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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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Game theory indeed should be a topic in the article MHP, and I will be trying to write a draft including Game Theory. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 09:13, 14 September 2011 (UTC)
== Monty Hall Disproof?? ==
The Monty Hall problem irritates me profoundly. Mathematics can be used to represent reality, but this is a deliberate con. The mathematics used, does not represent reality.
I see it this way. (And assuming that it is a 'fair test' and Monty does not deliberately mislead the contestant etc etc )
In mathematics, one must always get the same result to a problem. Two people can not get different answers.
Accepting that there were initially three doors. Once one door has been opened all of the previous statistics can be thrown out of the window. In the real world, once one door has been opened, there is now a certainty of what was behind that door. The situation now, is a completely new problem.
There are two doors, one has a car behind it, the other a goat.
Now, what would the odds be if you, or I walked off the street, into the studio and onto the set, and we chose a door. We would stand a 50/50 chance of getting a car or a goat. If mathematics means anything, the original contestant must have the same odds.
I say again, draw the problem at the final denouement. Two doors, one prize, 50/50 odds.
[[Special:Contributions/78.86.43.71|78.86.43.71]] ([[User talk:78.86.43.71|talk]]) 17:07, 20 October 2011 (UTC)
:your answer is the entire point of why this is a problem/fallacy. The intuitive answer is completely wrong. It has been proven multiple times by thousands/millions of times of experimentation, as well as logically proven. You are not just picking between two doors (which would indeed be 50/50). You are picking between 3 doors, with information provided that allows you to get the right door 2/3 of the time. [[User:Gaijin42|Gaijin42]] ([[User talk:Gaijin42|talk]]) 18:18, 20 October 2011 (UTC)
::There never were 3 doors to begin with. The game show host always eliminates one. So there are always 2 doors. To illustrate, supposing you were presented with 10 doors, with a car behind one of them. You select a door. Before opening it, eight unselected doors are opened showing goats. Does switching to the remaining unselected door now increase your chances of winning the car to 90 percent? Of course not, because there were never 10 different selections to begin with. Eight selections showing goats were always going to be discarded. The game is intended to be reduced to a 50/50 chance of winning or losing. On the bright side, the many elaborate arguments here might apply to a Deer Hunter problem where 5 prisoners at a table are given a gun with only 1 bullet in the 6 chambers and are asked to play Russian Roulette and they may select which chamber to fire according to a numbered chart on the wall. The first prisoner is asked to select one of the 6 chambers and he will take his turn last. The other four prisoners select their chambers one-by-one from the remaining unselected chambers. One-by-one the other four prisoners fire their designated chambers and survive their turns. Now the head of the prison camp gives the first prisoner a choice. The prisoner can fire the chamber he initially selected, or he can switch to the sixth unselected chamber. Would you switch? [[User:Safetyweek|Safetyweek]] ([[User talk:Safetyweek|talk]]) 05:18, 27 November 2011 (UTC)
:::I suggest you try an experiment. Take an entire deck of 52 cards. Let the ace of spades represent the car. Shuffle. Pick one and put it aside (as the player). Now shift roles. As the "host" look at the remaining 51 cards and choose 50 to discard that aren't the ace of spades - which will always reduce the number of choices to two. I believe you're saying it's now 50/50 whether the player's card is the ace. Flip a coin. If it's heads lets say the player keeps her original choice and if it's tails lets say the player switches. OK. Now look at the player's card. Record what the player's card was (ace or not), and what the result of the heads/tails choice was (keep or switch), and whether this choice "won". Repeat, say, 20 times. Please report back how many times the player's card ended up being with the ace, how many times the player switched, and how many times switching won. We'll wait. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 16:43, 27 November 2011 (UTC)
::::I see. I got it backwards. Switching in the Monty Hall problem increases your chances to 2 out of 3. Switching in the Deer Hunter problem makes no difference, as your chances of survival have gone from 5 out of 6 to 50/50. The key to my correctly understanding the Monty Hall problem is realizing that the initial selection prevents the host from revealing what's behind that particular door. This limits the all-knowing host's options in eliminating a goat. And so you're right, there's a 2 out of 3 chance the car is not behind the door initially selected, and if we know that one of the remaining doors would certainly reveal a goat, then the other remaining door will reveal the car 2 out of 3 times. On the other hand, in the Deer Hunter problem, the prisoners are not all-knowing. No prisoner knew if their selected chamber had the bullet, and so the odds of the first prisoner selecting an empty chamber go from 5 out of 6, to 1 out of 2. It makes no difference if he switches. Thank you Rick for the lack of derision. [[User:Safetyweek|Safetyweek]] ([[User talk:Safetyweek|talk]]) 05:07, 28 November 2011 (UTC)
:::::If you actually did this experiment it might be useful for anyone else reading this to see your results (they can, of course, do it themselves). Not to belabor the point, but there is a big difference between the odds between two alternatives being 50/50 and the odds of a ''random choice'' between two alternatives being 50/50 (this is the point of the coin flip). The odds of a random choice between two alternatives will always gives you a 50/50 chance of winning, regardless of the actual odds between the two alternatives - so, in the MHP, if you flip a coin to decide whether to switch or not you'll have a 50/50 chance of winning the car. However, if you stick with your original choice you only have a 1/3 chance and if you switch you have a 2/3 chance. The composite given a random choice of whether to switch is 1/3*1/2 + 2/3*1/2, which of course works out to 1/2.
:::::All of this assumes the host chooses evenly when given a choice (in the case the player originally picks the car) - without this assumption there's a whole other can of worms. Suffice it to say that this remarkable little problem has much more to it than meets the eye. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 05:32, 28 November 2011 (UTC)
::::::Actually, I did the experiment this way: my gorgeous hostess dealt 5 cards face down before me, 2 of them were cars and the other 3 were goats. I selected one and kept my hand over it. Of the remaining cards, my hostess eliminated a goat and teasingly offered to let me change my mind and select from one of her 3 remaining cards, which of course I did. And we did this 60 times, and each time I succumbed to the temptation to select one of her cards. (This was a precondition to our opening a bottle of Chateau Mouton Rothschild which we had acquired for this occasion.) As it turned out, I won the car 32 times, so my overall chances were about 50/50 by always switching. And so with this, my hostess moved my attention to the difference between mystery and mystique.[[User:Safetyweek|Safetyweek]] ([[User talk:Safetyweek|talk]]) 07:49, 2 December 2011 (UTC)
== Monty Hall (super simple) ==
Its just this simple people, chill with your bla bla bla math and 100 dollar words! If you are not going to switch after the host opens the door then you are forever LOCKED into the original odds of winning at the time of your choice, end of story. God All Mighty himself cannot change that number! If its three doors its 1/3, if its two doors its 1/2, if its eight doors its 1/8, do I need to keep going? Everyone keeps harping on the "switch", but it has nothing to do with "not switching" odds!!!! LOL all you "SMART PEOPLE" are amazing!!! LOL!! [[User:Davecross1|Davecross1]] ([[User talk:Davecross1|talk]]) 09:07, 24 November 2011 (UTC)
:Actually, there are two probabilities that may or may not have different values - i.e. the probability of picking one door out of three hiding a car, and the [[conditional probability]] the car is behind the door you've originally picked given that the host has opened the door that he opened (for example the conditional probability the car is behind door 1 which you originally picked given the host has opened door 3). Whether these two probabilities have the same value is the crux of the question, and the answer is it depends on exactly what rules govern the host's behavior. For example, say you pick door 1 and the host forgets where the car is and opens door 3 and luckily reveals a goat. The probability of your original pick being correct was 1/3. But, the conditional probability the car is behind door 1 given the host randomly opens one of door 2 or door 3 and reveals a goat is 1/2 (this is why a player at the end of "Deal or No Deal" with the grand prize still unrevealed has only a 50/50 chance of winning it).
