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::Remember that at the outset we knew that the car was behind one of two doors, that we didn't know which two, and that either of the two was as good as the other. Now we know "which two", and choosing one is ''still'' as good as choosing the other.
That being the case, there's no advantage to "switching". As I said, this is an abbreviated argument that I know incorporates certain subtle errors, but can anyone refute its conclusion in this same idiom? Apologies if this has been proposed before, especially if I'm wrong in my conclusion. ;-) Many thanks for reading, – <
:[[User:Ohiostandard|<
::Thanks for replying, [[User:Martin Hogbin|Martin]]. I didn't know if anyone would. I did read through the article, of course. But I thought it would be quicker to express this positive argument than to try to make a negative one, i.e. than to use the idiom of the current article's examples and show why I suppose they're in error. I'll give that a try, though, and would be pleased to hear what you think about that effort. I'll also post, of course, if I persuade myself that the solution examples presented there ''are'' correct.
::Likewise, if you feel inclined, I'd be pleased to know where you think I've gone wrong in what I presented above. No need to do so before I present my critique of existing solution examples, of course. But I wonder, for example, whether there would be general agreement with the proposition: "The first 'choice' ( i.e., before the contestant is offered the chance to 'switch' ) has no impact on the final outcome. Only the stay/switch choice is relevant." Best, – <
:::I would be happy to explain where you have gone wrong but this should probably be on my or your talk page as these pages are intended for discussions on how to improve the article. If you do not find the solutions in the article convincing then in my opinion the article need improvement. You will see from the article that most people get the answer wrong but one thing you can be sure about is that you have.
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::::I'll be curious to see how re-reading the article, and reading Gerhard's post, below, will affect my view of the problem. I purposely haven't done either yet, because I wanted to be able to express how I came to this with my presentation of it uninfluenced by doing so. Just in closing, I'll mention that one of the things I love about the way I came to see this, to the extent I presently do, is that it depends on the "reformulating the problem according to an extreme case" approach. Using a "gazillion" shells or doors, in other words, demonstrates Polya's statement to the effect that if you can't solve a problem, that there's also a simpler one (or an extreme case, as I'd put it, in this instance) that you can't solve, and that you should find that simpler problem, solve it, and use that knowledge to move on to the more puzzling one.
::::I see I no longer have time just now to read Gerhard's post with the attention it deserves, or to re-read the article carefully, either, but I'll do so soon, and will reply further. I expect the small degree of vague unease I still feel over the "three door" or "three shell" reductionist case, as opposed to the "gazillion" case, will be dispelled by the understanding I expect I'll gain in doing so. Thanks so much; this has been loads of fun so far. It really is a delightful and fecund problem! Thanks, – <
:::::Ohiostandard, I am interested in how you came to understand the correct solution as I think i will help improve the article. The way in which you understood the answer also explains another aspect of the problem that some people find troubling, which is that it matters whether Monty knows where the car is and does not just pick an unchosen door at random which happens to reveal a goat. I find your description of the transfer of information from Monty to you an interesting way of explaining why this is so.
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::::::I guess my major criticism of the article in its current state is that it focuses too little on helping mathematically unsophisticated readers get the "intuitive flash" they need. As I'm sure you're all aware, many, many mathematicians have written about the disparity between the insight that motivates a proof and the proof itself, and have observed that the proof isn't always or even usually the most efficient means of communicating that "flash".
::::::I love well expressed proofs dearly, of course, and I understand why it's absolutely crucial that they be painstakingly exact in the ideas they present and in the language and symbols they use to present them. It's just that the insight that motivated them in the first place is often times submerged, especially for the non-mathematician, in the necessary rigor.
