This is an archive of past discussions about Monty Hall problem. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.
Rick and Glopk, one hypothetical question, I'd like to know what you think: In case some remark would have been added to Marylin's answer or even to the well-known statement of the problem that has been published in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (as per Whitaker/vos Savant 1990): "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to switch your choice?" added with the following or a similar remark (for clarification that this was what she meant to be the basis for her answer): "Supposed/given the car is initially placed uniformly at random and the host chooses a goat door uniformly at random, if he has the choice from two goat-hiding doors, and he always offers to switch ... ' In case this clarification had originally been added by vos Savant, and if she had said: "No news, he always can open one goat-hiding door and it does not matter which one if he has the choice between two goats, because he will chose uniformly at random, so you should switch, yes it always will be wise to switch", would conditionalists nevertheless say "her answer is numerically right, but it does not reflect the conditional problem that asks for the probability P(win by switching AND player initially picks door Y AND host opens door Z)"? I would be glad if you could give your comment to this (my hypothetical) question. Thank you! Regards, Gerhardvalentin (talk) 12:21, 5 August 2010 (UTC)
Gerhard, I answer mu. I can read your sentences above (with the hypothetical addition) as a restatement in words of the "Mathematical formulation" section already in the article. In which case it is the "conditional" answer to the K&W statement + uniform initial placement. On the other hand, if you parse MVS's sentence as applying "regardless" of the choice of doors, I would say it is correct, but that it is a stronger statement than the original question asks for. I use the term "regardless" to imply a marginalization, so that MVS's statement would mean , where . So it boils down to our usual point of contention: the "simple" solutions can be formulated correctly, but in the sources they normally aren't. The requirement of complete symmetry is at their core, and it is not an obvious one, so omitting it is a logical mistake that makes their conclusions non sequiturs. And in any case the simple solutions do not answer directly the MHP question, because they compute a marginal probability, whereas the question asks for a conditional. glopk (talk) 15:08, 5 August 2010 (UTC)
I don't actually know whether anyone would have made a big deal out of it or not. As Glopk says the simple solution is still formally answering the wrong question. Since the heart of the paradox is whether there is a difference between P(win by switching AND player picks door 1) and P(win by switching AND player picks door 1 AND host opens door 3), it seems having the answer based only on P(win by switching AND player picks door 1) without saying anything at all about P(win by switching AND player picks door 1 AND host opens door 3) is missing the point.
Rick, you say that Falk is wrong?[1] Quite the contrary. — Assuming an unbiased host — the point is NOT whether there is a difference, but that there is NO difference between P(win by switching AND player picks door 1) and P(win by switching AND player picks door 1 AND host opens door 3). But admittedly this is only a sideshow and belongs to a separate section for people interested in maths, believing in a biased host outside of vos Savant's Monty Hall paradox. — Whereas the "real heart" of the MHP is that the "constant-ratio belief" of more than 90 % of people is the assumption that when the host has ruled out one door of three by opening it, the ratio of the probabilities of the two remaining closed doors should be the same as the ratio of their prior probabilities (before 1/3:1/3 and now 1/2:1/2). This constant-ratio belief, seeing no reason to switch, is obviously not valid since it leads to the wrong answer 1/2 : 1/2, whereas the correct answer (assuming an unbiased host) is 1/3 : 2/3. And even if the host (outside of vos Savant's Monty Hall paradox) is extremely biased, it would always be wise for the contestant to switch, and no maths has ever been able to change that. Gerhardvalentin (talk) 14:43, 14 August 2010 (UTC)
Gerhardvalentin, this would make an excellent topic to add to the mediation. It has been the subject of edit warring, it's interpreted in the article in a perversely incorrect manner, and you have a great reference & argument. Please consider adding this to the "Opposition's viewpoint" section on the Mediation page. Thanks. Glkanter (talk) 16:55, 14 August 2010 (UTC)
Thank you, Glkanter, will be going to try to be accepted, but I guess the strong belief that the question was not "Should the contestant stick or switch? (See Shaughnessy & Dick, 1991)", but clinging to answer completely different questions that some participants made their own, as "What's the exact Bayesien probability to win by switching, under some assumed telltale bias", ignoring vos Savants basis that the overall probability applies to each and any game if no additional info/hint about the actual current constellation and distribution of the objects is given by showing that there only can be ONE car, without telltale bias evidence, will be ongoing to prevail. Any treacherous "bias-hint" automatically means the simultaneous opening of more than only one door, presenting a new and closer hint and a new condition, then. We can tell about strange variants and add Bayes, but that belongs to a completely different and very clearly secreted topic. Regards, Gerhardvalentin (talk) 18:19, 14 August 2010 (UTC)
Well, that's what we're trying to highlight in the mediation. My portion calls out the insistence that every mention of simple solutions is accompanied by the nonsense you describe above. Martin calls out the insistence that only conditional solutions are true. 2 of those guys have latched onto 'it's not fair to put the conditional solution last among equals'. Your portion can highlight the bunk that people are confused by something other than the 50/50, that you just described. The arguments are clear, as is the outcome. Your contribution will help to ensure that all those reliably sourced contrivances are dealt with in the manner appropriate for a Wikipedia article, even after the mediation has ended. Glkanter (talk) 18:55, 14 August 2010 (UTC)
IMO, the basic reason the answer is counterintuitive is because the problem nearly forces you to think about the conditional case where there are two closed doors and one open door (and the host opening a door affects the two remaining closed doors unevenly). This is fundamentally conditional because the doors are distinguishable. It's possible to posit an unconditional version as an urn problem (I've described this before), which might be as counterintuitive (or might not). If I were an experimental psychologist looking to publish a paper (I'm not), I might actually run an experiment to see if an unconditional version of the problem is as difficult for people to solve correctly. My guess is it would be difficult - but not as difficult. -- Rick Block (talk) 15:56, 5 August 2010 (UTC)
Repudiate is too strong. He said in his second letter (you've quoted this) that "the basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random". This is what forces the conditional and unconditional answers to be the same. His original solution presents the unconditional answer. By acknowledging the assumptions that force these to be the same, he's acknowledging that the solution of interest is the conditional one. He then goes on to present a conditional solution as an "alternate". Seems pretty clear to me that he's asking the conditional question. -- Rick Block (talk) 03:20, 6 August 2010 (UTC)
So your answer is "No, I do not have a source that says Selvin has repudiated his simple solution to the MHP he posed. I have concluded that he did so based on my interpretation of his 2 letters." That's OR. And your whole argument rests on it. Your statement that the simple solutions, of which Selvin's is one, solve the wrong problem is illogical and unsourced, as far as Wikipedia is concerned. Glkanter (talk) 04:25, 6 August 2010 (UTC)
No, Selvin did not. The only published source that offers a change to it's original position is Morgan, as prompted by Martin's letter pointing out their math error. You refuse to acknowledge that they have admitted a mistake, yet claim Selvin has. Remarkable. Glkanter (talk) 20:36, 5 August 2010 (UTC)
All this talk of 'mathematically correct' solutions is nonsense. There are many ways to solve most maths problems. For the standard MHP it is perfectly correct to note that the answer to the conditional must, by symmetry, be equal to the answer to the unconditional problem and then proceed to solve the unconditional problem to get the answer to the conditional problem. This sort of thing is done in mathematics all the time. Martin Hogbin (talk) 18:20, 5 August 2010 (UTC)
Martin - this has been said so many times it seems rather pointless to repeat it. Again, for the 4,000th time, we're not talking about solving the conditional problem indirectly as you're suggesting and I don't think anyone here is saying that doing this would not be a valid approach. What we are talking about is presenting an unconditional solution, without noting the problem must be symmetrical and without saying anything about whether the conditional and unconditional answers must be the same - which is what nearly all sources presenting "simple" solutions do. -- Rick Block (talk) 18:50, 5 August 2010 (UTC)
Martin, we can solve some math problems quite well, thank you. On the other hand, you be careful touching upon mathematical correctness and nonsense - given that in less than a week you have authored two fine examples of nonsense on these very pages (see "simple solution solves the problem of calculating P(C=1|H=2 or H=3)" and "P(Ws| I) = P(Ws | I and S=s and H=h and C=c) for all legal values of h,s,and c"). What makes you think that you are qualified to interject in a conversation with Gerhard in the general area of probability theory? Please do not disrupt. glopk (talk) 20:14, 5 August 2010 (UTC)
A symmetry argument is OK?