:You might argue that in the MHP the host MUST open a door and must reveal a goat - but even this is not enough to say that the conditional probability has the same value as the original chance of picking the car from the three doors. For example, say the host always walks across the stage from door 1 to door 3 and opens the first door he runs into that is not the door you picked and not the door hiding the car. In this case, if you pick door 1 the conditional probability the car is behind door 1 given the host opens door 3 is 0 (not 1/3).
:For these two probabilities to have the same value, the host has to open a door, must reveal a goat (is not opening a door randomly), and if the player has originally picked the door hiding the car must randomly choose which of the other two doors to open. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:44, 24 November 2011 (UTC)
== Think about the doors as the players. They all have equal chances ==
The fallacy rests with the assumption that after the host opens one door, we are still playing the same game. In reality, that is a new game, a game with two doors, not three. While in the beginning there were three doors with one in three chances to pick the right door, after one of the doors in eliminated, the game has only two doors, with one out of two chances to get it right. If changing your pick would give you two out of three chances to get it right, then your new pick could have been any of the other two doors, including the eliminated one, which is not possible as you know that one is the wrong one.
Another way to look at this is considering three contestants holding three boxes, with only one having inside a prize. Every contestant has equal chances to be holding the price. The host eliminates one of the contestants with the empty box. The price is now within the other two contestants. In this instance, they still have equal chances to be holding the prize and that cannot be changed by any external observer’s personal gamble.
In probability, regardless if you have 2 or 1000 choices, every choice is equal. By eliminating one of the three doors, while you are making it more probable to get the right door, it is still equally probable to get it right, regardless on which of the two doors you will pick.
[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 12:36, 13 December 2011 (UTC)
:The MHP is different than your scenario with three players. In your scenario, the two remaining players are indeed equivalent but any one of the players ends up a remaining player only 2/3 of the time. For example, look at the player who holds box 1 over 300 trials of this game. In (about) 100 trials this player's box will have the prize and in these 100 trials this player must be one of the remaining two. In (about) 200 trials this player's box will be one of the two that does not have the prize and - everything else being equal - this player will be eliminated (about) 100 times. The player is in the game (at the end) only 200 times and in 100 of these this player's box has the prize. The analysis holds for any player, so any player in the game at the end has a 50% of winning (their box has the prize 100 out of 200 times).
:In the MHP, the one and only one player is ''always'' in the game at the end. To make this analogous to your 3-player game, one player would be designated "off limits" to the host and one of the other two would be eliminated. Think about 300 trials of the MHP (where the player has picked door 1). Please answer the following questions:
:1) In how many of these 300 trials will the car be behind door 1?
:2) In how many of these 300 trials will the host open door 3?
:3) In how many of these 300 trials will the host open door 2?
:4) Thinking about only the trials where the host opens door 3, in how many of these is the car behind each door (1, 2, and 3)?
:5) Thinking about only the trials where the host opens door 2, in how many of these is the car behind each door (1, 2, and 3)?
:Hints: The sum of the answers to #2 and #3 should be 300. The sum of the times the car is behind each door in #4 and #5 should be 100, and the car is behind door 3 zero times if the host opens door 3. "Everything else being equal" means that the door 1 answer for #4 and #5 is the same (if the car is behind door 1 the host must open one of the other two doors, picking which one at random).
:At the final stage of the MHP there are definitely two doors. And if you pick randomly between them you'll have a 50% chance of winning. However, this doesn't mean that the chances the prize is behind the two remaining doors is the same. I could (for example) roll a 10-sided die and put the car behind door 1 if the die comes up 7 and behind door 2 otherwise, and then give you the choice between these two doors. There are only two doors, but the car has a 10% chance of being behind door 1 and a 90% chance of being behind door 2. The MHP is very similar. Two remaining doors, but unequal chances. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:22, 13 December 2011 (UTC)
::My problem with your theses is that you still calculate the variables from the first choice into the second, which for me it appears to be false.
::The idea of looking at the problem as those three doors are the players is a way to point out that every time, there is equally chances between the available doors/players to have the prize. The variables are exactly the same; it is the same problem with the only difference being the way of looking at the problem.
::My theory is that after the one change on the variables of the game (one door opens) we end up with a new game and therefore the variables from the first game (2/3) cannot be applied in the later (1/2): the maths will be wrong. To make it obvious, add another irrelevant variable to the game, i.e. the players sometimes arrive to the studio with a taxi and sometimes they arrive with their own car, or half the times they wear black and the other half they wear their lucky shoes. If you try to calculate in to the probabilities those variables, you end up with superstition, not real maths, because it is irrelevant their mode of transportation or what they wear with which door has the prize.
::The confusion is that in this problem the variable that changes is not something outside the studio, like the mode of transport, but something that appears to be in the game, the door. But the variable that changes is not the door itself; it is the available sum of the doors.
::Another way of looking at my theory is to consider a second player being introduced into the game after the door opens. Initially, player one has 1/3 chances. If we stop there, he has 1/3 chances to win. After the door opens and before the player makes the second choice, we bring another player and ask to pick one of the two remaining doors. If they both pick the same door, then according to the wiki theory, the first player will have 2/3 and the second players will have ½ chances on the same door. But their individual choices and past experience is irrelevant to the fact that there are two doors and one car, so ½ chances.
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 20:31, 13 December 2011 (UTC)
:Did you work through the questions I asked? What are your answers? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 20:46, 13 December 2011 (UTC)
:: "My problem with your theses is that you still calculate the variables from the first choice into the second, which for me it appears to be false." Therefore, I believe that your formula is wrong from the start, regardless the outcome. I question your variables, not the logic after you apply them.
:: Here is another diagram. In this one, we first calculate every available option, including the host revealing the door with the car (12 different scenarios), in an effort to separate the two games, the "first" game with 3 doors and the "second" game with 2. Then we exclude the options where the host reveals the car and we see that there are equal chances on every door.
::[http://4.bp.blogspot.com/-ATj11G1ofzo/TufOhkS0_II/AAAAAAAAAIA/DDImWCQxAwk/s320/MHP.png Diagram]
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 22:19, 13 December 2011 (UTC)
:The game does not start anew after the host opens a door - it continues from the initial start. Let's take this in steps. The rules are:
:1) the host hides the car behind one of three doors (picking which door to hide the car at random)
:2) the player picks a door (which remains unopened)
:3) knowing what's behind the doors the host opens one (not the door the player picked), always revealing a goat - and if the player's initially picked door is hiding the car the host randomly picks which other door to open
:To make sure we're on the same page here, do you agree that one iteration of the game involves all three of these steps and that the question is what is the probability the car is behind the initially picked door as opposed to the other (unpicked, unopened) door after following these steps - for example, if you pick door 1 and the host opens door 3 we're interested in the probability the car is behind door 1 and the probability it is behind door 2 (having followed the steps above)? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 00:38, 14 December 2011 (UTC)
:: Yes, I understand that there are three stages and I propose that the probabilities change when the variables change, so when the door opens, the game/variables/probabilities are changing. Just because the problem is one sentence, it does not mean that all the proposition in that sentence are related to each other. When my hand throws the dice on the table and it brings a six and then I do it for the second time, that is my same hand throws the same dice on the same table, the probabilities are exactly the same to bring a six as it was the first time. No more or less.