::::::With a view toward trying to help address that, I've posted to the article's main talk page with a suggested re-write for the "increasing the number of doors" approach, and would be very pleased to learn what you all think of that. Thanks again, to everyone here, for your generosity and patience, and refusal to bite! Cheers, – <
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Will that table help the readers to see that switching will double the chance to win from 1/3 to 2/3? Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 18:16, 1 April 2011 (UTC)
:Gerhard, I'm extremely grateful to you for taking the time to respond so carefully here: It's very generous in you. You may notice the response I top-posted above, after Martin's reply to me. I hope you won't be offended that I (still) haven't read your own reply; I certainly will do so, but it would be disrespectful not to do so with the care it merits, and I have to fly out the door just now. I'll read through it carefully soon, though, and will reply. Again, many thanks for taking the time to post this. Best regards, – <
::Gerhard, I've read carefully through your post, now, and would like to again thank you for it. As I wrote above, though, I'm afraid I have to admit that I'm too dense for tables to be of much help to me. I certainly admit they can ''demonstrate'' a solution but, for me, they're not much assistance in understanding why that solution is as it is; they're not much help for me in trying to ''understand'' the problem, that is. There's just too much for me to look at, too much to take in ... I think it was Poincare' who wrote that you don't really understand a proof until you see it as a single idea. I don't know if I'd go quite so far as to say that for all problems, but I think it does apply to this one. Sorry to disappoint; I'm sure there are others who would read through your presentation above and at some point in their reading get that "Aha!" moment of understanding. I'm just not one of those, I'm afraid. Thanks very much for your presentation, however; as I said before, I'm very sensible of your generosity in making it. As I mentioned to Marin, up above, I've posted a section to the article's main talk page about a possible re-write based on the "increasing the number of doors" idea. I'd be pleased to know whether you think it has any value. Best regards, – <
:::But I am interested to hear from you whether it is plausible that the above table clearly shows that the probability to ''win by staying "1/6 + 1/6 = 1/3"'' is only one half of the probability to win by switching of "1/3 + 1/3 = 2/3". Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 23:34, 2 April 2011 (UTC)
::::Maybe it does show that, Gerhard, but I'm afraid I can't say that it does so ''clearly'', for me. Perhaps it would do so for someone who had taken a probability class recently, though, and who was thus already familiar with the interpretation of such tables. For my part, I spent 30 minutes before I understood what the "(1/2)" in the top cell of the third column was intended to mean, and then only did so by reading the text of [http://en.citizendium.org/wiki/Monty_Hall_problem#Explicit_computations the Citizendum page] that includes basically the same table. And without previous acquaintance with Bayes' rule ( "posterior odds equals prior odds times likelihood ratio", as Citizendum expresses it ), the "Joint probability" column wouldn't make sense to the uninitiated, and the rest of the table would be equally opaque, as well, I believe. To speak very candidly, I think the table would be very likely to confuse most readers, even assuming they'd take the time to try to understand what it's intended to communicate. Sorry to disappoint; perhaps other readers would find it easier to interpret than I do. Best, – <
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:::Thanks so much, Rick, for your generosity in replying like this, "in my idiom", so to speak. This is a really clear explanation of where I went wrong, and it was great fun to read and understand. I agreed with everything you'd written up to this point, but when I read, "Where you're going wrong is thinking that because the original probabilities of these doors was equal, then the conditional probabilities of these doors must be equal as well" I thought to myself, "that's a beautiful and concise explanation!" Before coming back to this I'd read James & James' definition/entry for "Probability", since it's been so long since I've thought about the subject ( and was never any good at computational or applied math, anyway ) and that refreshed my memory re the distinction between mathematical or (same) ''a priori'' probability, and conditional probability, an important distinction, of course, and one that the wikilink you provided points out, as well.
:::Again, I appreciate your kindness and effort in presenting this explanation, Rick, very much. You'll see that I've also responded at some length to Martin, above (sorry for the top post), although my reply there really is meant for you and for Gerhard, as well. As I said (as if) to Martin, I'd also be pleased if you'd care to look at the "Increasing the number of doors rewrite?" section I added to the article's main talk page, and see whether you think it has any value. Perhaps it just represents the way my own peculiar brain came to the "flash" or "grok" or "Aha! moment" of seeing the solution; I'd be pleased to hear objective opinions about whether it might be more universally helpful. Best regards, – <
Btw, since this is so long a thread, I'll have no objection if regulars here want to collapse it, or parts of it, or just move it manually to archives at some point when everyone has been able to reply. I'll leave that up to all y'all. Thanks again, everyone, for your very generous comments so far! – <
===A different table===
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:Thanks, Rick. I sincerely hope it won't offend anyone if I say so, but this is the first table I've seen that's simple-enough and clear-enough that I could understand it quickly, and see how it demonstrates the 2/3 versus 1/3 advantage re switching. I'm getting the impression that most regular contributors here have their favorite way of illustrating that, but I have to say that I think this is a really good one. It's much (!!!) easier for me to understand than either of the tables currently in the article, or than the decision tree either, for that matter.
:I know that someone must have put a great deal of work into the graphics for the illustrated table especially, and I certainly don't intend to disparage that effort: Quite the contrary; I honor it. If I'm to answer candidly, though, I'm afraid I have to say that while I imagine the decision tree has value to the uninitiated, I can't support the same statement for the two tables presently in use in the article, and I'd prefer to see the article simplified by using the one above, instead.