Let me get this right Rick, you are saying that the argument that the answer to the conditional must, by symmetry, be equal to the answer to the unconditional problem and then proceed to solve the unconditional problem to get the answer to the conditional problem is perfectly correct and that no one here disagrees with it? (Note that I am talking here about the correctness of the approach, not whether we can put it in the article.)
What about the argument that, by symmetry, the door specific goat opened by the host cannot make any difference even in the conditional case? Martin Hogbin (talk) 19:47, 5 August 2010 (UTC)
Are you saying there's some confusion about this? Really? I think Nijdam has said this literally about 100 times. I've said it at least another 100 times. Kmhkmh has said it maybe 50 times. Glopk has said it probably 20 times. Coffee2theorems has said it about 5 times. Please tell me you're not just now understanding this because if it takes 5 different people telling you something a total 200 times before you understand it this is going to take a VERY long time. -- Rick Block (talk) 00:01, 6 August 2010 (UTC)
Maybe I have misunderstood you all. You are confirming that everyone here accepts that by symmetry, the specific goat door opened by the host cannot make any difference to the probability of winning even in the conditional case? Martin Hogbin (talk) 08:22, 6 August 2010 (UTC)
If you show the problem is symmetrical, i.e. that P(win by switching AND player picks door i AND host opens door j) is the same for all i,j, then all of these are the same as P(win by switching). If you show the problem is symmetrical with respect to which door the host opens after the player picks say door 1, i.e. P(win by switching AND player picks door 1 AND host opens door 2) = P(win by switching AND player picks door 1 AND host opens door 3), then these both must be the same as P(win by switching AND player picks door 1). How about if you paraphrase what you think I've been saying?
If it helps, I'll paraphrase what I think you've been saying. Many published solutions effectively say that since the average player has a 2/3 chance of initially selecting a goat and always ends up with a car if switching, every individual player has a 2/3 chance of winning by switching regardless of what door is initially picked and what door the host opens. Because this can be shown to be a true statement if the problem is fully symmetrical (and the host must open a door showing a goat), the article should primarily present this and similar solutions - as "fact" without any further explanation or comment, in particular without saying anything about symmetry or any justification whatsoever for shifting between talking about the initial average 2/3 chance to the chances faced by every individual player such as one given in the problem statement who has picked door 1 and is deciding whether to switch after seeing the host open door 3.-- Rick Block (talk) 14:23, 6 August 2010 (UTC)
You have nearly got one of the reasons that I support showing the simple sources without a health warning. My reasons are:
The reason you give above, bearing in mind that we define the problem to be fully symmetrical in the article, thus all that is missing is a simple, obvious, and intuitive symmetry argument.
It is far from clear that Whitaker actually wanted to know the answer to the conditional problem. Seymann makes this point.
The door numbers did not form part of the original Whitaker question, they were added by vos Savant. Whitaker did not mention specific doors.
Hundreds of sources, including academic 'psychology' sources give a simple solution. To pick just the ones that you agree with is POV
The MHP is one of the world's hardest simple brain teasers. Some people will never understand it if we force them to read through an essay on conditional probability.
I agree that none of the above is completely bomb-proof. If that were the case I would be pressing to have the complex solutions completely removed as unnecessary verbiage. As it is, we have a somewhat POV editorial at the start of the 'Probabilistic' section followed by several complex and fully conditional solutions. The interested reader is free to read these. Martin Hogbin (talk) 17:17, 6 August 2010 (UTC)
I must admit that I thought you , and others, had been saying that the argument that the specific goat door opened by the host cannot make any difference to the probability of winning even in the conditional case was incorrect.
Now I guess your argument is this: 'A symmetry argument shows that the simple solution to the unconditional problem must be a solution to the conditional problem. We have sources that give a simple solution to the conditional problem and a source which says that the symmetry argument is impeccable but we have no one source which gives a simple solution and states the symmetry argument therefore we cannot show the simple solution a a valid solution to the conditional problem'. Martin Hogbin (talk) 19:01, 6 August 2010 (UTC)
The statement that the specific goat door opened by the host cannot make any difference to the probability of winning is incorrect unless it is accompanied by an actual argument. I'd rephrase what you said my argument is more like this: A symmetry argument can be made showing that the unconditional probability computed by the simple solutions must be the same as the conditional probability. We have numerous sources that give simple solutions, and a source that says that the symmetry argument is impeccable, but we have no one source which gives a simple solution and actually makes the symmetry argument and therefore we cannot add this argument to the simple solutions in the article (because of .
BTW - there is no "unconditional" and "conditional" problem. There is simply "the problem". In all the time we've been discussing this I don't think you've ever produced any source that claims to be interpreting the problem as asking about the unconditional probability (if you think Seymann does, please read this source again - what he's actually talking about is the assumption that the host pick randomly between two goats which does make the problem symmetrical but doesn't magically change the probability of interest to be the unconditional probability). We have ALL agreed, numerous times, that the problem involves the player picking a door, and then the host opening a door, and only then is the player deciding whether to switch. Not before the initial door selection and not before the host opens a door. And not without the knowledge of which door was initially picked and which door has been opened by the host. Surely you understand by now that this means the probability of interest is the conditional probability. -- Rick Block (talk) 00:26, 7 August 2010 (UTC)
I understand your WP:SYNTH point but think the symmetry argument is so obvious and natural in the standard case that it scarcely needs stating. In any case I strongly believe that this is a case where we should bend the rules a little to make the article as useful as possible to the people who are going to read it. The 'conditional' issue is likely to be of interest only to a very small fraction of readers.
I disagree that there is no unconditional problem. You are still reading Whitaker's question like a question in an examination. It was a question from an interested member of the public who wanted to know the answer to a question. Just ask yourself this: 'Why would someone want to know the probability of winning by switching only in the case that they had chosen door 1 and the host had opened door 3?'. Whitaker was not actually on the show when he wrote in to Parade. He surely wanted to just know 'Is it best to stick or swap on the show?'.
You could argue that Whitaker meant to ask about what is the best thing to do in the case of a player who goes on the show, chooses a door at random then sees the host open a specific door to reveal a goat. This is a more complex question that is not properly addressed by any of the solutions given. To make my point these are the answers to that question
If we know that the host has a non-uniform but undisclosed policy for opening doors and we pick a door uniformly at random to start with, the probability of winning by switching is 2/3.
If we know the host's exact door opening policy for every possibility and we choose door 1 and see the host open door 3, we can calculate the probability of winning by switching from the host's policy parameter for that specific case, we know the answer will be from 1/2 to 1.
For cases in between the two I have quoted above we need to state exactly on what basis we are to answer the question and 'exactly' what assumptions we are to make before the question can be answered. No form of mathematical treatment is to any avail until we have decided on the 'exact' question we want answered. This fact is made clear by a reliable source (Seymann), but unfortunately no source gives us any answers. No solution given in the article addresses this point properly either. Martin Hogbin (talk) 11:16, 7 August 2010 (UTC)
^Falk: "The constant-ratio belief is the assumption that when some alternatives are ruled out, the ratio of the probabilities of the remaining alternatives should be the same as the ratio of their prior probabilities. This belief is obviously not valid since it leads to the answer l/2, whereas the correct answer (assuming an unbiased host) is l/3."