:: I think the reason why the MHP is still under debate is not because some get it and others don’t. It is because the problem itself is not well defined, so one can make different assumptions from others, coming to a different conclusion.
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 10:04, 14 December 2011 (UTC)
:I agree the probabilities are changed by opening a door - for example the probability the car is behind the opened door is now (clearly) 0! Lets try another approach. I assume you agree the probabilities the car is behind each door after step 2, for a player who has initially picked door 1, are 1/3, 1/3, and 1/3. There are only two potential outcomes from this position - the host opens door 2 or the host opens door 3. So there are a total of six probabilities that might be interesting: the probability the car is behind door 1 AND the host opens door 2, the probability the car is behind door 1 AND the host opens door 3, the probability the car is behind door 2 AND the host opens door 2, etc. Lets call these P12, P13, P22, P23, P32, and P33. We know that the sum of these must be 1 (they cover all the combinations of where the car might be and what door the host opens), and we know (following step 2) P12+P13=1/3, P22+P23=13, and P32+P33=1/3. With me so far? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 15:56, 14 December 2011 (UTC)
:: Sure, 6 options, coupled together by where the car is = 1/3 each... and if there were 100 doors (or a large enough number for ones memory capacity), the 1/100 chance will be competing with the 99/100 collated chances if you choose to switch. I understand, no need to explain it. I have read the article and many others. The point I am making is that those calculations are assuming variables that are not true or have not a calculated connection. We calculate doors and cars along with human cognition on new information. For that one choice, the answer is predetermined, regardless of what you will do.
:: Maybe this paradox has been created because stats and probabilities have a meaning on events that have many occurrences, like going to the casino and playing the game many times. When there is only one occurrence, like when you can only be once in your life a contestant on the Monty Hall game and you can make that choice only ones, all (equal) options available have exactly the same weight. This would explain why people using common sense believe that it doesn’t matter if you change, while mathematicians, who look at repetitive applications of the paradox, think that it is better to switch.
:: [[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 20:58, 14 December 2011 (UTC)
:Ah, so you do understand where I'm going with this but you apparently don't believe it. You're essentially saying you don't believe in [[conditional probability]], since clearly where I'm going with this is
:1) P12+P13+P22+P23+P32+P33 = 1
:2) (regrouping by the door the host opens) (P12+P22+P32) + (P13+P23+P33) = 1
:3) (assuming it's equally likely the host opens door 2 or door 3) P13+P23+P33 = 1/2
:4) (the host can't open the door the car is behind) P22 = P33 = 0
:5) So, P13+P23 = 1/2
:6) (the probability the car is behind door 2 before the host opens door 3 is 1/3) P22+P23 = 1/3
:7) (from #4, P22 = 0) P23 = 1/3
:8) So, P13 + 1/3 = 1/2
:i.e. P13 = 1/6, meaning the probability the car is behind door 1 (given the host opens door 3) is exactly half the probability the car is behind door 2 (also given the host opens door 3). Expressed as conditional probabilities, this means after the host opens door 3 the probability the car is behind door 1 is (1/6) / (1/2) = 1/3 and the probability the car is behind door 2 is (1/3) / (1/2) = 2/3.
:If you're not going to believe this (which has the form of a formal proof using extremely elementary probability theory and simple math) I'm not sure where we can go with this discussion. Do you understand the playing card analogy - i.e. use the ace of spades to represent the car and two other cards (say, the red twos), shuffle, draw one, and looking at the remaining two discard one that is not the ace? If so, do you agree the 52-card version (shuffle, draw one, look at the remaining 51 and discard 50 that are not the ace) is equivalent? In both of these, after the action of discarding there are only two cards left. Are the chances 50/50 in both of these? If you truly believe this, I strongly encourage you to actually try it (say, 10 times) and let us know how it works out. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 05:12, 15 December 2011 (UTC)
:: Yep, Rick, I do understand your logic, as I stated above. I have read the article and many others. Have you got any comments on my last paragraph, where I say that the MHP "paradox has been created because stats and probabilities have a meaning on events that have many occurrences, like going to the casino and playing the game many times. When there is only one occurrence, like when you can only be once in your life a contestant on the Monty Hall game and you can make that choice only ones, all (equal) options available have exactly the same weight. This would explain why people using common sense believe that it doesn’t matter if you change, while mathematicians, who look at repetitive applications of the paradox, think that it is better to switch"
:: I haven't seen anyone not acknowledged the maths. I have seen many with different points of view and different calculations. My last assumption is that stats and probabilities are irrelevant on a single application. They only have meaning if they are applied multiple times, which is not what the MHP is about. What do you think?
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 07:58, 15 December 2011 (UTC)
:The difference between [[frequency probability]] and [[Bayesian probability]] is what you're talking about. In a pure sense, a single, non-repeated (or non-repeatable) event cannot be reasoned about with frequency probability - for such an event a frequentist can't say anything about the odds, because by naming any odds what you're saying is that if this event were repeated the proportion of various outcomes will (in the limit as the number of repetitions approaches infinity) have a particular value. Indeed, by asking a question whose answer is presumably based on the probabilities of various outcomes the ability for any frequentist to answer at all presumes repeatability. The only "correct" answer for a strict frequentist who is assuming the MHP is a non-repeatable, one-time only offer is "My notion of probability has no relevance to this question, so I can't say whether you should switch or not". Note that this is completely different from saying it's a 50/50 choice because there are two doors left (saying it's 50/50 is saying "if this were repeated over and over, half the time the car will be behind door 1 and half the time the car will be behind door 2"). However, this is not a view of the MHP I've seen published anywhere (I've read much of the available literature).
:On the other hand, Bayesian probability allows reasoning about probabilities of single events based on pure logic. As it turns out, these two seemingly wildly different interpretations of probability are much closer than you might imagine. In particular, the rules for computing conditional probabilities (for repeatable sequences of events for frequentists, and using Bayesian logic) are identical. In a sense, a Bayesian analysis provides a "prediction" that will be born out by experiment - i.e. if your understanding of the rules governing a particular situation is correct the Bayesian prediction will match the experimentally observed (frequentist) probability. Regarding the MHP, vos Savant's nationwide experiment is an example of this. In response to persistent claims that switching does not matter in the face of a logical explanation, she described an experiment (similar to the three card version) which many school classrooms did, which showed switching wins twice as often as staying with the original choice. Perhaps unfortunately, neither her explanation or experiment address the specific case the question asks about (player who has picked door 1 and has then seen the host open door 3) - but many people completely overlook this (by assuming the results must be the same for either door the host opens).
:Is this last point perhaps the actual source of your concern, i.e. many solutions do not address the situation ''after'' the host opens a particular door? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:13, 15 December 2011 (UTC)
:: Thanks for the links. I was not aware of those theories; I will have a look at them a bit more.
::It seems to me that there is a reason why this is a contested issue. The arguments and the variable as not well defined, too many assumptions and that is why not all agree. After all, every game and riddle is a construction of illusion in order to make the answer unclear and thus fun to solve. Is it possible that the reason for this is that MHP appears, not in a philosophical paradox form, but as a well visualised practical situation, a TV quiz, where people perceive it as a one off scenario? Are people presented with a situation that is an example of the frequency probability theory which makes everyone become a frequentist? Nowhere is MHP does it say that the problem will be repeated, that you will have the option to choose again from the start.
:: Of course, this takes away all the fun, but strictly speaking, it could be correct. After all, the MHP was not proposing that you are in a casino playing this game multiple times, what will your strategy be for the night? It explicitly (and intentionally confusingly) describes a one-off problem. The answer to the MHP could be "I can't say whether you should switch or not", so .. whatever!! It doesn't matter if you switch or not, which is same as 50-50. I don’t think this is an extreme theory. This is how dice and roulette works: The shape of the dice and the way our hand throws it is so random that makes it practically a one-off event. Same with the ball on the roulette. That is why stats on those two games are more superstition than anything else.