:I've top-posted this, above Guymacon's flush-left/outdented comment below, btw (sorry, Guy) because his outdenting made it impossible for me to otherwise indicate which post I was responding to. Thanks Rick, for this table. I think it's extremely helpful, and that it provides the strongest assist I've seen so far to help the average reader who has no training in probability or statistics understand the solution. – <
::Thanks for correcting the outdent mistake. It really is a quite good table. [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 22:59, 4 April 2011 (UTC)
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:* N = the number of doors at the outset,
:* P<sub><small>Contestant</
:* P<sub><small>Monty</
:Then it's clear that
:* N = 100
:* P<sub><small>Contestant</
:* P<sub><small>Monty</
:* P<sub><small>Monty</
:This last statement or formula isn't strictly necessary to solve the problem, of course, but it might help you understand it better: It says that Monty's probability of choosing correctly is 1 minus the contestant's probability of choosing correctly, because the contestant might have got lucky and picked the correct door before he could, thus eliminating Monty's ability to do so.
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I don't intend to introduce this proposed explanation into the article myself: That would seem the height of presumption to me, given that some of you have been herding cats here for years! ( Thanks for that! ) But I do think it presents the clearest chance that a person who's relatively unsophisticated in math has of getting the "flash of insight" that's necessary to understanding the problem. I don't claim it's the best way to ''prove'' the solution, or even that it's a proof at all, though, just that it's likely to be pretty accessible to a typical reader. What does everyone else think about that? – <
:<small>( On re-reading this, I have some misgivings over saying that Monty has only a probability of .99 in choosing the "right" door in the case where we start with 100 doors at the outset, or .66 in the case where we start with three. Obviously his probability rises to unity, to 1, if the contestant's first pick was incorrect. But I won't (yet?) complicate the presentation I made above by correcting it to accommodate cases based on whether or not the contestant's first choice is correct. I know that might be helpful, of course, because Monty's "chances" of picking the correct door drop to zero if the contestant gets lucky on his first pick. But if people think the approach has value, maybe we can work together to clarify the two cases, and dot the "I's" and cross the "T's". Best regards, all, and thanks for your ongoing work on this article, very much. Btw, special thanks to Martin, Gerhard, and Rick, for their patient and generous help on the "Arguments" subpage. Cheers, – <
:''"Monty is effectively choosing the door the car is behind, by keeping that one door closed [...] impossible for him to do anything else."'' — Incorrect, because if you should have chosen the winning door, the host will not dispose of the car. So what is written here is just incorrect. And if the host "should be biased to just ''never'' open the door he left closed, if any possible", then the chance that the still closed door hides the car could be only 1/2 at least (as per Ruma Falk), even in your gazillion example. And btw: "Then it's clear that ..." is never enough, the content of the article must be sourced. Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 23:10, 2 April 2011 (UTC)
::Thanks, Gerhard, but I think you missed the condition that the boldfaced part of the sentence is preceded by, viz. ''"Assuming you didn't guess correctly the first time (a reasonable assumption, given your one-in-a-gazillion chance)..."'' Also, I'm afraid I don't understand your objection about the possibility of the host's bias. As I understand the problem, Monty has no opportunity for bias; as I understand it he has no choice but to open all but one door, excluding your own. Could you explain further?
::I've moved your comment to just above, btw, from being interleaved with the presentation I made in the "callout box". If everyone did that, then the original intent of that post, and what I'd written versus what everyone else contributed, would become obscure. Also, if we can all agree on an aid to understanding (not a proof) then I don't see any reason why that would have to be be "sourced". Or was that a part of the arbitration requirements? – <
::: Good point, @Ohiostandard. Host bias (or not) is irrelevant to the simple solutions. Those who would always switch will win with unconditional probability 2/3. For frequentists, this holds whether or not the host is biased, and for subjectivists this holds whether or not their opinions about possible host bias are symmetric regarding the direction of any bias. Gerhard Valentin refers to Ruma Falk for the host-biased conditional result but it was already in Morgan et al, but more easily derived with Bayes' rule (odds form of Bayes theorem), for instance, cf. Rosenthal's paper and book. But you are talking about an argument for the simple (unconditional) result. It's indeed a valuable aid for understanding why the "naive" argument "no need to switch because Monty has not given you any information about your initial choice" is faulty. It's given by many sources. In the unbiased case, Monty has given you no information about your intitial choice, so the probability that that was right is still 1/3. There is still 2/3 chance left and that stays where it was, with the two other doors, one of which Monty has kindly shown to you does not hide a goat. As Gerhard mentions, in the case of a biased host, the identity of the door which Monty opens can contain information. In the most extreme cases, the chance your initial choice was right can rise to certainty, or fall to 1/2. But it is never unfavourable to switch, since there is no way to improve the 2/3 overall success chance of "always switching". Editor @Lambiam recently found an alternative proof that 2/3 cannot be beaten, and hence that all conditional chances of wining by switching are at least 1/2, by showing that a known host bias can only be advantageous to the player. Yet even with a maximally biased host, it's easy to see that 2/3 (overall) can't be improved. I wrote it out in [http://www.math.leidenuniv.nl/~gill/mhp-statprob.pdf]. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 10:47, 3 April 2011 (UTC)
:::: Thanks, Richard. Your kind response reminds of one of the many things I love about Wikipedia: There are experts here in just about any subject that could interest a curious person, and they're usually willing to share their expertise, even with untrained persons, quite generously and patiently. However it's communicated, I do wish it could be made explicit in the article that Monty is actually making a choice, too, but one that's based on better information, and that his doing so has information value for the contestant.