We should name the goats
We should name the goats
I believe that it is essential to differentiate the goats and treat them as separate entities in order to enumerate all of the equally likely possibilities that face the player. Let’s call one goat Billy and the other one Nanny. The list of possibilities for placement of car and goats is shown below. Designate door 1 as the first choice of the player. The last two columns show what happens for each distribution if the player switches or keeps his original choice. It doesn’t make any difference.
Door 1
Door 2
Door 3
Switch
Stay
Car
Billy
Nanny
Nanny
Billy
Car
Nanny
Billy
Billy
Nanny
Billy
Car
Nanny
Nanny
Car
Billy
Nanny
Car
Car
Nanny
Nanny
Car
Billy
Billy
Car
Nanny
Billy
Car
Car
Billy
This analysis follows the same logic structure as the “Marilyn” solution but it notes that the goats are different.
It seems like the switch and stay columns aren't quite right. If the player initially picks door 1, then I think it should look like this (and I've added a Notes column):
Door 1
Door 2
Door 3
Switch
Stay
Notes
Car
Billy
Nanny
Nanny or Billy
Car
Host can open door 2 or door 3, which one the host opens determines which goat the switcher ends up with
Car
Nanny
Billy
Nanny or Billy
Car
Host can open door 2 or door 3, which one the host opens determines which goat the switcher ends up with
Billy
Car
Nanny
Car
Billy
Host must open door 3 (showing Nanny)
Billy
Nanny
Car
Car
Billy
Host must open door 2 (showing Nanny)
Nanny
Car
Billy
Car
Nanny
Host must open door 3 (showing Billy)
Nanny
Billy
Car
Car
Nanny
Host must open door 2 (showing Billy)
In 4 out of the 6 total cases the switcher ends up with the car. This table makes it difficult to tell what happens if the host opens door 3, but of the last four cases clearly only 2 apply. Both of the first 2 do as well, but since the host has a choice in these cases these cases are not as likely (if the host opens door 3) as the other two. Assuming the host picks evenly between Billy and Nanny in these cases, these two cases are (together) as likely as only one of the other two cases (each case in the table has probability 1/6 - but for the first two cases there are subcases where the host opens door 2 or door 3 making the probability of these 1/12 if we also know the host opens door 3), so even if we're only talking about what happens if the host opens door 3 the switcher has a 2 to 1 advantage. -- Rick Block (talk) 15:06, 10 September 2010 (UTC)
Rick,
Thanks for your comments. I cannot say that they led me directly to agreement, but they did force me to start over and try to justify what I did, It is an interesting process. Eventually I came to the realization that the table of permutations and combinations does not give equally probable events after the host has opened his door.
Exactly. If you want to look at what happens after the host opens a door, a tree diagram like this one (that shows everything that can happen after the player picks door 1) is the way to go. From this tree it's clear if you've picked door 1 and the host has opened door 3, the probability the car is behind door 2 is twice the probability the car is behind door 1 (because the host must open door 3 if the car is behind door 2, but opens either door 2 or door 3 - presumably each one half the time - if the car is behind door 1). If you care about the names of the goats the first column of this table (labeled "Car location") ends up with 6 equally likely configurations - car/billy/nanny, car/nanny/billy, billy/car/nanny, billy/nanny/car, nanny/car/billy, nanny/billy/car - and from the top two of these there's a fork where the host opens door 2 or door 3 and from the bottom 4 there's no fork (the host only has one choice). You can then look at either all the cases where the host opens door 3, or just the cases where the host opens door 3 revealing billy (or only the cases where the host reveals nanny). -- Rick Block (talk) 00:57, 13 September 2010 (UTC)
Marking the goats as separate entities makes the solution to switch far simpler to grasp, this should be included on the main page near the top. —Preceding unsigned comment added by 75.158.2.195 (talk) 11:05, 11 January 2011 (UTC)
According to Krauss and Wang (cited in the article), viewing it from the host's perspective and ignoring the door 1/door 3 example given in the problem description makes it far simpler to grasp. The "simple solutions" ignore the door 1/door 3 example. Is there some source that says making the goats as separate entities makes it easier to grasp? -- Rick Block (talk) 14:09, 11 January 2011 (UTC)
This is an archive of past discussions about Monty Hall problem. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.
I just archived this page so everybody can make a fresh start after the Arbitration. I encourage all involved editors to put away any hard feelings from the past and to make a fresh start. Please remember that this article is now on probation and that editors making disruptive edits may be blocked by an administrator. I would strongly encourage all editors to stop and review WP:NPOV, WP:OR, WP:V, WP:OWN, WP:CIVIL, WP:NPA, and WP:EW. Even if you have read them before, read them again, and treat them as your road map for editing this article, its talk page, any any discussions about this article on your user page. Thanks! Guy Macon (talk) 04:12, 25 March 2011 (UTC)
One of two doors
Here's a quick way to argue against the current article's conclusion. It makes what appear to me to be some of the same subtle errors that I believe occur in the arguments for the currently-presented solution, but it has the virtue of being brief, and ( as it appears to me ) of producing the correct result. Let's ourselves be the contestant here. Assuming as much, then:
At the outset of the problem, before we've made any choice at all, consider that the car is behind one of two doors.
Yes, one of two. We just don't know which two, but it certainly is behind one of two. And of those two, based on what we know at this point, the chance that it's behind one is as good as the chance that it's behind the other.
Next, just to humor Monty, we "pick" one of the three available doors, and we try to keep his spirits up by pretending that our "choice" at this point matters. But we know it doesn't matter, because the outcome of the ensuing sequence will be the same regardless of which of the three doors we select at this point. After our apparent "choice", nothing relevant to the outcome has changed. The car is still behind one of two doors, and we still don't know which two those are.
Then Monty opens one of the unchosen doors, and now something has changed: We now know "which two".
Monty says, "Do you want to switch?" which translates into, "You now have the opportunity to pick between two doors. The car is behind one of them. Which one do you pick?" This is the first real choice we have, the first choice we can make that's relevant to the final outcome.
( There's zero chance that the car is behind the now open door, and no reason to believe that the now-dead chance that it was should be magically and preferentially transferred to any remaining door. There's a very subtle flaw involved here, imo, that I won't try to elaborate on, but that I think is due in part to a semantic confusion and in part to not properly accommodating time and instantiation as factors in this reasoning. Or maybe it would be more fun to observe that there was never any chance that the car was behind the now open door, but that we just weren't aware of that until now. ;-)
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, – OhioStandard (talk)07:52, 1 April 2011 (UTC)
OhioStandard your argument above shows exactly why the Monty Hall problem is so famous. Marylyn vos Savant got the answer exactly right, you should switch and you double your chances of winning if you do so. This is not in dispute. Please have a read through the article and see if anything there convinces you otherwise. Do you see anything wrong with any of the solutions given? If not, why are you not convinced by them? We would all be very interetsed to see how this article looks to a new reader. Martin Hogbin (talk) 09:21, 1 April 2011 (UTC)
Thanks for replying, 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, – OhioStandard (talk)13:37, 1 April 2011 (UTC)
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.
Have a look at the section beginning, 'Another way to understand the solution is to consider the two original unchosen doors together' and tell me why you do not agree with it. Martin Hogbin (talk) 20:47, 1 April 2011 (UTC)
What a delightful problem! I read quickly through the article a couple of days ago, and was unconvinced. Then, last night, without re-reading it, but after rapidly scanning through Marilyn Savant's presentation and remaining unconvinced, I posted the above. I was really very sleepy, though, both when I initially read the article and again, last night, when I scanned Savant's explanation and posted the above. ( I find Savant's self-promotion in very poor taste, and I suppose I'm quite prejudiced against her ideas. ) But when I awoke this morning it was the first thing on my mind, and I "grokked" it within the next couple of minutes. But how to express that "grok" concisely? Are you interested in hearing about my mental process to get there? I assume you might be, since you said you were curious about how the article would look to a new reader.