::[[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 20:29, 15 December 2011 (UTC)
:Yes, in its usual form the MHP is not well defined. However, most people who read it assume the same things making it well defined and repeatable (see the Krauss & Wang paper referenced in the article). In particular, the "game show" context provides a strong implication of repeatability and most people assume (even if these aren't explicitly stated)
:1) the initial ___location of the car is (uniformly) random
:2) the host always opens a door showing a goat (never opens the player's door)
:3) if the host has a choice between two doors to open (the player happened to initially select the car), the host chooses which door to open (uniformly) randomly
:4) the host always makes the offer to switch (the offer is not made more or less often depending on whether the player initially selected the car)
:With these assumptions (what the article refers to as the "standard problem") the game is immanently repeatable resulting in a probability of winning by switching of 2/3 regardless of whether you're talking about the overall outcome across all players (regardless of which door they initially picked and which door the host opens) or you're talking about the probability in a specific case (such as player picks door 1 and host opens door 3), and regardless of whether you choose to interpret probability in the frequentist or Bayesian sense. The psychologists who have studied this kind of problem (for example, Ruma Falk or Fox and Levav) say the issue people have with it is not that it is ill-defined but that the result goes against a very strongly held probabilistic intuition (two unknowns, therefor the probability must be 50/50). Piatelli-Palmarini's quotes are very apt: "no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." They're not misunderstanding it because it's sloppily or ambiguously presented - they're simply getting the wrong answer. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 00:24, 16 December 2011 (UTC)
:: Well, These are the assumptions I was talking about. We assume that most people who read it assume the same things making and you said that the "game show" context provides a strong implication of repeatability. Maybe those who don't assume these are those who do not agree with these answers. In any way, shouldn't this be stated in the article?
:: In “Statistical methods in experimental physics”, by W. T. Eadie, Frederick James, page 11, it states that: “ Frequentists probability can only be applied to repeatable experiments. This means, for example, that one cannot define the frequentist probability that it will rain tomorrow, since tomorrow will only happen once and other days are no identical to tomorrow.”
:: Also, professor of Computer Science at Olin College, Allen Downey, analysing a variation of the problem, he said “We can't reject the null hypothesis, so if we play by the rules of conventional hypothesis testing, I guess that means we can't take advantage of Monty's tell. If you are a committed frequentist, you should stop reading now.” http://allendowney.blogspot.com/2011/10/blinky-monty-problem.html
:: This should be enough doubt to grant a comment in the MHP wiki page that in order for the problem to have a meaning, we should assume that this is not one off, and that the contestant, unlike real life, can have multiple goes. That could ease many doubters, like me.
:: [[Special:Contributions/121.44.184.159|121.44.184.159]] ([[User talk:121.44.184.159|talk]]) 09:41, 16 December 2011 (UTC)
:That most people make the standard assumptions is what the reliable sources that discuss this say (i.e. it is not "our" assumption). See, in particular, the papers by Falk, or Fox and Levav. Mueser and Granberg explicitly tested responses to different versions of the problem and found no statistically significant difference between versions where 1) the "standard" assumptions were explicit, 2) no assumptions were provided (like vos Savant's version), or 3) the host explicitly opens a random door, and only happens to reveal a goat (in which case the probability is 50/50 for the remaining two doors). Anyone can change the article however they'd like, but if you want to make a change suggesting that there are a significant number of people who get the wrong answer because they're strict frequentists and view the problem as a non-repeatable event I'd encourage you to find a reliable source that says this (as I read it, Downey's blog is not saying this). -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:11, 16 December 2011 (UTC)
::To 121.44.184.159: First of all, this is not "Let's make a deal", so nothing is known about tapes of previous shows that - if available - could eventually give additional hints on the actual constellation, resp. on the actual ___location of the car behind those three doors. And nothing is said about this actual show to already having been repeated or will ever be repeated. You can take it as a singular show if you like, or to be repeated in future, if you like.<p>But just think of this one variant: '''There are one hundred doors''', with just only one single car behind anyone of those one hundred doors. Just one car. Each door with equal chance of 1/100. And the contestant may choose just one door. Only one. Then the host asks the contestant "do you want to stick to your one chosen door and open your only one door, or would you like to swap to the residual 99 doors, you may open the rest of all of them at once and they all will be yours, if you like." Irrespective of the host having opened one or more empty doors, or not having opened any door in this variant, or having opened none or one door in the original MHP, the question is: Should the contestant stick to her '''one''' chosen door or should she prefer to take all the residual doors altogether, instead? Hard to believe, but I am sure there will be some saying that there is no difference, and it doesn't matter whether only 1 door – out of 100 – or 99 doors – out of 100 – have been chosen. Or in the original MHP one door or two doors have been chosen.<p>Please consider carefully that basic interpretation of the [http://en.citizendium.org/wiki/Monty_Hall_problem#Explicit_computations Monty Hall problem] in Citizendium, do you feel that this concise report on odds of 2/3 : 1/3 eventually could help to guess and to recognize the underlying basics of the MHP? Regards, [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 20:32, 16 December 2011 (UTC)
:::: It is an one off for the contestant of the show: Monty is not picking doors.
:::::121.44.184.159: apart from eventually expected "additional hints", for the basic proportion of "99 : 1" in the aforementioned variant and for the proportion of "2/3 : 1/3" in the MHP, it does not matter whether the host has shown a goat or not. Read it again, please. [[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 14:12, 17 December 2011 (UTC)
:: Rick, you are seriously thinking I can find a reliable source that says "that there are a significant number of people who get the wrong answer because they're strict frequentists and view the problem as a non-repeatable event"?. I wonder if you can find the opposite. Anyway, the souce from the "reliable" Eadie is here: http://books.google.com.au/books?id=QbBm2VhV5TQC&lpg=PA11&dq=Frequentist%20probability%20can%20only%20be%20applied&pg=PA11#v=onepage&q=Frequentist%20probability%20can%20only%20be%20applied&f=false and I think his claim is applicable to the MHP.
:: After 5500+ words on our discussion, I would have thought that we had come to an understanding that there is a position/opinion/variation which does not appear in the article, a frequentists view on this problem that practically will occur only once, but for argument sake we assume that it will happen multiple times.
:: You appear more knowledgeable than me and as you have assume the position of the protector of this article, maybe that gives you the power to say NO to any change that is not your cup of tea, but maybe it also gives you the responsibility to spend some time searching on an extra +5500 words to find a source that would be good enough to grant a mention in the original article of a position that you yourself brought up. I am just passing by and by no means do I want to spend more time on “the ongoing farce that Monty Hall problem has become”. I think W. T. Eadie's source is good enough, apparently it is not.
:: Nevertheless, thank you for helping me prove (to myself) that there is a problem with Monty.
:: [[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 20:51, 16 December 2011 (UTC)
:::“There is no doubt that with respect to a large enough sample of Monty Hall games the player should switch. But what if we look at a single game (cf. Moser and Mulder 1994; Horgan 1995)? I want to argue that we run into serious problems if we apply probabilistic notions and arguments like the one above to a single Monty Hall game. The application of such notions and arguments to a single case (a single game) does not make sense; hence, there is no answer to the question what the rational player should do in an isolated case, at least no probabilistic answer.”
