assumtion the simple solutions can be wrong. Say the host chooses 20/80 in favour of door 3? Martin Hogbin (talk) 20:49, 20 November 2010 (UTC)[reply]

That premise doesn't exist in my 'start with Selvin and K & W and remove the uniform host premise' problem. Nor is it in any reliable source for the MHP. I don't have to address it. Also, unless the contestant has that knowledge, it's not part of the model. Glkanter (talk) 20:54, 20 November 2010 (UTC)[reply]

You know, Martin, its kind of funny, or ironic. You are using the same technique of creating a 'variant' to disprove my argument which applies exclusively to the Selvin and K & W MHP problems. Which I've always vehemently rejected from the other editors, and now you, as well. Glkanter (talk) 21:00, 20 November 2010 (UTC)[reply]

But you yourself said, 'my argument removes the 50/50 host premise'. What do you mean by that if not that the host does not choose equally between the two doors? Martin Hogbin (talk) 00:10, 21 November 2010 (UTC)[reply]

I mean nothing more than there is no knowledge that he has any bias. I'll discuss host bias of 50/50 vs no known host bias, but I will not discuss variants such as 0 & 1 or 80/20 which introduce new premises and are not the Selvin/ K & W MHP. Glkanter (talk) 00:22, 21 November 2010 (UTC)[reply]

Then I will say, all we can do is build a (mathematical or logical) model of the puzzle. With the simple solutions, I know that each unchosen door has a 2/3 likelhood of being a goat. Which is what Morgan says I need for the simple solution to solve the conditional solution. I don't need to concern myself with anything beyond that. Why go looking for trouble? Hey, I don't know know how the contestant will do in real life, but I know that I've built a proof (solution) that is logical, supported by the facts, and returns the same results as the other MHP solutions. In my mind, until someone shows an error or incongruity of some sort, I'm done. But a made up premise doesn't accomplish that. Hey, Morgan wrote a whole paper about how an unstated premise made a bunch of simple and conditional solutions false. Why should you, or anybody, be allowed to introduce an unstated (because its non-existent) premise to prove me wrong? Glkanter (talk) 00:32, 21 November 2010 (UTC)[reply]

I will also say that without any host bias premise, there can be no conditional decision tree or Bayes' theorem solutions like the ones in the article. Unless one devises an 'unstated one' with some values, say 50/50 (or 0 & 1, or 80/20 I suppose). But when vos Savant (and others) tried that, she was taken to task pretty bad. Glkanter (talk) 01:07, 21 November 2010 (UTC)[reply]

Glkanter, you are right again. You not only "need not" struggle with any such unstated values, say 50/50 or 0&1 or ..., but moreover: You never may struggle with those assumed unstated values", as per Ruma Falk: Only if any bias "is given (!) and IS KNOWN TO BE GIVEN (!)", you may struggle with such "known-to-be-given-value". But within the MHP it never can be known to be given. Period. So, if you – outside the MHP(!) – should decide for "0&1 or 1&0 or any other", then - by a closer look – you can plainly see that Pws in two out of three can be 1/2 at least, but never less!, and in one of three can be at most 1/1, but never more! :)
Even in such assumed case, the "simple solution" tells you that - for any given game - staying never can be better than to switch, and so you should switch anyway, in any given situation, and in any given game. With assumed extreme bias or "without" any bias. So you are right, that famous "core of the MHP" is totally irrelevant for the MHP and for the decision asked for. And no "conditional solution" can and will ever be able to tell you any better result. And so your objection is valid. Gerhardvalentin (talk) 16:28, 21 November 2010 (UTC)[reply]

Glkanter, I understand what you are saying but to give a definitive single value answer the conditional problem you have to propose a value for the probability that the host will choose door 3 when the car is behind door 1. Without doing this you cannot get a single answer to the conditional problem (as you well know the best you can do is say the answer is >= 1/2). What you have done is to say that because you (as the player) do not know this probability you must take it to be 1/2 or 50/50 as you put it. I agree with you that this is the best thing to do but you should know that by doing this you are applying the principle of indifference. Which is what I said in the first place. Martin Hogbin (talk) 16:38, 21 November 2010 (UTC)[reply]

No, Martin. The 1/3 & 2/3 symmetry already exists when the contestant first sees the 3 doors. I am saying there are no premises of any sort that cause that to change. And that making them up is bullshit, just like Morgan says. This solution avoids any requirement for a host bias. Like I said, why do I (this solution) have to go looking for trouble? Honestly, I think you've gotten a little too close to the Kool Aid. That concerns me. Glkanter (talk) 16:49, 21 November 2010 (UTC)[reply]

How do you know that the host does not choose door 3 80% of the time the car is behind door 1? This would break the symmetry. Martin Hogbin (talk) 17:12, 21 November 2010 (UTC)[reply]

The same way Morgan 'knows' its not 50/50. And I don't have to deny the existence of indifference, logic, or the definition of a game show to do so, as Morgan needed to do. I just refuse to acknowledge that which does not exist. Glkanter (talk) 17:17, 21 November 2010 (UTC)[reply]

Martin, let's keep in mind that all we hope to accomplish is the creation of a model that assists us in making logical/benficial decisions about hypothetical events. We are not creating a photo-realistic rendering of one portion of human existence. My solution fully meets these expectations, and withstands any of the criticisms advanced so far. Glkanter (talk) 17:38, 21 November 2010 (UTC)[reply]

Glkanter, the argument of Morgan et al. to get from the simple solution to the conditional solution uses symmetry, ie it assumes 50/50 for the host choice, and 1/3:1/3,1/3 for the location of the car. Conditional decision trees and Bayes' theorem give the same answer in this situation, as they must. Richard Gill (talk) 14:02, 21 November 2010 (UTC)[reply]

No Richard, you are wrong. They say it is because p = q = 1/2. That *can* come from the 50/50 host (and also requires the initial 1/3 & 2/3, doesn't it?), but it *does* come from the initial 1/3 & 2/3 for all 3 doors. There are no premises, implied, stated, or unstated, that cause the 1/3 & 2/3 of the unchosen doors to change other than the host driving one of those values to 0. Morgan rejects the unstated 50/50 premise being used to prove the simple solutions, and based their entire paper and conclusions on this. As it is unknown, I reject ANY unstated/contrived host bias premises (host bias of 0 & 1 or 80/20 or whatever) being used to disprove this solution for the same reasons. With NO HOST BIAS premise present, those values can't change when the host opens a door. I stand by my analysis. Glkanter (talk) 16:32, 21 November 2010 (UTC)[reply]