When I awoke, I found I did so with an image of a "shell game" present to mind, but one modified to include a gazillion shells with still just one pea. Obviously, I had combined one of Savant's suggestions with the presentation I vaguely recall from the article ( I'll look at both again when I'm through writing this ) having something to do with a million doors. I imagined Monty sweeping all the other shells but two right off the table, one of those two being the one I'd previously chosen, of course. Then it "hit" me: I'd been ignoring the information value of Monty's removal of ( a gazillion minus 2 ) shells.
I'd been preoccupied with my own subjective knowledge as a contestant, and missed the fact that Monty, transferred from his studio to the setting of my shell game, wasn't making his removal of all but one of the remaining shells at random, that his doing so was informed by intelligence I didn't have access to. I realized that, in effect, by sweeping all the other shells off the table, Monty, also, was making a choice as to which shell the pea was under, being constrained only by the (almost) negligible restriction that he couldn't choose the one I'd already chosen. Like me, he'd manifested a choice as to which shell the pea was under; he just had better information to inform his choice than I'd had.
I'm not certain how to formalize that insight without so-encumbering it with explanation that the "intuitive flash of recognition" it presents gets too submerged in notation or words or tables to be easily grasped. Decision tables don't seem to do it for me; it's one thing to grasp an idea, but another thing entirely to prove or demonstrate it, of course, and it's not often the case that the proof communicates the grasp, communicates that initial "intuitive flash of recognition" very economically, I believe...
I suppose one could explain that in the "gazillion shells" scenario, if my probability of making the correct choice was one-over-a-gazillion, that Monty's probability of doing so would equal 1 minus one-over-a-gazillion, i.e. that his probability wouldn't equal 1 only because he couldn't select from among all the gazillion to make his choice. But I'm not sure this would leave the typical reader very much enlightened, even if he were to accept the truth that the same formula must apply as the number of initial selections is (gradually?) reduced in successive example trials from a gazillion to just three. But it seems a promising avenue to me; is that in the article already? I'll be interested to see if it is, in that form. It wouldn't surprise me at all if I were recollecting it from there and assuming I'd come up with it originally.
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 (talk)04:55, 2 April 2011 (UTC)
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.
If you read the explanations in the article you will probably find them all convincing now you have grokked it, that is why I was so interested in your thoughts earlier as I want to find out how effective this article is for newcomers and how it might be made more so. Martin Hogbin (talk) 08:47, 2 April 2011 (UTC)
Thanks again, Martin, very much, for being patient with yet another newcomer who initially thought he knew better. :-) I can entirely understand why this article has resulted in such a shouting match; a person would need to have the patience of a saint to keep trying to explain, over and over again, for years on end.
I have now re-read the article, and Gerhard's and Rick's very kind explanations, as well. I'm very grateful for the work everyone has put into the article, and especially for Gerhard's and Rick's generosity, and for yours as well. But without the least possible slight toward any preceding explanation, I have to admit that none of those really do it for me. They're fine as proofs or demonstrations, but for me ... well, they don't really nudge me along toward that "flash of insight" (aka "grok") that seems so necessary in this case.
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, – OhioStandard (talk)21:29, 2 April 2011 (UTC)
@Ohiostandard: Will that table in your opinion help to show that the contestant will win by staying in 1/3 of cases, but will win by switching in 2/3 of cases, no matter at all which door she should have chosen (1, 2 or 3), and no matter which other door whatsoever the host has opened, showing a goat:
Supposed door A was chosen by contestant:
"Another" door was opened by the host, showing a goat:
contestant's final decision:
Initial arrangement behind doors A, B, C (probability)
Open Door A (probability)
Open Door B (probability)
Open Door C (probability)
Joint probability
Win by staying
Win by switching
Car Goat Goat (1/3)
No
Yes (1/2)
No
1/3 x 1/2
Yes (1/6)
No
No
No
Yes (1/2)
1/3 x 1/2
Yes (1/6)
No
Goat Car Goat (1/3)
No
No
Yes (1)
1/3 x 1
No
Yes (1/3)
Goat Goat Car (1/3)
No
Yes (1)
No
1/3 x 1
No
Yes (1/3)
Note that the host has no choice when the contestant in 2/3 of all cases fails and chooses a goat, as in these cases the host is just bound to show the second goat.
Win by staying 1/3
Win by switching 2/3
And note: If, in only 1/3 of cases, the contestant luckily should have chosen the car, only then the host has a choice indeed and can open either of his two doors, both containing goats.
Some sources now said that the host eventually could be "biased" to some extent to open just only one special "preferred" door if ever possible, avoiding to open the other door, avoiding his "unwanted" door.
But for the decision asked for, to stay or to switch, the "door number first chosen" by the contestant and "the door number opened by the host" is of no relevance at all, as you do not exactly know about such "bias" and about its "direction", and even if you exactly knew about such bias, its extent and its direction, it would never "help better" to make the right decision. Let us assume even an extremely biased host:
If, in 1/3 of cases, the host got both goats, then he can and will open his preferred door, and switching will loose. Probability to win by switching: Zero. And if, in another 1/3 of cases, he got the car and one goat, and the goat being behind the door he usually prefers to open, then he can and will also open his preferred door, but in this case switching will win, probability to win by switching: 1.
So, whenever in these 2/3 of cases an extremely biased host opens his preferred door, probability to win by switching will be at least 1/2. And that means staying never can be any better than to switch, so you should switch.
If however, in the last 1/3 of cases, he got one goat and the car, but the car being behind his preferred door, then he exceptionally will be opening his "avoided" door and, by doing so, be showing that the car is very likely to be behind his preferred but now still closed door, and switching is very likely to win the car, and switching is imperative.
You never can nor will know any "assumed closer is better" -probabilities for any "actual game". Conditioning on "just assumptions". Unneeded diligence. Results of "conditional probability theorems" will forever remain within the strict bounds of at least 1/2, but never less, to 1. Irrevocably. You know that already from the outset. So conditional probability calculus is quite irrelevant for the question asked for, even if one should exactly know about "host's bias, its extent and its direction", as staying never can be any better than always to switch, and you know for sure that switching will win the car in exactly 2/3 of cases. Full stop. Irreversible. No further "proof by conditional probability calculus" needed.
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, – OhioStandard (talk)04:55, 2 April 2011 (UTC)
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, – OhioStandard (talk)21:29, 2 April 2011 (UTC)
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, 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 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, – OhioStandard (talk)02:20, 3 April 2011 (UTC)
@OhioStandard - Responding to your original request (in the same idiom as your original argument), if we follow your argument through it's fairly simple to point out where it goes wrong.
At the outset of the problem, before we've made any choice at all, consider that the car is behind one of two doors. We don't know which, and the chances that it is behind one is the same as the chances it is behind the other.
Perfectly true. It's even true that if you pick ANY two doors at this point the chance that the car is behind either one is the same. However, the only way to make this true is to consider the probability that the car is behind any of the three doors is the same, i.e. p(door 1) = p(door 2) = p(door 3), and they must add up to 1, so they must all be 1/3. If you pick any two, the chances are even at 1/3:1/3 which only adds up to 2/3. This reflects the uncertainty that you've picked the right two. The chances that the car is behind a randomly selected two doors at this point is only 2/3.
Next, just to humor Monty, we "pick" one of the three available doors, and we try to keep his spirits up by pretending that our "choice" at this point matters. But we know it doesn't matter, because the outcome of the ensuing sequence will be the same regardless of which of the three doors we select at this point. After our apparent "choice", nothing relevant to the outcome has changed. The car is still behind one of two doors, and we still don't know which two those are.
Also perfectly true. The key is we still don't know how to pick two doors guaranteeing the car is behind one of those two. So the probability of each of the three doors is still 1/3. The probability of any two (at this point) is 2/3.