::: Peter Baumann: Single-case probabilities and the case of Monty Hall: Levy’s view: SYNTHESE, Volume 162, Number 2, 265-273,
::: http://www.springerlink.com/content/652828555l629711/fulltext.pdf
::: [[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 21:22, 16 December 2011 (UTC)
::::Paul K Moser and D Hudson Mulder of Loyola University of Chicago, PROBABILITY IN RATIONAL DECISION-MAKING : Philosophical Papers Vol. XXIII (1994). No. 2
::::Page 112 : “ Why should we hold that the results of switching over a hypothesized long run of trials automatically bear, in any relevant way, on the advisability of switching in an isolated single case, or even in an isolated short run of consecutive trials? The view that certain hypothesized long-run results bear on an isolated single case needs explanation and argument, as that view is not self-evident. Typical results in a hypothesized long run are not automatically matched by results in an isolated single case or even in an isolated short run of consecutive trials. A typical long-run result does not influence the propensity of outcome in an isolated single case or in an isolated short run of trials. In particular, a typical long-run result has no causal influence on a single case, although sometimes long-run results reflect the causal powers at work in a single case.”
::::Page 126: “The rational advisability of switching in an appropriate long run of Monty Hall games depends on a presumable statistical correlation that will hold only in runs of games where the rate of success on the initial choice is about 1/3. This statistical correlation is irrelevant to an isolated individual case, because it makes no sense to talk of a rate of success x, where 0<x<1 , for the initial choice in an isolated individual case. It is impossible for one to have a 1/3 rate of success on the initial choice in an individual play of the game; one’s rate of success here can be only 1 or 0. The rational advisability of switching depends essentially, however, on a presumable 1/3 rate of success for the initial choice. Such a rate of success is not only impossible in an individual case but also highly unlikely in a short run of cases. It becomes progressively more likely only as the length of the run gets progressively greater (and where there is no decisive contravening evidence, of course). Switching, then, is not rationally advisable in an isolated single case, and it becomes only gradually more advisable, under certain conditions, as the length of the run of games one considers increases. If one is offered 10 games, switching is less strongly rationally advisable than if one is offered 100 games. If one is offered only one game (as one is in the Monty Hall game of section I), switching is not at all advisable, given the bonus gained by staying.”
:::::Kahneman and Tversky 1972, pp. 434-36 : “People expect that a sequence of events generated by a random process will represent the essential characteristics of that process even when the sequence is short . . . Thus, people expect that the essential characteristics of the process will be represented, not only globally in the entire sequence, but also locally in each of its parts. A locally representative sequence, however, deviates systematically from chance expectation: it contains too many alternations and too few runs. “
:::::[[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 21:50, 16 December 2011 (UTC)
:: And there is where my argument stands. MHP is not presented as a PHD thesis in statistics, but as a simple, visual problem that anyone with a TV can relate. MHP is placing us exactly in a one case scenario, a “what would you do if you were so lucky to appear in a game show” ? Everyone thinks that the prize could be behind any door, so it does not matter if you change or not. And the logic makes sense because everyone’s head is focusing in this one case scenario. MHP has no suggestions for multiple applications; it a question for YOU to answer for this one event. For the sake of the argument we should indeed take it to another place, where multiple events take place, like a casino etc. But wiki should point this out and should have a paragraph for this theory, the one off event. It is not that weird or too unique: Dice and roulettes are exactly the same, they are one off events.
::[[Special:Contributions/121.44.231.208|121.44.231.208]] ([[User talk:121.44.231.208|talk]]) 22:07, 16 December 2011 (UTC)
:First of all I am in no sense the "protector of this article". I said above "Anyone can change the article however they'd like". Not only did I actually mean this, but it's literally true as well. If you want to change the article, do it. If you want to propose a change, propose it on the [[Talk:Monty Hall problem|article's talk page]] (not here). Second, please look at the info box at the top of this page. This page is for discussing mathematical aspects of the MHP, not for suggesting or discussing changes to the article. I thought we've been discussing the math behind the problem. You seemed (still seem) to be somewhat confused. I've been trying to help. If your stance is it's a 50/50 choice and you're going to ignore anything anyone has to say that contradicts this, that's certainly your choice. However, I can assure you there is absolutely no academic controversy about this - including the references you've quoted above (which are not saying that it's a 50/50 choice). What these sources are saying (and, by Wikipedia's standards they'd be considered [[wp:fringe|fringe]]) is that if the MHP is viewed as a singular event the (frequentist) probability of winning is indeterminate - not 50/50, not 1/3:2/3, not anything. I'm somewhat surprised [[User:Gill110951]] hasn't commented on this thread yet. He's a professor of mathematics with an interest in the MHP. If you're finding this conversation with me to be not to your liking I'm sure he'd be willing to talk about it with you (on his talk page at [[User talk:Gill110951]]). In any event, I apologize if I haven't been helpful. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 08:45, 17 December 2011 (UTC)
:: Hi, sorry I have not been watching these discussions. I have been taking a break from MHP and got deeply into the [[Two envelopes problem]] instead. But sure I am still very interested in good old Monty Hall. <p> I see that the discussion is about frequentist versus Bayesian approaches to MHP. I agree that this is an interesting issue. I believe that the typical assumptions of a Bayesian and of a frequentist for MHP would be different. The meaning of probability is different, so this is no surprise. Because their assumptions are different, their solutions are different too. They both decide to switch, but for different reasons. I have written about this issue in a published article which you can also find on my university home page, together with some other MHP writings; the link is http://www.math.leidenuniv.nl/~gill/#MHP. <p> And for what it's worth, here's my conclusion. The frequentist will realise that we know nothing about how the prize is hidden, nor how the host chooses a door to open when he has a choice. So he will take fate into his own hands by choosing his own door, in advance, uniformly at random, and then proceed to switch, whatever door is opened by the host. For him, the chance that he'll get the car, given his initial choice and given the door opened by the host, is at best unknown, at worst undefined. But he doesn't care. He does know that he has a 2/3 unconditional chance of getting the car, since his initial choice is correct exactly 1/3 of the times; he also knows that one cannot do better (by the minimax theorem from game theory). This solution is the unique minimax solution: it minimizes (by the player's strategy of combined door choices) the maximal unconditional chance of not getting the car (maximal with respect to possible car-hiding and door-opening strategies of the host). <p> On the other hand, for the subjectivist, all doors are initially equally likely to hide the car since he has no information on whether to prefer one door to another. Similarly, for him the host is equally likely to open either door when he had a choice, because he (the subjectivist player) has equal reason to believe a possible bias of the host would work in one direction or the other. Consequently the subjectivist will initially choose door 1, say, because 1 is his lucky number, and switch because after the host has opened, say, door 3, the other door has a 2/3 chance of hiding the car. The reasoning behind 2/3 is because initially door 1 has chance 1/3 of hiding the car. The host is going to show a goat behind another door anyway, so when he does this, the chance that door 1 hides the car doesn't change. Finally, *which* door is opened by the host does not give any new information to the player, by symmetry. So given the door chosen by the player and the door opened by the host, the subjectivist gives 2/3 chance to the car being behind the other door, and switches. <p> Two meanings of probability, two quite different solutions, both perfectly acceptable I think, and all this is written in my paper if anyone finds it notable enough to use in the article. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 17:28, 6 January 2012 (UTC)
===The doubting continues===
Initially there are 3 possible outcomes. Each door has a 1/3 likelihood of having the car.
Once a door has been opened, there are only 2 possible outcomes.
The car is behind door 1 or the car is behind door 2. The odds of the car being behind 1 or 2 were the same at the beginning, 1/3 each. That relationship hasn't changed because door 3 had a goat. The odds are now 1/2 for each of doors 1 and 2. Switching does not make a difference.