I think I am right, Glkanter. Morgan et al's p and q are the probabilities that the host will open door 2 or door 3 when he has the choice. So p=q=1/2 is the same as the 50/50 assumption. Morgan et al everywhere also assume all doors initially having probability 1/3. So p=q=1/2 gives full symmetry and that makes the conditional probabilities all equal and hence equal to the unconditional 2/3. When there is no host bias, as you say, the fact the host opens door 3 doesn't change the 1/3 - 2/3 of "door 1" versus "doors 2 and 3", hence we end up with 1/3, 2/3, 0. If there is host bias then we will end up with something else - it can be anything between 0,1,0 and 1/2,1/2,0 (ie odds q:1 where q is anything between 0 and 1). Your intuition is splendid but it is so good (it seems to me) that you sometimes don't notice when you skim over logical intermediate steps which people with less intuition might appreciate being filled out. Richard Gill (talk) 20:36, 21 November 2010 (UTC)[reply]

Yes, you are right about the p = q = 1/2 as Morgan uses it. Of course the math rule itself relates to the probability that each door will be opened with equal likelihood. Which can't be proven by the conditional methods without the 50/50 host bias, whether stated outright or 'assumed' due to indifference or however. But the simple solutions don't need that 50/50 host bias, as the doors start out equally likely, 1/3 & 2/3 for each, and there are no premises, stated or assumed, that change that. Right, Morgan, et al?

It's not my problem if those conditional decision tree-favoring guys picked the wrong tool to solve the (conditional) MHP. The simple solution guys didn't make that mistake, as Morgan points out. Over time, I've learned that identifying, and then properly applying, the best tool for the job is perhaps the most important attribute of a skilled craftsman. Don't you agree, Richard? Glkanter (talk) 11:48, 22 November 2010 (UTC)[reply]

Please note, I made changes to the above paragraph that one could easily overlook if just going through the diffs looking for red text. Glkanter (talk) 14:51, 22 November 2010 (UTC)[reply]

Yes I agree that identifying, and then properly applying, the best tool for the job is perhaps the most important attribute of a skilled craftsman. In this case I would say within the subjectivist probability solution to MHP, the best tool is the simple solution to get the unconditional probability plus symmetry to argue that you needn't condition on anything more. Within a frequentist probability solution I would say that the best tool is randomization plus switching (ignoring the door numbers deliberately) and getting the game-theoretic and frequentist optimal solution with rock-hard guaranteed properties. I would say that the super-craftsman is able to craft different elegant solutions for different consumers. Sometimes he builds a farm-house kitchen table. Sometimes he builds a posh town-house dining table. He'll use quite different tools for the two jobs and both will prove his sublime craftsmanship. Twentieth century applied mathematics shows that we shouldn't be afraid of randomness, we should harness and exploit it. Don't be passive. The subjectivist solution says "since I don't know anything I should switch". The game-theoretic solution says "since I know nothing I'll randomize and switch". See my recent paper Gill (2011). Richard Gill (talk) 11:28, 23 November 2010 (UTC)[reply]

Extreme MHP

Let's see if this works... I claim with the puzzle beginning 'Suppose you're on a game show...', the only premise that must be stated is 'The host will reveal a goat behind one of the unchosen doors'. With this one stated premise, I can solve the door 1 and door 3 conditional problem.

The contestant can choose from 3 doors. Has no idea which one hides the car. Our simple model can only use uniformity across the 3 doors, giving each door a 1/3 likelihood of having the car and a 2/3 likelihood of having a goat. There is nothing to contradict this assumption.

The contestant chooses a door. Each of the 3 doors still has a 1/3 & 2/3 probability as described above. In Whitaker/vos Savant, the contestant has chosen door 1.

The host is equally likely to reveal either of the unchosen doors, as they each still have a 2/3 probability of having a goat. The host in Whitaker/vos Savant opens door 3 to reveal a goat.

As per Morgan's rejoinder, when the 2 unchosen doors are equally likely to be opened to reveal a goat, the simple solution solves the conditional door 1 and door 3 problem.

The revealed door's probability goes to 0, and the remaining unchosen door's probability goes to 2/3. The chosen door is unchanged at 1/3.

Switching increases the contestant's chances of winning from 1/3 to 2/3.

No OR. One premise. Glkanter (talk) 22:15, 20 November 2010 (UTC)[reply]

You're saying (I think): we use a subjectivist notion of probability. Because our prior information is complete neutral regarding the door numbering, we are initially competely indifferent as to which door hides the car and as to which other door the host will open if we happen to pick the car-door initially: uniform probability distributions. Since our initial choice is correct with probability 1/3, the other closed door hides the car with probability 2/3. Symmetry shows that the conditional probability that the other door hides the car, given player chose, say, 1 and host opened, say, 3, must equal 2/3 too, since the unconditional probability is the average of the conditional probabilities (law of total probability) and all the conditional probabilities are the same (by symmetry, they don't depend on the specific door numbers).— Preceding unsigned comment added by Gill110951 (talk • contribs) 13:57, 21 November 2010 (CEST)

Response requested

Hi - Can you please respond to this? Thanks. -- Rick Block (talk) 21:08, 21 November 2010 (UTC)[reply]

Mathematics vs Reliable Sources

I don't get why we needed Morgan, or some other reliable MHP source to link the simple solutions to the door 1 door 3 conditional problem, and couldn't just refer to whatever math rule that Morgan applied? Its no more than what all we simple proponents have been saying endlessly, and that I think Boris proved in some fashion, maybe a year ago.

I mean, there has to be a (million) text book(s) somewhere that say exactly the same thing, right? Glkanter (talk) 23:53, 22 November 2010 (UTC)[reply]

Boris mentioned it too a year of so ago, I was delighted when I saw that, since it had never occurred to me before that you can do it that way. The "simple proponents" have not been saying it endlessly, neither before nor after Boris' contribution.

If you want to know the mathematics rule which does this, it is called the law of total probability.

Let me do it in the case the player chooses Door 1. All probabilities below, till further notice, are conditional on that first choice of the Player.

By the law of total probability, since the host always either opens Door 2 or opens Door 3,

Prob(Switching gives car) = Prob(Switching gives car | Host opens Door 2) x Prob(Host opens Door 2) + Prob(Switching gives car | Host opens Door 3) x Prob(Host opens Door 3)

By symmetry

Prob(Switching gives car | Host opens Door 2) = Prob(Switching gives car | Host opens Door 3),

Prob(Host opens Door 2) = Prob(Host opens Door 3) = 1/2

By the simple solution,

Prob(Switching gives car) = 2/3

Therefore

Prob(Switching gives car | Host opens Door 2) = Prob(Switching gives car | Host opens Door 3) = 2/3

One can give the same line of argument without conditioning on the player's initial choice, splitting up over the six configurations of initial door and door opened.