Then Monty opens one of the unchosen doors, and now something has changed: We now know "which two".
Yes, now something has changed. But it's more than you think. The probability that the car was behind the two remaining doors that used to be 2/3 is now 1, and the probability the car is behind the open door that used to be 1/3 is now 0. How this is possible is that we're now not talking about the "original" probabilities - but something different. Specifically, we're now talking about the conditional probabilities in the case that the host has opened (say) door 3. This doesn't mean the "original" chances that the car was behind the two remaining doors was 1 or that the original chances that the car was behind the door that has been opened was 0. The original chances were (and still are!) 1/3 for each door, 2/3 for any two doors, and 1 for all three doors. If you pick (say) door 1, when the host opens a door these original chances are split into two cases, one where the host opens door 2 and the other where the host opens door 3. The probabilities in each of these cases are conditional probabilities. The original probabilities of door 1 and door 2 were indeed equal, at 1/3 each. And now, if we're in the case where the host opened door 3, we know its (conditional) probability is 0. We also know the conditional probabilities of the two remaining doors will add up to 1 (just as they do in the case where the host opens door 2). 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. If the car is behind door 2 the host MUST open door 3. However, if the car is behind door 1 the host chooses whether to open door 2 or door 3. Assuming this is a fair choice, this means the host opens door 3 only half the time the car is behind door 1. Half the time behind door 1 and all the time the car is behind door 2 means you have a 2 to 1 advantage if you switch.
If you pick door 1 the original 1/3 + 1/3 + 1/3 = 1 splits (unevenly) into
1/6 + 1/3 + 0 = 1/2 (if the host opens door 3)
and
1/6 + 0 + 1/3 = 1/2 (if the host opens door 2)
Expressing these as conditional probabilities, you get 1/3 + 2/3 + 0 (considering only the case where the host opens door 3), or 1/3 + 0 + 2/3 (considering only the case where the host opens door 2).
The essential confusion is the difference between the original unconditional probabilities and the conditional probabilities in effect after the host opens one of the doors. -- Rick Block (talk) 05:40, 2 April 2011 (UTC)
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, – OhioStandard (talk)21:29, 2 April 2011 (UTC)
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! – OhioStandard (talk)21:40, 2 April 2011 (UTC)
A different table
OhioStandard - can you comment on this table showing how many times we'd expect the car to be behind each door in a sample of 300 games, and the outcomes when switching (assuming the player picked door 1 initially)?
Situation BEFORE the host opens a door
Situation AFTER the host opens a door
Door 1
Door 2
Door 3
total cases
host opens Door 2
host opens Door 3
cases
result if switching
cases
result if switching
Car
Goat
Goat
100
50
Goat
50
Goat
Goat
Car
Goat
100
0
N/A
100
Car
Goat
Goat
Car
100
100
Car
0
N/A
The point is to show that by opening either door the host splits the original 300 cases into two subsets. The (conditional) probability of winning by switching (if the host has opened, say, door 3) is evaluated in the context of one of these subsets, not in the context of the entire original sample of 300 shows. -- Rick Block (talk) 16:17, 3 April 2011 (UTC)
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. – OhioStandard (talk)15:11, 4 April 2011 (UTC)
This is the "Arguments" page. IMO, here (but not on the main talk page) discussions are explicitly permitted to drift more toward "truth" - the topic is mathematical arguments concerning the MHP as opposed to content of the article. -- Rick Block (talk) 14:25, 4 April 2011 (UTC)
Ah. Of course. (note to self: next time smoke crack after posting to Wikipedia...) Guy Macon (talk)
Talk about notation
The present article contains in the conditional probability section a fully written out proof of Bayes' theorem in the context of MHP. (IMHO, totally superfluous, since also present in the wikipedia article Bayes theorem). But anyway there were discussions about the proper notation to be used in this proof. I have written out an alternative here (on my talk page) and written a general essay about notation here also in my user area. Richard Gill (talk) 09:51, 5 April 2011 (UTC)
Increasing the number of doors rewrite?
I think the "Increasing the number of doors" approach that's currently (permalink) in the article is probably the one that's most likely to be accessible to our widest swath of readers. But I also think it needs to be rewritten to accomplish that. Here's how I'd use the idea to explain the solution to someone who knew only simple algebra:
Look at it from Monty's perspective
To understand this, you need to set aside your natural preoccupation with your own perspective as a potential contestant, and look at the situation from Monty's perspective instead. Specifically, you need to recognize that you're not the only one who expresses a choice about which door the car is behind. It's easier to understand that if you imagine there's some impossibly large number of doors, at first; let's say a "gazillion".
So you recognize that when you first choose one door, your chance of doing so correctly is really small, one-in-a-gazillion, right? Then switch perspectives, and put yourself in Monty's shoes: Ask yourself "What is Monty's chance of choosing the correct door, the one with the car behind it?" "But wait!", you say, "Monty doesn't choose anything!!"
Monty does make a choice, though, and this is the key to understanding the problem: When Monty opens all the other doors but one, he eliminates all those doors because he knows the car isn't behind them. Assuming you didn't guess correctly the first time (a reasonable assumption, given your one-in-a-gazillion chance) then Monty is effectively choosing the door the car is behind, by keeping that one door closed and opening all the others. The rules he has to operate under make it impossible for him to do anything else. Further, his expressing that choice, even though he does so indirectly, discloses information to you that you'd be smart to take advantage of when he eventually asks if you want to switch from your original choice to his.
The mathematics of this insight, and what they show for the three-door case
The mathematics of this become pretty simple to understand if you reduce the number of doors at the outset from a "gazillion" to just 100. In that case, when you first choose a door, it's pretty clear that you have a .01 or (same thing) 1% chance of choosing the right one. But then ask yourself, "What chance does Monty have of choosing the right door?" ( Recall that he has to express his choice not by picking the door directly, of course, but by opening all the other ones, except yours. )
Since Monty knows where the car is, the answer would be that he'd choose correctly 100% of the time if he were not constrained by your having eliminated one of his possible choices when you picked first. So that constraint means that he has only a .99 or (same thing) 99% chance of choosing the door with the car behind it. His chances of picking the correct door are, in this example, 99 times better than your chance of having done so.
To express that insight in a formula, if we let
N = the number of doors at the outset,
PContestant = the contestant's probability of choosing right the first time
PMonty = Monty's probability of choosing the right door, i.e. of leaving closed the door with the car behind it
Then it's clear that
N = 100
PContestant = 1 / N = 1/100 = .01 = 1%
PMonty = 1 - ( 1/N ) = .99 = 99%
PMonty = 1 - PContestant
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.
Now these formulae continue to hold if, instead of starting with a gazillion doors, or even 100 doors, we reduce the number of doors at the outset to 50, or 25, or even to three. So with three doors at the outset, we have:
N = 3
PContestant = 1/3
PMonty = 2/3
It may seem contrary to common sense in this case, when we've reduced the number of doors to just three, but the same reasoning and the same formulae do continue to hold, and if you switch to the door Monty "chose" (by leaving it closed), you do in fact double your chances of winning.
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? – OhioStandard (talk)19:50, 2 April 2011 (UTC)
( 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, – OhioStandard (talk) 22:12, 2 April 2011 UTC )
"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, 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? – OhioStandard (talk)00:00, 3 April 2011 (UTC)
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 [1]. Richard Gill (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 (permalink) that Rick Block posted to the Arguments page under the heading, "A different table". Best regards, all. – OhioStandard (talk)17:25, 4 April 2011 (UTC)
OhioStandard, I am a little concerned that you are deferring to other editors based upon them being here longer. Please read WP:NVCWP: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.) Guy Macon (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. – OhioStandard (talk)10:16, 10 April 2011 (UTC)
Probability concepts
IMHO, one can see from any written source about MHP whether the writer is a subjectivist or a frequentist in his or her use of probability. Often of course without even knowing about the distinction. I think that the present page contains quite a few subtle biases to one or the other concept of probability. For instance the just mentioned proof of Bayes' theorem explicitly put randomness in the physical procedure used by game-show team and the host to hide the car and open a door respectively. Probability is in the real world. Ontological probability (about things). Whereas the "usual assumptions" of equal probabilities really (IMHO) only make sense for the subjectivist (aka Bayesian) viewpoint, whereby probability describes our information or lack thereof. It's in our mind. It describes our personal ignorance or knowledge. Epistemological probability (about our knowledge). I wrote a little essay about this on my talk page, here, it was originally a response to a correspondent. Read Probability interpretations on wikipedia.