The vos Savant chart is flawed. Once door 3 has been opened, there are only 2 possible outcomes. Outcome A has the car behind door 1 and outcome B has the car behind door 2. Door 3 is no longer relevant. The odds of the car being behind either #1 or #2 have changed, from 1/3 to 1/2, but the relationship of the relative likelihood is still the same as it was at the outset.
Similarly, if door 2 is opened, there are only 2 possible outcomes -- car behind #1 or car behind #3. [[User:Kwinzenried|Kwinzenried]] ([[User talk:Kwinzenried|talk]]) 01:04, 7 January 2012 (UTC)
:Are you interested in talking about this, or are you absolutely convinced you're right and nothing anyone says can change your mind? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 01:56, 7 January 2012 (UTC)
This is an interesting problem. I am certainly open to obtaining a clear presentation of the correct answer, whatever that might be. Per my comments above, I don't see that switching makes any difference. Additionally, the problem of symmetry hasn't been dismissed if switching is anyways the preferred answer -- if switching were always preferable, then the initial choice must matter, but since the initial choice is random, symmetry can't apply if switching is the best answer. <small><span class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Kwinzenried|Kwinzenried]] ([[User talk:Kwinzenried|talk]] • [[Special:Contributions/Kwinzenried|contribs]]) 03:31, 7 January 2012 (UTC)</span></small><!-- Template:Unsigned --> <!--Autosigned by SineBot-->
:OK. First - I'll agree with you that vos Savant's solution is flawed. It's definitely part of the story - but she changes the question from "is it better to decide to switch (having picked door 1) ''after'' seeing the host open (say) door 3" to "is it better to decide to switch (having picked door 1) ''before'' seeing which door the host opens". Do you see these as two different questions, and do you agree vos Savant's solution addresses the second one? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 04:34, 7 January 2012 (UTC)
:: I do not agree that vos Savant's solution is flawed, but that's another matter. <p> Anyway let's agree on what we mean by probability, so we know what we are talking about, and what we can assume, and what we can't assume. Let's take probability to have its subjective meaning. Let's agree that a priori, as far as we are concerned, the car is equally likely to be behind any of the three doors. Let's also agree that whether or not the host has any bias in opening doors when he has a choice, for us the door numbers are mere labels, completely uninformative. So in particular, if we happened to choose the door hiding the car, so that the host had a choice, we find it equally likely that he chooses either door. OK so far? <p>
:: Now here follows a solution which ends with the 2/3 '''subjective''' probability of the car being behind the other door, '''given''' our choice (Door 1), and given the door opened by the host (Door 3). <p> Go back to the moment when we have chosen Door 1, but the host hasn't opened a door yet. At that moment we believe the chance is 1 in 3 that the car is behind our door. We know that the host is going to open Door 2 or Door 3 and show us a goat. Keep your eyes shut, let him open Door 2 or Door 3. We hear a goat bleating, but we don't know yet which door was opened. Nothing has changed. This was going to happen anyway. The chance is still 1 in 3 that the car is behind our door, Door 1. Finally, we open our eyes. We see that Door 3 is the one which has been opened. '''Nothing changes regarding Door 1'''. Because, '''whether or not''' the car was behind Door 1, '''we considered it equally likely, that Door 2 or Door 3 would be opened''', by symmetry. <p> Consequently, the probability the car is behind Door 1 still remains unchanged at 1 in 3. But we do know now that the car certainly isn't behind Door 3. So we would bet 2 against 1 that the car is behind Door 2, we will accept the offer to switch. <p> If you want to go through this argument with full mathematical details, I recommend you use Bayes' rule for updating probabilities on obtaining new information in the form: posterior odds equals prior odds time likelihood ratio; where "likelihood ratio" means the ratio of the probabilities of the new information, under each of the two alternatives considered. Take as alternatives: Car is behind Door 1, Car is not behind Door 1. Given that we initially pick Door 1, the initial odds that the car is behind Door 1 are 2:1 against. In either situation, the host is equally likely to open Door 2 or Door 3, by the symmetry of our prior beliefs. No need to do any calculations. So the likelihood ratio, for and against the hypothesis "car is behind Door 1", coming from the information "Door 3 got opened" is 1:1. So the posterior odds remain 2:1 against. <p> Vos Savant's argument can be considered to be completely correct. In essence she is ignoring the specific door numbers from the start, by symmetry. This is a completely intuitive step of many people thinking about the problem, and is also suggested by the problem statement's use of the words '''say''' Door 1, '''say''' Door 3. The words are used to help us visualise the problem, but also to trick us into '''seeing''' the problem at the stage after we have chosen a specific door and a specific door has been opened, forgetting the procedure whereby one particular door got opened, which does depend in fact both on our own choice and on the ___location of the car. The host does not have a free choice, and the probabilities of what he does, depends on the situation at that moment. We need to keep in mind the whole procedure. We are mislead into discarding the relevant information ('''how''' we got to this situation), and remembering only the final situation. And then we intuitively come to the wrong decision. The right information to discard are the actual numbers written on the doors in the case at hand! <p> Here is another, rigorous, complete, solution, on the lines of vos Savant: Because the numbers are completely arbitrary, one can discard them in advance. The only relevant things are the '''functions''' or '''roles''' of the doors. Forget the numbers painted on them! There are three doors: a door chosen by us, a door opened by the host, and a door remaining closed. One of the three hides a car. The question is, what are the probabilities that the car is behind each of these three doors. '''These''' subjective probabilities are determined in advance, and they never change, till the very end of the game. The probabilities are 1/3, 0, 2/3. That's obvious, isn't it? From this point of view Vos Savant's solution is completely correct, and there are reliable authorities enough who support this point of view. Forget the door numbers, they are irrelevant! <p> MHP is a brain teaser, a trick, a joke, a surprise; the formulation is cleverly chosen so as to entice you into the wrong line of thought. Just like a good joke. One needs to step back, rethink, look at the whole problem from a new point of view, and only then will intuition give us the right answer, and you see the joke. As the Joker said in the film "[[The Dark Knight (film)|The Dark Knight]]", "not so serious!". The good intuition is: '''the door numbers are irrelevant. The initial choice has 1/3 chance to hit the car. Hence switching gives the car with probability 2/3. End of story''' (at least, that's my opinion, but also the opinion of quite a few reliable sources, including many professional mathematicians). See http://www.math.leidenuniv.nl/~gill#MHP for further contributions by me. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 09:44, 7 January 2012 (UTC)
I said I could change my mind and I have.