The law of total probability can be used in exactly the same way to show that the overall success probability of any mixed strategy must be less than 2/3. Simply write the probability of success as the sum over configurations of the probability of success given each configuration, times the probability of that configuration. In order to maximize the overall probability of success you must maximize the conditional probability of success in each configuration. Richard Gill (talk) 07:26, 23 November 2010 (UTC)[reply]

Modeling With Unstated/Unknown Values

In the Whitaker/vos Savant version of the MHP, no host bias when faced with 2 goats was mentioned. Selvin only mentions that it is 50/50 in his 2nd letter, saying his solution(s) is based on this. Then he gives a conditional solution which relies on this info, then he applauds Monty Hall's simple solution.

Morgan only knowing of Whitaker/vos Savant calls 4 different simple solutions false and a verbal presentation of the conditional decision tree false, because they rely on the unstated/non-existant host bias of 50/50. In their paper, they also say the simple solutions are false, because they are answering the unconditional MHP. Later, in the same issue, in their rejoinder, they say that with the 50/50 premise, doors 2 and 3 are equally likely to reveal a goat, so the simple solutions solve the conditional door 1 and door 3 problem, as does the conditional decision tree.

So, because it wasn't stated outright as a premise, Morgan rejected the use if the 50/50 host bias, and concluded that >= 1/2 was the only value that could be calculated for switching.

My question then is, when creating a model for decision making purpose, how does the 'person of interest's SoK' handle unstated, and therefore unknown probabilities? In Morgan's paper, the unstated/unknown original distribution of the car, from the contestant's point of view is uniform across all three doors, as 1/3 for each. For the host's bias, equally unstated/unknown the 50/50 is prohibited, but conjecture of other unstated values is permitted as a means of demonstrating why the conditional solutions are better in certain situations, but not the MHP with the 50/50 host. Many text books and Rosenhouse (?) do this, as well. In both the simple solutions and the conditional solutions in the Wikipedia MHP, each door is originally equally likely to have the car with a likelihood of 1/3. This is a stated premise that K & W version includes.

1. Since the 50/50 host is a K & W premise, the simple solutions are capable of solving the door 1 and door 3 solution. So the simple solutions themselves are not 'flawed' or 'false'. They are equivalent in this problem to the conditional solution in answering the question, 'should the contestant switch?'

Why then, do editors argue that the simple solution solves the 'wrong problem', etc? Why are the (limited) sources that say this given such prominence?

2. The actual mechanics of using the simple solutions do not rely on the 50/50 premise. Selvin's original simple solution, and Whitaker/vos Savant used simple solutions without a stated 50/50 host bias, or any host bias. Critics claim this is unacceptable, because the host *might* have a bias other than 50/50, and the contestant would be aware of this. Thus, without a stated host bias, the 2/3 equal likelihoods of being a goat = 2/3 cannot be assured in some cases after the contestant has chosen a door. This causes, they say, the simple solutions, without a stated host bias of 50/50 to be false.

My question is, since the host bias is unstated and therefore unknown, how is the contestant to 'fill in' this missing info? Or, since its unstated and unknown, and not needed for the actual math of the simple solutions, why is the contestant required to know of a host bias at all? Doors 2 & 3 are 2/3 goats when the contestant selects a door, and there is no information whatsoever provided to cause, or enable, the contestant to recalculate those values when the host opens door 3. So, why don't those values stay as 2/3, allowing the contestant to make a calculated decision of 2/3 & 1/3, and since the doors are equally likely, solve the door 1 and door conditional problem, rather than declaring the simple solution false?

3. Surely there's a rule when making probability models on how to handle unknown distributions like these? How about in this case, where the values aren't needed for the calculation? Is there really a requirement for a 'premise that precludes any unstated/unknown premises from potentially disqualifying the model', which is all the 50/50 host bias does for the simple solutions? Glkanter (talk) 05:07, 23 November 2010 (UTC)[reply]

Wouldn't the contestant reason that the unstated and unknown host bias *might* have any values from 0 & 1 through 1 & 0, and that they are all equally likely? Then the average host bias would be determined to be uniform at 50/50, remedying any criticisms that the 50/50 host bias is missing? Doesn't the above serve as justification for modeling using uniformity for an unstated/unknown value? Isn't this exactly what vos Savant does for the producer's 1/3,1/3,1/3 unstated premise for the original car placement? Morgan didn't have a problem with *this* identical 'potential, but unknown/unstated human bias' premise, using it themselves in their solution. Isn't there a math book out there that supports this approach? Glkanter (talk) 05:36, 23 November 2010 (UTC)[reply]

Yes. Host bias can be anything, but the player's opinions about it are symmetric. It is just as likely for him to be below 1/4 as above 3/4, just as likely to be below 3/8 as above 5/8, etc etc etc. Hence the subjectivist expectation about a single play is 50/50.

So, my 'What if' section above is correct? The 50/50 host bias does not need to be explicitly provided in the premises in order to use the simple solutions to model the conditional door 1 and door 3 problems solution? Glkanter (talk) 10:02, 23 November 2010 (UTC)[reply]

Please distinguish "solution" and "argument" and "answer". If you want to model MHP using subjective probability (not everyone agrees with this, but it is a common approach and I won't say it is unreasonable), then it would be conventional to state that because your prior knowledge is indifferent to the numbering of the doors, the following two deductions can be drawn: (1) your prior beliefs about the location of the car are 1/3, 1/3, 1/3 and (2) your prior beliefs about which door the host would open, and in each of the three possible cases that he would have a choice, are 50/50. From (1), logic tells us that the unconditional chance of winning by switching is 2/3, since the switcher wins if and only if the stayer loses. Next, the symmetry of our prior beliefs (which is of course also encapsulated in statements (1) and (2)) tell us that the conditional chance of winning given door chosen and door opened is independent of the specific pair of door numbers. Next, the law of total probability tells us that the conditional chance of winning by switching must therefore be equal to the unconditional chance, 2/3. This is of course confirmed by an explicit computation using probability trees, Bayes theorem in the odds form, the definition of conditional probability, or whatever you like. In my opinion, a simple and rigorous logical argument is as good as, if not better than, a direct computation.

Moreover, if you want to model MHP using subjective probability, as you and many others do, but by no means everyone, then it seems to me that you need to find the probability that switching gives the car given all the *relevant* information available to the player, or equivalently, given *all* the information available to the player (in this context, the equivalence is a tautology: the definition of "relevant" is exactly, everything that influences the probability). You can do this via computing the probability given a part of the player's information as long as you explicitly add a rigorous argument why including the remainder of his knowledge won't change the probability. You have to say why what remains is not relevant.