So far editors of the page have shied away from this issue. Maybe they are scared of opening a can of worms? Or they have a POV and don't know any other? I think the issue can't be avoided in any decent discussion of MHP. You can only draw conclusions by making assumptions and where do those assumptions come from? The man in the street is a subjectivist, the academic literature is biased to the frequentist position. Richard Gill (talk) 10:00, 5 April 2011 (UTC)
I am very happy to discuss the subject with you. It has indeed caused a lot of confusion in recent arguments.
I do not think the distinction is as great as you suggest. The two approaches can always be made to agree by making sure that the set-up of the frequentist approach matches the state of knowledge in the Bayesian approach. If you disagree, please give an example of where the two approaches give different answers. Martin Hogbin (talk) 22:03, 7 April 2011 (UTC)
Better shoes argument?
If it won't annoy people, may I ask whether any sources have tried to give the reader the necessary "flash of insight" by asking him to put himself not in the position of the contestant, as is usual, but in the position of the host, instead? As in the following, for example:
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, – OhioStandard (talk)00:34, 8 April 2011 (UTC)
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. -- Rick Block (talk) 06:24, 8 April 2011 (UTC)
In Selvin's second note, this is long before Marilyn Vos Savant, Selvin quotes from a letter sent by Monty Hall himself, who did solve the problem in one sentence from the point of view of the host. For the host, the location of the car is known, and the way he will choose a goat-door if necessary is his own business. The only thing random for him, is which door the player chooses, which from his point of view has a odds of 2 to 1 against of being correct. Richard Gill (talk) 11:09, 10 April 2011 (UTC)
Will the real probability please stand up
One of the continuing problems here, in my opinion, is the assumption by some editors that there is some 'real' meaning of the the term 'probability' that everybody knows and understands. The truth is much closer to the reverse; there are two definitions of probability in common use and nobody understands what the word 'really' means.
Richard Gill and I have tried several times to discuss this important subject but to no avail. Might I suggest that it would avoid many pointless arguments if people would state which definition of probability they are using and stick to it throughout their argument. It is generally unhelpful to talk the results of 'repeating the experiment' when discussing a subject from a Bayesian perspective. Similarly, states of knowledge are of little relevance in calculating a frequentist probability.
Now, my personal opinion, at least, is that it is always possible to make the two interpretations of probability agree by carefully making sure that the frequency distributions in the frequentist interpretation match the state of knowledge in the Bayesian interpretation. However that is a subject for another time and place.
What would greatly help here would be for editors to clearly state which interpretation they are using and then stick to it. Unfortunately, not many sources do this either. Martin Hogbin (talk) 08:57, 9 April 2011 (UTC)
"The terminology of this topic is rather confusing, in part because probabilities are studied within so many different academic fields. The word 'frequentist' is especially tricky. To philosophers it refers to a particular theory of physical probability, one that has more or less been abandoned. To scientists, on the other hand, "frequentist probability" is just what philosophers call physical (or objective) probability. Those who promote Bayesian inference view 'frequentist statistics' as an approach to statistical inference that recognises only physical probabilities. Also the word 'objective', as applied to probability, sometimes means exactly what "physical" means here, but is also used of evidential probabilities that are fixed by rational constraints, such as logical and epistemic probabilities."
Guy, I am not sure what you are trying to tell me. Your quote above confirms my point above that there is no general agreement on exactly what probability 'really' is.
1. Frequentists talk about probabilities only when dealing with experiments that are random and well-defined. The probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment. Frequentists consider probability to be the relative frequency "in the long run" of outcomes.[1]
2. Bayesians, however, assign probabilities to any statement whatsoever, even when no random process is involved. Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, or an objective degree of rational belief, given the evidence.
My point is that in discussing probability issues regarding the MHP it would be very helpful and would avoid pointless disagreements if editors would adopt a particular perspective and stick to it for the duration of the argument.
My personal opinion is that the Baysian approach is the most appropriate approach for the MHP but I do wish to try to force that opinion on anyone else and I am happy to discuss the problem from a frequentist perspective.
I am not sure from your post whether you agree with me or not Guy. Are you asserting that there are other important interpretations of probability or do you agree that there are only two?
Do you agree that in a discussion about probability it is best to stick to one interpretation for the duration of a logical argument rather than change the meaning of the word part way through a discussion? Martin Hogbin (talk) 15:39, 9 April 2011 (UTC)
Yes. I fully agree that we need to pick a definition and stick to it. The definition I use as a working engineer is not the definition a philosopher uses, and I suspect that a statistician will have a third definition. Any of the different definitions will work if everybody uses them , but having different people use different definitions is an invitation to misunderstanding and strife. Guy Macon (talk) 20:16, 9 April 2011 (UTC)
Yes the words are confusing and are used to mean the opposite by different communities. Maybe we should just stick to using the adjectives "epistemological" and "ontological" ;-) Richard Gill (talk) 06:56, 10 April 2011 (UTC)
Responding to Martin's original post in this thread - IMO, what we mean by probability is not even remotely the issue here. What IS the issue is what probability we mean (per my response to Martin in the thread above). Bayesians and frequentists are both completely capable of distinguishing P(win by switching) from P(win by switching|player picks door 1 and host opens door 3). If we want to reduce the confusion here, we should clarify which one of these we're talking about. Are we interested in the probability of winning of a strategy of (always) switching, or the probability of winning given the player has picked a particular door (say door 1) and has seen the host open some other door (say door 3)? Which of these we're talking about is and always has been the issue - not the meaning of probability. -- Rick Block (talk) 23:59, 9 April 2011 (UTC)
Rick, that was not a response to Martin's point at all, I think. Of course everyone ought to be able to agree on a distinction between P(A) and P(A|B). And both kinds of probability satisfy the same rules so the arithmetic is the same too. But the words around the arithmetic and the distinction of concepts *is* important. And the literature is confusing since different writers do clearly have different intended meanings of probability in mind. Moreover, the assumptions which some people want to add into the problem in order to be able to solve it their way have to be justified or explained, and the justification or explanation is different per different understanding of what the word probability is supposed to mean.
So an editor of the page his well advised to be aware of the distinction and he or she has to be master of using formulations which are probability interpretation free. It can be done but it requires careful and clear thinking. Never talk about "how many times". Aways talk about "so many times more likely". Basicly it forces you into the language of odds, not of probabilities. Richard Gill (talk) 06:51, 10 April 2011 (UTC)
Rick I am not saying that the issue we are arguing about is related to the different interpretation of probability, just that not making clear what interpretation you are using makes it very hard to discuss the subject without continual misunderstandings. I am not even asking you to stick permanently to one approach, all I am asking is that we talk about them one at a time. I have not been able to respond to your reply to me above because you seem to continually flit between one approach and another whilst using concepts from one approach in the context of another.