The misdirection in the wording of the problem (and several of the explanations of the answer) is the focus on the participant rather than the host. The host captures the benefit of the 2/3 odds of having the car in the first stage since he gets two of the three choices. Once he has seen the first stage outcome, he always discards a loser and then, in the 2 of 3 first stage outcomes where he has the winner, he keeps it. So in 2 out of 3 first stage outcomes, the host has kept the winner. The participant, not knowing the first stage outcome, can elect to swap his 1 in 3 chance of having won the first stage for the host's 2 in 3 chance of having the first stage winner.[[User:Kwinzenried|Kwinzenried]] ([[User talk:Kwinzenried|talk]]) 16:19, 7 January 2012 (UTC)
: That's right. Under the minimal assumption that the player's first choice is right 1 times in 3, he should always switch: that way he gets the car 2 times out of 3. Moreover, if *each* door has a 1 in 3 chance of hiding the car, the player can't do better. That is because whatever strategy he uses (two stages of door choices) there is always a prize ___location for which his strategy will lead him to lose. <p> One does not have to do conditional probability calculations in order to solve MHP adequately. Careful considerations of strategy do the job, too. [[User:Gill110951|Richard Gill]] ([[User talk:Gill110951|talk]]) 17:12, 7 January 2012 (UTC)
:Yes, you can change your focus from the situation ''after'' the host has opened a door (and you can see which door the host has opened) to the situation ''before'' the host opens a door (or, as Richard does above, to the situation ''after'' but keeping your eyes closed so you don't know yet which door the host has opened) - this is indeed what vos Savant does. If the show is on, say, 300 times and the players all picked door 1, if all of these players keep their original choice about 1/3 will win the car and if all of these players switch (to whichever of door 2 or door 3 the host does not open) 2/3 will win the car. The flaw is that this doesn't actually answer the question about a player who sees the host open door 3 without the additional argument that the cases where the host opens door 2 or door 3 must be the same. This argument is where the trick actually is! You can choose to completely gloss over this and beat people on the head to change their focus, but it's completely unnecessary to do this. Just fill in the blanks in the following (imagining 300 shows, i.e. each group of answers sums to 300):
:1) ___ Number of times the car is behind door 1
:2) ___ Number of times the car is behind door 2
:3) ___ Number of times the car is behind door 3
:4) ___ Number of times the host opens door 2
:5) ___ Number of times the host opens door 3
:6) ___ Number of times the car is behind door 1 when the host opens door 2
:7) ___ Number of times the car is behind door 1 when the host opens door 3
:8) ___ Number of times the car is behind door 2 when the host opens door 3
:9) ___ Number of times the car is behind door 3 when the host opens door 2
:The answer for #5 must be the sum of #7 and #8 (if the host opens door 3 we know the car must be behind door 1 or door 2), and the answer for #1 must be the sum of #6 and #7 (if the car is behind door 1 the host must open either door 2 or door 3). With a little bit of thought, it's quite obvious the answer to #2 and #9 (and #3 and #8) must be the same (any time the car is behind door 2 the host must open door 3, and vice versa).
:From these answers we can then look at the proportion of times the car is behind door 2 when the host opens door 3 (#8 out of #5) and the proportion of times the car is behind door 1 when the host opens door 3 (#7 out of #5).
:There is no trickery involved whatsoever. The problem is people leap to the wrong conclusion based on faulty intuition. As Carton (referenced in the article) says: ''We often cringe when our students, or members of the general public, make rudimentary mistakes in probability. But even qualified scientists and mathematicians can make such mistakes, although theirs are sometimes less obvious. The misapplication of conditional probability, the haphazard use of “equally likely” outcomes, and the non-use of Bayes’ Rule can lead to all manner of incorrect answers.''
:The point is that people's intuition is what screws them up here. Appealing to a different intuition, even one that happens to produce the correct answer in this case, simply isn't helpful. A very good test of this is to ask the followup version - is it better to switch if the host blindly opens door 3 (without knowing where the car is) and happens to reveal a goat? Fill out the same table above. Think very carefully about the answers to #8 and #9. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 18:01, 7 January 2012 (UTC)
== Where all the confusion stems from -- explained simply. ==
The reason some people are having problems with this is that they are failing to see the difference between a host who ''knows'' where the car is (and will never choose that door), and a host who ''doesn't know'' where the car is (and randomly chooses a door). The two situations are '''completely different''', and yield different probabilities. ''One extra note: the contestant must also be aware of what the host knows and how the host will choose.''[[User:Hellbound Hound|Hellbound Hound]] ([[User talk:Hellbound Hound|talk]]) 05:40, 15 January 2012 (UTC)
:I agree the host knowledge is crucial. However, if the question is "'''should''' the contestant switch?" (which it generally is), what the contestant does or does not know is irrelevant. Specifically, if the host knows what is behind the doors and deliberately shows a goat (and picks between the remaining two doors randomly if the contestant's initial selection is the car) whether the contestant is aware of what's going or not, the contestant '''should''' switch and has a 2/3 chance of winning by doing so. Conversely, if the host opens a door without knowing what's behind the doors and happens to reveal a goat, the contestant's chance of winning by switching is 1/2 (whether the contestant realizes it or not).
:BTW - your conjecture that people arrive at the wrong answer because they are confused about whether the host is acting randomly or with knowledge of what's behind the doors has been experimentally tested (see the paper by Mueser and Granberg cited in the article). They found no significant difference between versions of the problem where this is explicitly clarified vs. those where it's not (e.g. the standard "vos Savant" version). The conclusion here is that it's not a misunderstanding of whether the host is opening a door randomly that confuses people - it's something else. Perhaps the definitive paper on the psychology of the problem is by Ruma Falk (also cited in the article). -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 06:57, 15 January 2012 (UTC)
== An alternative problem ==
At school, my teacher was talking about this problem, and I thought about something... If the candidate first selects a door, then the show host opens a door, the candidate switches door, he will have 66% chance of winning. What if someone else comes in after the host opens, and chooses between the two doors that are not opened yet, and then the original candidate switches doors. They both pick one of the 2 doors, but the other person who came in after the first door was opened, without knowing the candidate's first choice, has 50% chance (he chooses between 2 doors) and the original candidate has 66% chance, because he switches doors. This also works when throwing a coin after the host closed a door, as this makes you choose randomly between the two doors, not considering the other door (works exactly the same as someone else coming in after the host opened a door).
If you want to test this alternative problem, I wrote a script that simulates the 2 situations of switching doors (guy 1) or randomly picking one of the 2 unopened doors (guy 2): http://jsfiddle.net/tszp7/3/show/ [[User:Joeytje50|Joeytje50]] ([[User talk:Joeytje50|talk]]) 10:40, 26 January 2012 (UTC)
:Exactly. Let's say the first player initially chooses door 1 and the host then opens door 3 (after initially randomly placing the car behind one of the 3 doors, knowing what's behind the doors, and choosing which door to open randomly if the player's initial choice happens to be the one hiding the car). What these two problems distinguish is that "probability" is a function of information. The host knows where the car is, so (for the host) the car has a 100% chance of being behind one of door 1 or door 2 and a 0% chance of being behind the other. Given the set up as described, the first player (and the audience) knows enough to conclude the probability the car is behind door 2 is 2/3 - so if the player switches to door 2 (and we repeat this over and over, with the player always switching) the player will win the car about 2/3 of the time. Lacking the information about the initial set up, the second player can only choose randomly between two doors. Even if there is other information available to others allowing them to determine the probability between the two doors is not the same, the probability of a random choice between two alternatives being correct is 1/2. This is easy enough to prove. Say the actual probability the car is behind door 2 is p (for example, we started with 2 doors and the host picks a random number between 0 and 1 and hides the car behind door 2 if this number is p or less). If it's behind door 2 with probability p, the probability it is behind door 1 is 1-p. If you pick randomly between these doors you have a 50% chance of picking either door. So, your total chance of picking the door hiding the car is (p x 50%) + (1-p) x 50%, which is (p+1-p) x 50%, which is 1/2. What this means is that if you pick randomly over and over, you'll win about 1/2 the time overall regardless of what anyone might know about the probability the car is behind either door - even, for example, if there are only two doors and the host ''always'' puts the car behind door 2.
:All three of these are simultaneously true:
:1) the car has a 100% chance of being behind one of door 1 or door 2 and a 0% chance of being behind the other (this is the probability from the host's perspective)
:2) the car has a 1/3 chance of being behind door 1 and a 2/3 chance of being behind door 2 (these are the probabilities given what the first player, and the audience, know)
:3) picking randomly between door 1 and door 2 you have a 50% chance of winning the car, but if you do this repeatedly the audience will see that you win the car 1/3 of the time if you happen to pick door 1 (the first player's original choice) and 2/3 of the time if you happen to pick door 2 (the door the first player could switch to).