Professionals distinguish between the number 2/3, the conditional probability switching gives the car given door numbers chosen and opened, and the unconditional probability switching gives the car. They also distinguish between the probability and the argument used to derive it. The solution (in the sense of the answer: 2/3, switch) could be correct even though the solution (in the sense of the logical/mathematical steps used to get from vos Savant's words to the answer) could be incorrect. In particular if you want a conditional probability but only compute a marginal probability and don't indicate how to get from the one to the other, or don't even indicate that you are aware of the conceptual difference, then your argument is wrong, even if your answer is right. Richard Gill (talk) 10:52, 23 November 2010 (UTC)[reply]

Please read Gill (2011). See [7] and especially [8]. I sent you a pdf several months ago, hoping for comments. In the meantime it has been enthusiastically accepted for publication in the next issue of Statistica Neerlandica so it is almost too late to take account of any comments you might have. It is written for a professional audience but I wanted to have the main messages accessible for lay persons. But for that I need feedback from people like you.Richard Gill (talk) 08:18, 23 November 2010 (UTC)[reply]

Richard, the last time I checked, you were telling me that if the car wasn't placed uniformly at random behind the 3 doors, unless the contestant selected his door randomly, you couldn't be certain that he had a 1/3 chance of selecting the car. So, if the contestant chose his door because 1 is his lucky number, you had no opinion on what the likelihood he selected the car is. Given that, why would I want to read your article, which I presume gets into more complex stuff, when I don't even agree with your conclusion on this elementary item? We're talking about models of hypothetical situations that aid in decision making, Richard, not photographs of reality. Glkanter (talk)

Which, by the way, as I read it, contradicts this comment from above that you left this morning:

"Yes. Host bias can be anything, but the player's opinions about it are symmetric. It is just as likely for him to be below 1/4 as above 3/4, just as likely to be below 3/8 as above 5/8, etc etc etc. Hence the subjectivist expectation about a single play is 50/50."

Which is not consistent with the lecture you delivered to me this morning on the mediation talk page:

"...They write "The standard version provides no information about Monty Hall’s strategy. Is the problem therefore mathematically under- specified and insoluble? The answer is no, because the standard version does not ask for a probability, but for a decision. The general Bayes’s rule for the standard version of the Monty Hall problem in the absence of information about Monty Hall’s strategy is ... Because the strategy-dependent probability p(M3 | C1) varies between 0 and 1, the conditional probability p(C2 | M3) can vary only between 0.5 and 1. Therefore, whatever strategy one assumes Monty Hall to use, the conclusion is that the contestant should switch." Kraus and Wang (psychologists, by the way) use Bayesian formalism throughout their paper. Richard Gill (talk) 08:43, 23 November 2010 (UTC)"

That's how I see it, anyways. Glkanter (talk) 10:32, 23 November 2010 (UTC)[reply]

Well, I don't see any contradiction, but then I distinguish between finding the right number and knowing what it stands for, and I distinguish between the answer and the reasoning used to get it. After all, that's my job as a mathematician and as a teacher. You're a practical guy and a fast thinker and you are not accustomed to splitting hairs in logical deductions. Sometimes, splitting hairs in logical deductions led to major advances in science, sometimes it held things back from years.

In the present case I think that being aware of the subtleties helps to understand everyone's point of view, and hence it could lead to concensus and to a constructive collaboration between editors of different backgrounds and with different priorities. I think that the diversity between the present editors' points of view and backgrounds reflect the diversity of our future readers' points of view and backgrounds. You won't be able to change the mindset of the present editors and more importantly, you can't prevent future readers with the same mindsets from editing the article again. So it would be wise to accomodate different mindsets from the start.

So: why don't you spend some time trying to understand my point of view by carefully reading my paper? I put a lot of thought and energy and time into it. Of course it will not be perfect and not everyone's bedtime reading. But I'm not asking you to agree with my points of view. There are certainly passages in there which are a bit technical, too, but don't worry about them, just read the plain text, bearing in mind my professional obligation to distinguish between the right number and what it stands for, and between the answer to a puzzle and the reasoning used to get it. Richard Gill (talk) 11:05, 23 November 2010 (UTC)[reply]

By the way, the contradiction which you see between various arguments I have given before is an apparent contradiction only: you left out the context. Within a subjectivist picture of probability, one can say ... , within a frequentist picture of probability, one would say something else .... Finally, you keep enlisting Kraus and Wang to your side, but the quotation I gave you and their extensive use of Bayesian formalism shows them endorsing the Morgan et al. unknown host bias - frequentist point of view which you find so stupid! It's not enough just to go cherry-picking for sentences which, out of context, appear to support your point of view. You have to try to understand what the writer is trying to say, like it or not. Richard Gill (talk) 11:14, 23 November 2010 (UTC)[reply]

Pseudo Science

Compare and contrast the following host bias premises:

1. Given as values = 50/50

2. Not stated, unknown: value = 50/50 (invoking the same reasoning as the producer's original placement of the car)

3. Not stated, unknown: value = cannot be assigned (invoking Morgan's reasoning)

4. Given as values = 1 & 0, or 80/20, or anything other than 50/50

So you guys are telling me I can only use the simple solutions to solve the MHP in cases #1 & #2, except #2 dWUsn't apply for some reason known only to the highest of the high priests?

I'm just trying to make the best decision available, given the information available. I think #1, #2, and #3 all allow me to make an equally valid decision using the simple solutions to solve the MHP with the Selvin/K & W premises. There's no difference in my confidence level, either, because I'm not using the value in my calculations.

Somehow this bs supports the 'simple flawed, only conditional is valid' POV that currently exists throughout the article and is being held sacred in the mediation? And you defend this? And tell me at length how I don't get it? And you ridicule me on the mediation page? Another reason why "You...people make me..." Glkanter (talk) 23:49, 23 November 2010 (UTC)[reply]

No, I am not telling you any of this, at all! Did you read my paper yet? Carefully, with an open mind? I did tell you indeed that #1, #2, #3 all allow you to argue with the simple solution plus the remark that the specific door numbers are irrelevant by symmetry, that the probability is 2/3 (conditional or unconditional makes no difference). Richard Gill (talk) 10:03, 24 November 2010 (UTC)[reply]

Richard, you wrote above "Host bias can be anything, but player's opinions about it are symmetric" - but I do miss this "expressive" argument in your paper. Of course, even an extreme host's bias is without effect for the asked "decision", for
in 1/3 he has two goats and can open his preferred door, switching will hurt: Pws=0
in 1/3 he has the car and a goat, the goat behind his preferred door, he also can open his preferred door, switching will win: Pws=1
So, in 2/3 he can open his preferred door, and in those 2/3 you will know that Pws=1/2
And in the last 1/3, having one goat and the car, but the car being behind his preferred door, he has to open his avoided door, showing that Pws=1

And so you know that Pws (prob. to win by switching) will always be at least 1/2, and never less (in 2/3), and can be max. 1 (in 1/3), and on average be 2/3. So: host's bias is of no relevance for your decision to switch anyway. You did say that at the end of page 7 of your paper? (unconditional prob. and conditional prob. are the same?) Regards, Gerhardvalentin (talk) 10:59, 24 November 2010 (UTC)[reply]

That's right Gerhard: host bias is irrelevant anyway, since it doesn't change the fact that according to your conditional probability of winning by switching, switching is never unwise. That was the "novel" contribution of the Morgan et al. paper. It corresponds to Glkanter's #3 and #4. (Whether we pretend we know a value of q, or whether we use algebra and leave it fixed but free, the answer is the same: switch).