Let me say again, I do not want to have a philosophical discussion about the meaning of probability on this page all I ask is that during any given discussion we adopt one interpretation and stick to it. Can there be any valid reason not to do that? Martin Hogbin (talk) 21:14, 10 April 2011 (UTC)
And, what I'm saying, is that for the purposes of any discussion we've ever had, which interpretation of probability we're talking about makes no difference whatsoever. In my reply above (that you haven't responded to) I don't "flit between one approach and another" but connect what we ARE talking about to both. The OP above was asking why his professor insisted on using conditional probability. The explanation (whether you're a frequentist or a Bayesian) is the same - i.e. that at the point of making the decision whether to switch, the player knows which specific doors are involved (and the doors are not "otherwise indistinguishable"). Whether you're a Bayesian or a frequentist, this means you're interested in the conditional probabilities. Can there be any valid reason for continuing to harp on a point that has no relevance to what we are talking about, and refusing to address the actual issue?
Richard says above "everyone ought to be able to agree on a distinction between P(A) and P(A|B)". So, Martin, are we talking about P(A) or P(A|B)? Harder question - do the "simple" solutions address P(A), or P(A|B)? If you think it's necessary, feel free to specify what interpretation of probability you're talking about. -- Rick Block (talk) 23:20, 10 April 2011 (UTC)
Rick, I am not harping on, I am asking you to make your arguments clearer by stating which interpretation of probability you are using. This is simply to help other editors understand exactly what you mean. You may know what you mean but Richard and I need to as well. I am quite ready to address the actual issue provide we agree at the start to use and stick one interpretation of probability. We may well arrive at the same conclusion whichever path we take but it will be an easier journey, for me at least, if we stick one philosophy throughout. Martin Hogbin (talk) 13:27, 11 April 2011 (UTC)
Regarding your question, let us agree to take a Bayesian perspective on Whitaker's question. You asked whether simple solutions calculate the value of P(win by switching)? I think we can all agree the answer is 'yes'.
Having said 'yes', I should point out that the simple solutions are in no way rigorous and include many assumptions that are not stated. As I think Boris said, there is a very rigorous for of mathematics in which even the most simple solutions become inordinately complex and completely unsuitable for anyone except mathematical specialists. For our expected readership and especially those who may read only the first part of the article, the simple solutions give, in my opinion, perfectly adequate calculation of P(win by switching).
If you ask whether the simple solutions calculate P(win by switching|player picks door 1 and host opens door 3) my answer would be the same except that they are marginally less good in that we do not actually state the obvious fact that as we have no information to allow us to distinguish between doors P(win by switching|player picks door 1 and host opens door 3) must be equal to P(win by switching). And as we have agreed that the simple solutions are satisfactory for calculating P(win by switching) we know that they also calculate P(win by switching|player picks door 1 and host opens door 3). It is obvious that, from a Bayesian perspective, P(win by switching|player picks door 1 and host opens door 3) = P(win by switching|player picks door 1 and host opens door 2) = P(win by switching). Martin Hogbin (talk) 14:00, 11 April 2011 (UTC)
Question to Glopk and to Rick and any others regarding success rates
Article understandable to the reader, desirably to grandma and grandson likewise. Therefore my question:
MHP – "decision to switch or to stay" – gains and losses as results of effectively taken decisions of frequentists
a) decision based exclusively on the so-called flawed simple solution, after the host had shown a goat and thereafter had offered to switch to his still closed door
b) decision based exclusively on the so-called only correct conditional solution as per updated actual probability calculus
Please can anyone help me to find some reliable source that ever has determined the respective results of decisions to switch (resp. to stay), that effectively had been taken on solely "a", compared to solely "b" and thereafter has presented the specific results, for any combination of door chosen by the guest and door opened by the host likewise?
Is it conceivable that decisions that had exclusively been made based on the only correct conditional probability are of any higher success rate?
Is there evidence that specific decisions for any combination of doors, that had been made exclusively based on the only correct conditional probability "b", correctly updated for every show, has a higher success rate, compared to the specific results of decisions that have strictly been made on the so called flawed simple solution as per Marilyn vos Savant, i.e. to switch in the respective "actual" game show after the host has made his offer? And if so, to which degree? Because that must necessarily be added to the article. And otherwise, the contrary as well.
It is about the possible superiority of the so called conditional solution. Is its success rate really indeed so significantly higher for any specific combination of doors, as it is claimed by some?
Please help, because afaik there is no better possible decision than just "to switch" in the specific actual game, irrelevant of door numbers. Is there reliable evidence in reliable sources that "updated b" is of remarkably higher success rate than "a"? Otherwise "updated b" impossibly can be presented as a relevant topic here, but just the quite other aspect that the MHP is a suitable method to train conditional probability theorems. Thank you. Gerhardvalentin (talk) 17:32, 11 April 2011 (UTC)
The simple solution shows that always switching is a much better strategy than always staying. The conditional solution shows that there is nothing better still. But of course nobody would imagine this, anyway. This is true if all doors are initially equally likely to hide a car. It doesn't matter whether or not the host has some bias. It's easy to see that the more biased the host is the more favourable it is to the player. And with a maximally biased host you can easily see (just work out the two possibilities in the Monty Crawl game) that switching is never unfavourable. Yet also in Monty crawl always switching only has success rate 2/3. This proves rigorously that there is no better strategy than always switching. (This argument was discovered by Lambiam, it is inspired by game theory, and it makes the explicit computations of Morgan et al. superfluous and allows you also to explain the biased host case to your Grandma or Grandpa. I have written it up in the longer version [2] of my article in the StatProb encyclopedia [3]. It's also written out on my University of Leiden home page. @Lambiam got a prize from me for finding this proof. [User:Gill110951|Richard Gill]] (talk) 16:48, 14 April 2011 (UTC)
Course on probability
Sorry, dind't know of this page. Moved to here.
In my course on probability the problem is treated and solved by computing with Bayes' formula the conditional probability on the car under the condition of the door number chosen by the contestant and number of the door opened by Monty. As far as I know this is done in all probability courses. Is the point of discussion here that there is a different approach possible? Handy2000 (talk) 22:38, 6 April 2011 (UTC)
Most probability courses do this, I believe, because the Monty Hall problem has become a popular example with which to exemplify Bayes theorem. It then becomes necessary to make assumptions which may or may not be well justifiable, and whose justifiability might well depend on whether you think of probability in a frequentist or subjective manner. Marilyn Vos Savant, who made the problem famous, and many others, including mathematicians, probabilists, statisticians, game theorists and mathematical economists, show that the player who always switches wins with probability 2/3, under weaker assumptions than the usual assumptions needed to apply Bayes. And with arguments which ordinary people can understand and appreciate. So in short: not only probabilists are interested in the Monty Hall problem. It is popular to a large number of people, many of whom do not have any education in probability at all.
From the mathematical point of view, showing no conditional probability favours switching is equivalent to showing that the uncondional win-chance of 2/3 (attained by always switching) cannot be improved upon. In the fully symmetric case favoured by most probability text books symmetry shows immediately that the unconditional and conditional probabilities must be the same. I think that ordinary folk who are not familiar with conditional probability can also appreciate this argument. Though most people would find it not worth the bother. Always switching gives 2/3, always staying gives 1/3. It's pretty inconceivable that a mixed strategy could do better than always switching. There is a short proof of this fact (in the situation where the three doors are equally likely to hide the car, but the host is not equally likely to open either door when he has a choice) which uses the fact that knowing the outcome of the (possibly biased) coin toss which Monty would make if he needed to make one, can't be unfavourable to the player. Knowing the outcome puts the player in the Monty Crawl problem: the problem where, if you choose door 1 and the car happened to be behind it, you know which door Monty would open. Now it's easy to see by inspection of the two possible cases (that he opens the door which you would expect in that case, or the door which you do not expect in that case) that also in the Monty Crawl problem it is never unfavourable to switch. But always swithcing in the Monty Crawl problem still only gives an overall success chance of 2/3. Consequently 2/3 can't be beaten.
In short, there are a lot of reliable sources who think that MHP doesn't have to be solved by going to Bayes theorem (including the lady who made it famous), a few reliable sources who think that people who only give the unconditional probability are wrong and Mrs Vos Savant is stupid, a lot of probability text books which do the conditional probability problem as an exercise to illustrate Bayes theorem, and a whole lot of other ways to get whichever results you like under whichever assumptions you find natural.