:::: Also, I don't know whether it can make it into the article or not ( I'll leave that to you who've contributed here for so long ), and without the ''least'' wish to disparage any other presentation, all of which I honor, I feel that candor requires me to disclose that the first immediately convincing presentation that has worked for me for the simple version is the one ([http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem/Arguments&oldid=422161032#A_different_table permalink]) that Rick Block posted to [[Talk:Monty_Hall_problem/Arguments | the ''Arguments'' page]] under the heading, "A different table". Best regards, all. – <
:::::OhioStandard, I am a little concerned that you are deferring to other editors based upon them being here longer. Please read [[WP:NVC]] [[WP:OWNERSHIP]] and [[WP:ODNT]]. (I do agree that it is best to discuss changes and and seek consensus rather than just jumping in and changing things, but we are all equal here.) [[User:Guymacon|Guy Macon]] ([[User talk:Guymacon|talk]]) 07:52, 10 April 2011 (UTC)
::::::That's a fair point, Guy, based on the language I used above. If it'll make you feel better, I don't mind disclosing that my apparent humility is much more formal than real. "Jungle manners" on entering new or disputed territory, as Marie Louise von Franz used to say. ;-) The links are helpful, nevertheless, but please let me assure you that if something comes up that seems important to me, I'm perfectly willing to make any amount of noise or to (figuratively) bloody my knuckles to be heard about it. I genuinely appreciate your remarks, though; thanks. – <
== Probability concepts ==
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:''Imagine the producers of the show let the host keep the car for himself personally if the contestant picks the wrong door at the outset and the host then (indirectly) "picks" the correct one, i.e. by choosing to leave that one of the remaining two doors closed. Now suppose the producers, at the start of the event, offer to let you play the role of the host instead of that of the contestant. The contestant gets to pick a door first, but you'd be told in advance which door the car is behind if you played the host. Which role would you prefer, and why?''
This seems pleasantly concise, and (to me) very likely to at least give the reader pause, to give him a salutary doubt that his immediate "50/50" assumption re the stay/switch decision is correct. It also has the pleasant consequence, I think, that if people think about it, it will bring them to the conclusion that it's more advantageous to know where the car is than it is to pick a door first. One could then go on to explain that by choosing "switch" one effectively puts himself into the host's shoes, and thus benefits by the host's foreknowledge of where the car is. I'm mostly just curious to know whether this has been argued before. Thanks, – <
:Krauss and Wang (reference in the article) set out to try to get people to arrive at the correct answer more frequently. Asking people to take the perspective of the host is one of the things they tried (experimentally). It indeed does help. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 06:24, 8 April 2011 (UTC)
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:::::::::I am asking you what other prior there could possibly be other than a noninformativce one, considering we are given no information. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 22:01, 9 May 2011 (UTC)
::::::::::So you're not talking about sources here, but you want my personal opinion? No thanks. You asked me to substantiate, from sources, a claim I made. I've done that, and you picked Morgan et al. to discuss further (not me). You are now apparently trying to change the subject to THE TRUTH. Will you please respond to my question to you. Are you claiming Morgan et al. says anything other than "the probability" (even the Bayesian probability) is 1/(1+''q'') for the vos Savant/Whitaker version (with vos Savant's clarifications) and that if one insists on a single numeric answer a Bayesian can get one by also assuming a specific prior (and the probability one arrives at depends on what prior is assumed)? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 04:30, 10 May 2011 (UTC)
:::::::::::Well, of course, we are free to talk about The Truth here on the talk page and indeed some understanding of probability would be assumed by Morgan for their intended audience. There is nothing new or unexpected in assuming a uniform prior in discrete probabilty theory where there is no information to indicate otherwise. It has been a standard part of probabilty theory since Laplace (seeTruscott, F. W. & Emory, F. L. (trans.) (2007) [1902]. A Philosophical Essay on Probabilities. {{ISBN
:::::::::::We can both read what Morgan write. They give only one numerical answer and this for the uninformative prior. They also discuss priors where the player may think that the host has a particular strategy (for example with a large weight near q=1). There is nothing in Whitaker's question to suggest that this may be the case, it gives us no information on the matter, hence the noninformative prior would be the standard choice and that is the one that Morgan actually use in calculating a single numerical solution. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 08:45, 10 May 2011 (UTC)
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