But these discussions are connected to your notion of probability. That is one of the reasons why I think Glkanter doesn't follow my arguments - he does not admit there to be any other way to define probability except for his way. He does not like it to be labelled, but it is a fact that this notion of probability is called subjectivist. There is another notion called frequentist. According to the frequentist view, probability is in the physical world. Then it could be reasonable to suppose that the host has a probability q of opening door 3 when the car is behind door 1 and the player chose door 1, and you do not know q. This just means that in many imaginary repetitions he would turn out to have some bias, just because that's the way his brain works. You have no way of knowing what his bias would be. Fortunatley, that does not stop you from knowing that you should switch. For all values of q except one, switching is better than not-switching; and in that special extreme situation, it is still true that switching is just as good as non-switching. So whatever q, switching achieves the best you can hope for, so you should switch.

If you see probability in your information about the world but admit that an individual host might have an unknown bias q, then you can say that because you don't know q, it could be anything. If your information about quiz-shows and quiz-masters is neutral as to the labelling of the doors, then your information about q is symmetric in exchanging door numbers 2 and 3. So the strength of your belief that q is between 1/4 and 1/2 is equal to the strength of your belief that q is between 1/2 and 3/4. And so on. Consequently in one game the host's choice is completely unpredictable to you. In many many games you expect that the host's choice for door 3 whenever he has the choice between 2 and 3 will average out at some number, q. You have no idea what q is. But for any bet you might make on the value of q, you'ld be equally happy to make the same bet on the value of 1-q. Your opinions about host bias are symmetric in the door numbers. Glkanter said that he'd take q to be uniform. He'd bet 3:1 against q being between 1/4 and 1/2/ He'd be prepared to bet at even odds that it is between 1/8 and 5/8.

This is, amusingly, exactly the point where Morgan et al. made their silly mistake, the one which Hogbin and Nijdam pointed out. They computed the probability of winning by switching in one single game, when q is uniformly distributed between 0 and 1. It must average out at 2/3 but they got something else by a silly calculus error. They should have realised by symmetry that the answer should have been 2/3. They should have realised that Glkanter's models #1, #2 and #3 are identical when we take a subjectivist view to the unknown q of viewpoint #3.

This also tells us something about the reliablity of peer review in well known scientific journals: it took 20 years before anyone noticed that mistake. Richard Gill (talk) 11:28, 24 November 2010 (UTC)[reply]

You are not consistent in your responses between this page and the mediation page.

I DO NOT ACKNOWLEDGE A HOST BIAS CAN AFFECT THE RESULTS IN THIS PUZZLE ABOUT A GAME SHOW. "It's 2/3, period." - Morgan.

You don't know that no one noticed an error for 20 years. You *do* know it took Morgan 20 years to acknowledge one of their errors.

What you wrote in your paper will become of interest to me only after your responses on the mediation page bear a resemblance to the notions of probability of the Wikipedia reader. Glkanter (talk) 12:26, 24 November 2010 (UTC)[reply]

How many reliable sources say that the K & W formulation of the MHP, which states the host bias = 50/50, allows for this nonsense?:

"This just means that in many imaginary repetitions he would turn out to have some bias, just because that's the way his brain works. You have no way of knowing what his bias would be."

...and that affects which modeling tools we can validly use for the MHP? That's all we, and that includes you, Richard, should be discussing on the mediation page.

Besides, (if we eliminate the stated 50/50 host bias) since he has no way of knowing what this non-existent hosts bias is, it can only be estimated by the contestant as 50/50, the average of all hypothetically equally probable hypothetical possible values. Which, for the simple solutions, ONLY MEANS some contrived reason for denying its use as a proper modeling tool for the K & W formulation (read: Wikipedia mediation page and article) has been removed. It does not affect the actual results derived from the tool.

In addition (but not relevant to the Wikipedia article), this means that the host bias DOES NOT NEED TO BE STATED AS 50/50, as I have been putting forth separately on this talk page in 2 recent sections.Glkanter (talk) 12:44, 24 November 2010 (UTC)[reply]

I believe it took 20 years before some-one pointed out Morgan et al's error in print.

Yes, you use this approach often, Richard. Your statement was:

"...it took 20 years before anyone noticed that mistake."

Now, you defend a different statement, and imply that my criticism was unfounded. I call that 'intellectual dishonesty', and I find its usage abhorrent. Glkanter (talk) 14:02, 24 November 2010 (UTC)[reply]

Not all wikipedia readers have the same notions of probability as Glkanter. I refer all to the wikipedia page Probability interpretations. If you could enforce the rewriting of the page in your way, new editors would start changing it immediately. If you want a well-written and stable and informative page you will have to find ways to respect other people's POV. Replacing one dogmatic POV by another dogmatic POV does not work.

Forget what I believe. Or what you *think* I believe. The Wikipedia article uses the K & W formulation as 'the MHP'. Does all this 'unknown host bias' mumbo jumbo have any meaning when the 50/50 hos bias premise had been unambiguously stated and included? Glkanter (talk) 14:02, 24 November 2010 (UTC)[reply]

My paper discusses how different notions of probability justify different ways to solve MHP. All solutions I give are very very simple. They all conclude that the player should switch.