:The intermediate condition in the MHP, where the player knows ''something'' about the ___location of the car, is a little unusual. People are accustomed to the notion of knowing nothing about a situation (like the second player), even a situation where there is a 3rd party who knows with certainty (I flip a coin and look at without showing you - even though I now know whether it's heads or tails with 100% certainty you know nothing). Evaluating probabilities where you know only ''something'' is the ___domain of [[conditional probability]]. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 20:23, 26 January 2012 (UTC)
== The Monty Hall Problem — solved visually. ==
===Legend:===
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:left;"
|-
! scope="row" style="background:#efefef;" | [W]
|Win (a car) if you choose this option.
|-
! scope="row" style="background:#efefef;" | [L]
|Lose (get a goat) if you choose this option.
|-
! scope="row" style="background:#efefef;" | [X]
|Option chosen.
|-
! scope="row" style="background:#efefef;" | [ ]
|Option not chosen.
|-
! scope="row" style="background:#efefef;" | [\]
|Not an option, because host of game show excludes it in stage 2.
|-
! scope="row" style="background:#efefef;" | <nowiki>:W</nowiki>
|Result of scenario is a Win for contestant.
|-
! scope="row" style="background:#efefef;" | <nowiki>:L</nowiki>
|Result of scenario is a Loss for contestant.
|}
Columns in tables below represent stages in the game show illustrating the Monty Hall problem. Stage 1, the contestant chooses an option (a door); Stage 2, the host excludes a remaining option which he knows is not the winning door; Stage 3, the contestant may or may not change their answer.
===Staying strategy:===
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:center;"
|-
! scope="col" style="background:#efefef;" | [W][L][L]
! scope="col" style="background:#efefef;" | [W][L][L]
! scope="col" style="background:#efefef;" | [W][L][L]
! scope="col" style="background:#efefef;" | :Result
|-
|[X][ ][ ]
|[X][\][ ]
|[X][\][ ]
|:W
|-
|[X][ ][ ]
|[X][ ][\]
|[X][ ][\]
|:W
|-
|[ ][X][ ]
|[ ][X][\]
|[ ][X][\]
|:L
|-
|[ ][ ][X]
|[ ][\][X]
|[ ][\][X]
|:L
|}
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:center;"
|-
! scope="col" style="background:#efefef;" | [L][W][L]
! scope="col" style="background:#efefef;" | [L][W][L]
! scope="col" style="background:#efefef;" | [L][W][L]
! scope="col" style="background:#efefef;" | :Result
|-
|[X][ ][ ]
|[X][ ][\]
|[X][ ][\]
|:L
|-
|[ ][X][ ]
|[\][X][ ]
|[\][X][ ]
|:W
|-
|[ ][X][ ]
|[ ][X][\]
|[ ][X][\]
|:W
|-
|[ ][ ][X]
|[\][ ][X]
|[\][ ][X]
|:L
|}
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:center;"
|-
! scope="col" style="background:#efefef;" | [L][L][W]
! scope="col" style="background:#efefef;" | [L][L][W]
! scope="col" style="background:#efefef;" | [L][L][W]
! scope="col" style="background:#efefef;" | :Result
|-
|[X][ ][ ]
|[X][\][ ]
|[X][\][ ]
|:L
|-
|[ ][X][ ]
|[\][X][ ]
|[\][X][ ]
|:L
|-
|[ ][ ][X]
|[ ][\][X]
|[ ][\][X]
|:W
|-
|[ ][ ][X]
|[\][ ][X]
|[\][ ][X]
|:W
|}
====Results:====
6 Wins / 6 Losses = 50% Success rate in scenarios using the staying strategy.
===Switching strategy:===
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:center;"
|-
! scope="col" style="background:#efefef;" | [W][L][L]
! scope="col" style="background:#efefef;" | [W][L][L]
! scope="col" style="background:#efefef;" | [W][L][L]
! scope="col" style="background:#efefef;" | :Result
|-
|[X][ ][ ]
|[X][\][ ]
|[ ][\][X]
|:L
|-
|[X][ ][ ]
|[X][ ][\]
|[ ][X][\]
|:L
|-
|[ ][X][ ]
|[ ][X][\]
|[X][ ][\]
|:W
|-
|[ ][ ][X]
|[ ][\][X]
|[X][\][ ]
|:W
|}
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:center;"
|-
! scope="col" style="background:#efefef;" | [L][W][L]
! scope="col" style="background:#efefef;" | [L][W][L]
! scope="col" style="background:#efefef;" | [L][W][L]
! scope="col" style="background:#efefef;" | :Result
|-
|[X][ ][ ]
|[X][ ][\]
|[ ][X][\]
|:W
|-
|[ ][X][ ]
|[ ][X][\]
|[X][ ][\]
|:L
|-
|[ ][X][ ]
|[\][X][ ]
|[\][ ][X]
|:L
|-
|[ ][ ][X]
|[\][ ][X]
|[\][X][ ]
|:W
|}
{|border="1" cellpadding="2" cellspacing="2" style="font-family:monospace;text-align:center;"
|-
! scope="col" style="background:#efefef;" | [L][L][W]
! scope="col" style="background:#efefef;" | [L][L][W]
! scope="col" style="background:#efefef;" | [L][L][W]
! scope="col" style="background:#efefef;" | :Result
|-
|[X][ ][ ]
|[X][ ][ ]
|[ ][\][X]
|:W
|-
|[ ][X][ ]
|[\][X][ ]
|[\][ ][X]
|:W
|-
|[ ][ ][X]
|[\][ ][X]
|[\][X][ ]
|:L
|-
|[ ][ ][X]
|[ ][\][X]
|[X][\][ ]
|:L
|}
====Results:====
6 Wins / 6 Losses = 50% Success rate in scenarios using the switching strategy.
===Notes:===
Scenarios where the <span style="font-family:monospace;">[W]</span> and the <span style="font-family:monospace;">[\]</span> align do not exist because the host would never reveal nor exclude the winning door as an option in stage 2.
Scenarios where the <span style="font-family:monospace;">[\]</span> and <span style="font-family:monospace;">[X]</span> overlap do not exist because the contestant cannot choose a door that has been revealed by the host in stage 2. ..Well cannot unless they want a goat, which is beyond the purpose of this exercise.
===Conclusion:===
It seems to me that the above exemplifies every scenario in the problem, yielding a 50% success rate. This supports what I originally thought of the problem intuitively. After the host eliminated one option, I viewed the situation as having a 50% probability since what happened in the first portion of the show seemed irrelevant. In any scenario of the first portion, the contestant chooses a door but doesn't learn anything about the chosen door and the host reveals a goat behind a different door instead of a winning car, leaving the contestant with two choices of doors. From what I can see, this happens every time and does not conventionally reveal any novel information, thus leaving you with nothing more than a choice between two unknown doors, to which you can apply a 50% probability of holding your desired prize.
===Comments:===
I have only today become aware of the Monty Hall problem and so this is just my initial reaction as an attempt at solving it. It could be all wrong, but so far it's what makes sense to me. Please think about it if you care to understand, and if you agree with the above conclusion, then please say so and perhaps do something to have the view more well represented on Wikipedia's article. Thanks.
--<span style="font-family:'Courier New',Courier,monospace;color:#000077;">— [[user_talk:sloth_monkey|sloth_monkey]]</span> 10:37, 13 August 2012 (UTC)
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