I agree that within a subjectivist (informationalist) notion of probability there is no need to talk about host bias. By symmetry of prior knowledge under renumbering of the doors, the door numbers are irrelevant, full stop. Richard Gill (talk) 13:33, 24 November 2010 (UTC)[reply]

Is the above statement relevant to the MHP mediation? If so, have you shared it there? Is Morgan the only source for such a statement? Its nowhere to be found in formal math literature? How can that be so? Glkanter (talk) 14:02, 24 November 2010 (UTC)[reply]

Yes it's relevant; yes I said it there many times; no, apart from Morgan et al. one can also refer to Gill (2010, 2011) but I'm not aware of any other place. I'm sure one can find reference to the use of symmetry in general, in the formal math literature, for instance wikipedia symmetry in mathematics and especially the section on randomness [9] And of course the law of total probability is part of the story here. The really pedantic who want to write it out in the obvious symmetry argument in all bloody detail will probably want to refer to that well known result too. Richard Gill (talk) 14:08, 24 November 2010 (UTC)[reply]

In light of the above response, and as part of the great unwashed who has been told, and continues to be told, by countless Wikipedia editors that the simple solution is not capable of solving the conditional door 1 and door 3 MHP, I do not understand why the 'symmetry is not enough' argument against the simple solutions is allowed to continue to prevail over the article and is still being advocated in the mediation. Glkanter (talk) 14:52, 24 November 2010 (UTC)[reply]

Who says symmetry is not enough?? Simple solution plus symmetry is the most beautiful way I know to rigorously answer the question "what is the conditional probability?". Simple solution alone is a beautiful and rigorous way to answer the question "what is the unconditional probability?". That's what I wrote in my article. It's clearly a matter of opinion which question ought to be answered. Experts disagree. The great unwashed don't see a difference or don't care about it.

Well, Rick is still saying its not enough. He's also saying, which I think is logically flawed, that a source who describes the door 1 and door 3 problem, then gives a simple solution, is not claiming to be solving the 'conditional' MHP problem. I don't care what you call it, they're answering door 1 and door 3 with a simple solution. Right, wrong, or otherwise, that's what that source is doing in his paper. Please read Rick's new comment on the mediation page. Glkanter (talk) 08:15, 25 November 2010 (UTC)[reply]

Vos Savant knows there is a difference, but is only interested in the unconditional probability. Morgan et al. knows there is a difference, and are only interested in the conditional probability. Selvin does know but doesn't care. Carlton and Rosenthal and Rosenhouse all do know and all care very much, they all strongly think that we need to find the conditional probability. Kraus and Wang are interested in how people solve problems, they are not interested in how this problem ought to be solved.

The great unwashed don't know and don't care, that's a wikipedia valid POV, but I am not sure how one should weight popular opinion and expert opinion in matters like this.

Nobody is allowed to impose a single POV on the article. Right now, only you and Nijdam seem to be trying to do so. I think that that stance weakens your own position. I have much more sympathy for yours than for Nijdam's, but you both seem incapable to see, if only briefly, through other people's eyes. Richard Gill (talk) 07:58, 25 November 2010 (UTC)[reply]

I think that comparison is a 'false equivalency'. I don't want the article to favor, exclude or criticize either approach. I continue to be amazed and disappointed that you do not understand this from my postings. Glkanter (talk) 08:15, 25 November 2010 (UTC)[reply]

My apologies. That is the impression I got. Then it's a false impression. How about you read my paper and try to understand my POV? Richard Gill (talk) 09:44, 25 November 2010 (UTC)[reply]

Your apology is appreciated but not needed. What would help me in the future is if you would you explain how I had given you the wrong impression? Glkanter (talk) 10:06, 25 November 2010 (UTC)[reply]

Well, I think it is your strongly held POV that the simple solutions are perfect and perfectly adequate and the conditional stuff is bullshit. I am not saying that you want this POV to dominate the article. But I do get the impression that you don't follow the arguments for the conditional solution because you don't distinguish between a solution and a logical deduction, and don't admit that other readers might have different notions of probability from you. I get the impression that you don't want to make these distinctions, or try to step into these alternative mindsets (if only briefly) because you don't want to give up your POV. But now you'll be angry with me because I am trying to read your mind.

I find it can help resolve disagreements by trying to understand how different peoples' minds work. It is sometimes considered insulting to let on that this is what you are doing. But inside science it is how we try to solve disagreements. Richard Gill (talk) 10:48, 25 November 2010 (UTC)[reply]

Thank you for the explanation. As I see it, on the mediation page, and the article's talk page prior to that, I haven't spoken about the conditional solutions at all. Other than to say that without the 50/50 host bias, Morgan says they, too are false. On this page, I have intentionally written things like, 'I am not advocating this POV for the article, however...'. You might benefit from paying greater attention to what I post, rather than how you infer my mind works. Maybe that works for you sometimes, but I've pointed out too many times, and you have previously acknowledged, how you have failed at it with me. In business, we stop using a methodology in a particular case when we find its not applicable and/or beneficial. Glkanter (talk) 13:25, 25 November 2010 (UTC)[reply]

That you could equate my arguments, which clearly make use of reliable sources, and clearly note the difference between reliable sources and my own opinions, to Nijdam's proclamations is preposterous, incredibly frustrating, and downright insulting. I wouldn't think it humanly possible if you hadn't done so.

Maybe you could comment on the mediation page whether the current article meets your standards for NPOV and accessibility to the reader. Because I, for one, have no clue. And its a pretty important question, don't you think? Glkanter (talk) 13:25, 25 November 2010 (UTC)[reply]

As for reading and commenting on your paper: I've made it quite clear in my postings on this page that I don't agree with, or understand the justification behind, your opinions on when the contestant's chances of selecting a car from behind 3 unknown doors is 1/3, 1/3 and 1/3 vs 'cannot say for sure'. I have described myself as a 'Maslowian' when it comes to analysis. Meaning I don't build upon a foundation unless its solid. Of course, you may have chosen not to notice these things, focusing instead on what I 'really' mean to say. Therefore, as I have written numerous times before, I have no interest in reading your paper. So you may as well stop asking me. Of course, if you *did* actually know how my mind works, this paragraph of explanation would not have been necessary, nor would you keep asking me in the first place. Because I don't flip-flop with my analytical tools, or the conclusions and POVs I derive from them, without a paradigm change occurring. And that sure hasn't happened in the 2 years + I've been arguing about the MHP article. Glkanter (talk) 13:54, 25 November 2010 (UTC)[reply]

That's an incredible amount of misconceptions! Let's take a break. Richard Gill (talk) 14:42, 25 November 2010 (UTC)[reply]

Aw, heck, Richard. No hard feelings. In fact, I never thanked you properly for making it unnecessary for me to actually post my own thoughts on the mediation page, since you did it for me!

"Glkanter: how dare you label me! How dare you claim to be able to read my mind! How dare you use long words which only arrogant mathematics professors know about! You ****ing people make me ****ing sick! This was not posted by Glkanter on this mediation page. Glkanter (talk) 13:47, 24 November 2010 (UTC) No it was posted by me. Richard Gill (talk) 14:23, 24 November 2010 (UTC), Prof. Kill"

Thanks, bro! Glkanter (talk) 19:42, 25 November 2010 (UTC)[reply]

Agreed. Since you know what I'm thinking despite what I post, its not necessary, or of any utility for me to try to influence you in my postings regardless of my arguments or technique. Glkanter (talk) 14:46, 25 November 2010 (UTC)[reply]

If a probability is completely unknown, its square is also completely unknown. If its square is uniformly distributed, the probability itself is not uniformly distributed. This is a famous counter example which has been around for 100 years or more. You just cannot justify taking all values of an unknown host bias equally likely.

Fortunately, you don't need that. You only need to assume symmetry of your prior opinions about host bias (symmetry with respect to relabelling of the doors) in order to get that the average value is 1/2. 19% is as likely as 81%. 22% is as likely as 78%. And so on. Richard Gill (talk) 13:43, 24 November 2010 (UTC)[reply]

I was noodling with some ideas over the weekend, while the mediation was 'at rest'. My 2 new sections here are for fun, as well as to hash out what I think are inconsistencies in the various criticisms, including your's. I'm going to return my focus to the mediation page. Glkanter (talk) 14:02, 24 November 2010 (UTC)[reply]

Fine. Richard Gill (talk) 14:08, 24 November 2010 (UTC)[reply]

The criticisms of the 1 car, 2 goats solutions

Without the 50/50 host bias premise, the 1 car, 2 goats solutions don't work.

This may or may not be true. What *is* true is that the conditional solutions fail without the 50/50 host premise. Funny that *that* little tidbit isn't in the article, and never gets discussed.

This 50/50 host bias premise is clearly stated in the K & W formulation, making the comment moot. This is a topic for a probability course, or the Wikipedia article on probability, not the solution section of the MHP article on Wikipedia.

The 1 car, 2 goats solution fails when premises are changed.

That is a true statement, but a meaningless 'criticism' of the 1 car, 2 goats solution, or *any* solution to *any* puzzle. The K & W premises don't change, making this comment moot. This is a topic for a probability course, or the Wikipedia article on probability, not the solution section of the MHP article on Wikipedia.

The 1 car, 2 goats solution is false.

This comes unambiguously from only 2 reliable sources (neither of them named 'Morgan'). Being such a minority opinion, and contradicted by the math and other reliable sources (one of them named 'Morgan'), it may not even be worthy of inclusion in the article. This is a topic for a probability course, or the Wikipedia article on probability, not the solution section of the MHP article on Wikipedia.

Glkanter (talk) 12:45, 26 November 2010 (UTC)[reply]

Shell Game

Glkanter walks up to a street hustler on the corner of an intersection in a busy city. The hustler has a card table, 3 shells, 1 pea, and a wad of cash.

There will be no moving of the shells, just a game of 'did you select the shell with the pea under it?'

There *is no* presumption that the hustler places the pea uniformly at random under the shells. Quite the opposite, actually.

Glkanter confides to Richard that he has enough $ for a single play, and just wants a fair chance.

Glkanter also confides to Richard that he will select the shell using the random number generator app from his smart phone.

What odds would Richard advise Glkanter he needed to have a 'statistically even' chance at the game?

All of the above story remains unchanged, except:

Glkanter doesn't own a smart phone, and instead will use his 'lucky number' to select a shell.

What odds would Richard advise Glkanter he needed to have a 'statistically even' chance at the game? Glkanter (talk) 13:03, 26 November 2010 (UTC)[reply]

If 'statistically even' is not a valid term, please correct it. I presume you understand my meaning. Glkanter (talk) 13:06, 26 November 2010 (UTC)[reply]

Both of our Bayesian probabilities that we'll get the pea by switching, for both versions of the game, are 2/3. Both before we pick a shell, also after we have picked a shell but the hustler hasn't turned over another yet, and after he has done this too. Please read also Probability interpretations. Richard Gill (talk) 13:15, 26 November 2010 (UTC)[reply]

Please read my paper, where this is also stated. There is still time to make minor edits so that it becomes more usable by the great unwashed. While you're at it, consult law of total probability and symmetry in mathematics. Richard Gill (talk) 13:19, 26 November 2010 (UTC)[reply]

I find your response to the 2nd scenario inconsistent with 'B' from this section above. Can you help me understand the difference? Thank you. Glkanter (talk) 13:34, 26 November 2010 (UTC)[reply]

Remark B was made with a frequentist interpretation of probability in mind. In fact it was part of a discussion with you in which I was trying to explain that there are different interpretations around. See Probability interpretations. Richard Gill (talk) 13:51, 26 November 2010 (UTC)[reply]

As you know, I remain intentionally ignorant of such matters. So, both answers ('B' from the previous discussion, and the 2nd scenario from this section) are unchanged, and both answers are correct? That's why I won't read your paper. Glkanter (talk) 14:06, 26 November 2010 (UTC)[reply]

Each is correct in its own context, yes. And you are proud of your ignorance. I think that plays into the hands of the conditioning fundamentalists. (I support the simplist side.) Richard Gill (talk) 18:08, 26 November 2010 (UTC)[reply]

My answers for what they are worth. Assuming the iPhone random number is random and secure I would give you a 1/3 chance of picking the pea using either the Bayesian or frequentist model.

If you chose your lucky number I would give you less that 1/3 chance of picking the pea because hustlers probably know more about lucky numbers than you do. From the Bayesian perspective of someone who does not know much about hustlers your chance of winning is still 1/3; these are the punters the hustlers are looking for. Martin Hogbin (talk) 12:09, 27 November 2010 (UTC)[reply]

Now, compare your answer to this question's part B, about halfway down the screen. Do you agree with Richard that the odds are not determinable? Can the answers to these 2 questions (the 2nd scenario above, and B in that other section) be different? If so, am I, or a Wikipedia MHP article reader supposed to grasp this? Is it material to me or that reader? Glkanter (talk) 13:15, 27 November 2010 (UTC)[reply]

Yes, I agree with Richard's answer. In the shell game, if you pick a shell using your luck number you probability of winning is no longer 1/3. In principle, it could be more or less that 1/3 but my money would be on the hustler having a better than 1/3 chance of guessing your lucky number (in advance, so to speak). Martin Hogbin (talk) 15:55, 27 November 2010 (UTC)[reply]

You didn't answer all my questions. What about Richard's lucky number answer if it hasn't been stated that the car is placed uniformly at random? One play of the game. How would you model the shell game in the 2 differing scenarios? Is there a new premise in your shell game along the lines of "...and you, being a typical punter are easy prey..."? What if we removed that premise? How would you model the 'non-uniform-car' door odds if you didn't know the contestant's door selection method at all? Glkanter (talk) 16:13, 27 November 2010 (UTC)[reply]

I am not sure which questions you are referring to. If the car/pea is initially placed uniformly at random then the probability of picking it is 1/3 however the player chooses.

Q1 = MHP, no knowledge of initial car distribution, lucky number, one play of the game (come to think of it, Whitaker/vos Savant's question + lucky number)

Q2 = shell game, no knowledge of initial pea distribution, lucky number, one play of the game

Q3 = shell game, no knowledge of initial pea distribution, random number generator, one play of the game

Q4 = either game, no knowledge of initial distribution, no knowledge of how the punter makes his selection, one play of the game (new Q)

Glkanter (talk) 13:04, 28 November 2010 (UTC)[reply]

Regarding the new premise, "...and you, being a typical punter are easy prey...", yes, I did add that premise, because of your use of the word 'hustler' and the fact that you were adamant that the pea was not initially placed uniformly. A reasonable assumption in those circumstances would be that hustler places the pea so as to maximise his profits. If I am not allowed to make that assumption then I have to agree with Gill, I am not able to give you odds that give you a 'statistically even chance' of winning. The same with doors and cars. This has always been my point, with the MHP there are only two consistent answers, 2/3 or 'don't know'. The latter answer is rather uninteresting and generally ignored.

The pea (or car) *may* be placed uniformly at random, I don't know, so I can't eliminate that, or any other possibility. Glkanter (talk) 13:04, 28 November 2010 (UTC)[reply]

The real problem is the exact meaning of the word 'probability', a subject you refuse to discuss. Although most people have an intuitive notion of what it means there is no universally agreed precise definition of what it means and I defy you to come up with one.

There two well-known meanings (maybe Richard knows some more) which are Bayesian and frequentist. For the frequentist definition to make any sense you have to at least contemplate the repetition of the 'experiment'. A good example would be one of your favourites, a casino. It makes sense ask what the probability of winning with a single bet on a 7 at roulette is because you know that the casino repeat the 'experiment' many times even if you do not. If you want to ask about a frequentist approach to the MHP then you have to ask what would happen if you repeated the show.

The other meaning of the word is Bayesian, which is based a state of knowledge. Now you have to decide what it is that you know. If you know nothing other than the pea was placed under one of the shells then the only option you have, apart from giving up, is to take it that it was placed withe equal probability under any shell. Thus people who know nothing about street hustlers are likely to lose money to them.

There is no 'everybody knows' definition of probability. In the end, how good your answer is depends on how well the question is defined, as Seymann said. Martin Hogbin (talk) 11:55, 28 November 2010 (UTC)[reply]

My most recent question to Richard was 'what you you advise Glkanter'? Maybe I should have added, '...Who is intent on playing the shell game'. But I already said '1 play of the game, didn't I?

My point is this, I asked Richard 'How would you advise...'. Shrugging one's shoulders is not advice. Saying, "All's we can do is surmise uniformity, so its 1/3, but maybe he's counting on us to think that way" *is* advice, and can be justified, I presume, no matter what type of probabilist one is. The idea that you or I *could* model the hustler *if* I use a random number generator, but not if I use my 'lucky number' continues to strike me as preposterous. Plus, Richard answered the 'lucky number' question 2 different ways in the MHP and the shell game scenarios. Which was my original point to you, not 'how does the hustler make a profit?'. I don't know why 'probability' would differ in those cases. Glkanter (talk) 13:04, 28 November 2010 (UTC)[reply]

The answer depends on exactly what you want to know. If you just want advice on the question as asked, that is quite simple, 'wear a cheap shirt, because you will lose it if you play'.

If you do not expect me to use any real world knowledge and your question was really, 'A pea has been placed under one of three shells...' then, you might take it that the distribution was uniform. On the other hand, you used the word 'hustler' and made clear that the pea was not necessarily placed with equal probability under each of the three shells. A hustler would likely know that the typical punter's choice of shell was not uniformly distributed and would therefore place the pea to take advantage of this, placing it more often under the shell least likely to be chosen by the punter, thus improving his odds and worsening yours. If you choose randomly then his plan fails. Martin Hogbin (talk) 00:02, 29 November 2010 (UTC)[reply]

Good discussion. My *advice* is clear. Choose your shell with a random number generator and switch. You can bet at odds 2 to 1 that you'll win. I'd advise not to use your lucky number. I have no idea whether smart hustlers could guess it or not. Richard Gill (talk) 01:56, 29 November 2010 (UTC)[reply]

Your Apples <> Pears Comment on Martin's Page

Yes, Richard, I understand the different sample sizes in the 1 car, 2 goats & the door 1 and door 3 interpretations.

And I instinctively understand that with the obvious symmetry the meaningfulness of this difference is akin to the difference between 1 blade of grass from my lawn vs any other blade of grass from my lawn. Door 1 & door 3, or door 2 & door 1...<yawn>.... Especially when the whole point of the various models is to aid in decision making. That they all validly come up to 2/3 & 1/3 *is* very meaningful to the unwashed. Glkanter (talk) 15:12, 26 November 2010 (UTC)[reply]

I don't say 2/3 <> 2/3. I say that 2 out of 3 apples is different from 2 out of 3 pears. The whole point of the various models is to aid in decision making. The advice is "switch". If you're going to play the game many times I would recommend you bring along a good random number generator to make your initial door choices.

Like you, I do not find the conditional approach very interesting. But I would like to see a decent and stable MHP page, so somehow we are going to have to accomodate ourselves to the wishes of people who think differently from ourselves. Including all those teachers of statistics at not top notch US universities who wrote books and articles about it and all those wikipedia editors who learnt statistics from those books. Richard Gill (talk) 17:46, 26 November 2010 (UTC)[reply]

You see, that adds no value. 'If you're going to play the game many times I would recommend you bring along a good random number generator to make your initial door choices.' What if I don't? And what if I play once? Is it 2/3 if I use my lucky number. Talking in circles, avoiding direct questions, going layers and layers deeper than the topic requires.

Not to mention outright lies:

"Now Glkanter can object that I can read words in Morgan et al. which he doesn't see there. Too bad. Nobody can have it all exactly their own way. Many editors will remain insisting that the conditional probability and the unconditional probability are different concepts, it doesn't matter a hoot whether you are a professional or a layman, everyone must realise that 2/3 of all the games is different from 2/3 of the games that the player chose 1 and the host opened. Richard Gill (talk) 12:06, 26 November 2010 (UTC)"

That's 2/3 & 2/3, not 'apples and pears'. And with that, I am done with you. Glkanter (talk) 17:56, 26 November 2010 (UTC)[reply]

Back to my talk page

User_talk:Gill110951