If you must do it with Bayes I'd recommend you teach your students Bayes' rule which not only gets the right answer even faster but also gives insight into why conditional and unconditional probabilities are equal (in the symmetric case) as well as being a powerful and internalizable tool and a beautiful result .. while Bayes theorem is just the definition of conditional probability twice and the law of total probability. It's easier to figure it out from first principles rather than remember it. Bayes' rule is on the other hand memorable and surprising by its utter simplicity. You just have to pick up the notion of odds. Richard Gill (talk) 23:11, 6 April 2011 (UTC)
Thanks for your extensive answer. I'm not teaching, but just a student. Our professor said that because you know the door numbers of the chosen and the opened door, you have to use conditional probabilities. How come? Handy2000 (talk) 21:48, 7 April 2011 (UTC)
He should tell you *why* you have to, or rather, why it could be wise to do so. There is no *must*. Mathematics advises. It is not about (legal) law, and it is not about morals, its about wisdom. The reason it would be wise to do so is because that way you will be sure of doing the best you can, overall. The chance of winning the car cna be broken down by the situation in which you find yourself. You have the best chance overall, if you have the best chance in each situation.
There are many many other ways of showing that 2/3 overall win chance is the best you can possibly get. If you use one of these other ways, it's a waste of time to use Bayes to do the same thing.
There is one interpretation of probability in which a sort-of "must" can be deduced. In the Bayesian world we are continually getting information and if we are self-consistent we are continually updating our beliefs about uncertain things according to the laws of probability. (If we don't update by Bayes we could be tricked with a Dutch book into accepting simultaneously different bets such that whatever happened, we'd lose money). A Bayesian first chooses a door. Probability 2/3 of missing the car since all doors are equally likely - according to his knowledge. The host opens a door and shows a goat, but the player doesn't take any notice of the number. Still probility 2/3 that there's no car behind door 1, because whether or not the car is behind door 1, the host is certain to do what he did. Next the player takes notice that it was door 3 the host opened. Still probability 2/3 that there's no car behind door 1 since whether or not the car is behind door 1, the host is for him equally likely to open door 2 or door 3, by symmetry of the player's knowledge.
Actually, a very smart Bayesian knows in advance that the door numbers will be irrelevant to whether or not his initially chose door hides the car, by symmetry. So he'll just pick any door and thereafter switch without taking any notice of the number of his door nor of the number of the door opened by the host.
This brings the smart Bayesian into harmony with the smart frequentist. The smart frequentist knows that he doesn't know the probabilities of the location of the car, nor does he know the probabilities by which the host opens a door. He does know game theory though. So he'll initially pick his door completely at random and thereafter switch, and he'll get the car with overall probability 2/3. He doesn't know about the conditional probability he got the car (if he did), and he doesn't care about it either. Both of them choose at random and switch because they know nothing. That's wisdom.
Sorry for another extensive answer. There are in fact many answers. And it's a rather important question.
I'm a little confused by Richard Gill's lengthy explanation. It seems he shows that the average contestant wins with probability 2/3 when switching. That's no surprise, and he might as well have proven that 1+1=2. Neither leads logically to the conditional probability being 2/3. What point am I missing? Handy2000 (talk) 22:06, 7 April 2011 (UTC)
I was actually referring to an explanation of my remark above but although I cannot speak for Gill, I will give you my take on his argument. The point is that, in the normal interpretation of the problem, the situation is perfectly symmetrical with respect to door number. The numbers on the doors tell us nothing, thus we can change 'the host opens door 3' to the host opens door 2 without changing the answer, because we have no information that allows us to distinguish between the two doors. Martin Hogbin (talk) 23:04, 7 April 2011 (UTC)
If you know the door numbers you can ask (for example) what is the conditional probability of winning if a player picks door 1 and the host opens door 3 - and this probability may or may not be the same as the overall average. You should use conditional probabilities in this case because you can do an experiment that will in the limit show this probability (randomly select a door, see which door the host opens, and throw this sample away unless the player picked door 1 and the host opened door 3). If you do this you are experimentally determining the conditional probability, NOT the average probability. If you think because the average is 2/3 that this experiment should come out with the same answer, you might be surprised. If you figure out the conditional probability, and see what it is sensitive to (the initial car location AND how the host picks if the player initially picks the car), you might run your experiment differently. The experiment vos Savant suggested in her third column did not control how the host selects in this case, so unless the results are undifferentiated (by specific case) the experimental answer can be anything from probability 1/2 to probability 1 of winning by switching. However if all 6 of the conditional probabilities are 2/3 the average (clearly) must be 2/3. Richard is inverting this - if the average is 2/3 and all the conditionals must be equal then all the conditionals must be 2/3 as well. What it means for all the conditionals to be equal is that the problem is symmetrical, so if you know the problem is symmetrical and you know the average probability is 2/3, then you know all the conditionals must be 2/3 as well (it's really just a minor short cut to figure out the conditionals). -- Rick Block (talk) 23:12, 7 April 2011 (UTC)
"Minor short cut to figure out the conditionals"! For some reliable sources, it's a major insight which makes even thinking about them completely superfluous! And Rick, you write that you "should" use conditional probabilities... where does this imperative come from??? Mathematics does not tell you what you *must* do. It can only give you advice on what might be wise. You can have good reasons to ignore the advice in many situations. Certainly, you are not obliged to use it. Richard Gill (talk) 08:09, 11 April 2011 (UTC)
If I do understand you correctly, from the symmetry you conclude that all the relevant conditional probabilities are the same, and hence have the same value as the average probability. But why do you want this conclusion if you on the other hand state that it is not needed to consider the conditional probabilities? On the other hand, if you want to use the symmetry for some purpose, you should have to mention this in your argumentation, and I get the impression you don't want to mention this. Can you explain this? I take the opportunity to make another remark. I discussed the problem with my fellow students and we have the idea that the solutions, mentioned in the simple solution section, with reference to some Devlin, are logically nonsense. Indeed do the two remaining doors have probability 2/3 on the car, but this is not only because the chosen door has probability 1/3, but also because each of these remaining doors itself has probability 1/3 on the car. It seems impossible to conclude from this that the remaining unopened door should have probability 2/3 on the car. Are we right? Handy2000 (talk) 20:33, 11 April 2011 (UTC)
Pardon for the intrusion: Handy2000, you are right to say that any of the three doors originally had a chance of 1/3, so the chance of the pair of the two remaining host's doors together is 2/3. And you know that at least one of those two host's doors definitely must hide a goat, as there is only one car. So you could say that if the host shows a goat behind one of his two doors, that's no news, and that it didn't influence the chance of your originally chosen door of 1/3, and that its chance has remained 1/3. That's right, so in 2 out of three the host's second door will hide the car, and the probability to win by switching is 2/3, as long as you got no "additional information". That's a fact. And in only 1 out of three there will be the second goat. That's a fact also.
But now comes "the crux" (conditionalists call it "the core") of the MHP: The host "could" have been telling you additional bootless info in showing you a goat, he could have been giving you some closer knowledge on the "actual" probability to win by switching: If, for example, he always uses to open just one special door if he has got two goats, say he always uses to open his door with the lowest number only, if ever possible, and never his door with the higher number, then - in exceptionally opening his "other, strictly avoided door with the higher number", then he will have shown you by that action that he was unable to open his preferred door with the lower number, and that it is very likely that his usually preferred door with the lower number "actually" hides the car! And that your chance by switching actually is exceptionally "higher than 2/3", and the chance of your originally chosen door dropped below 1/3, as a consequence. And vice versa. That's the argument that conditionalists use to present: that it could be possible that the host just could have given us "additional information" on the actual location of the car. But as the host forever will be out of position to indicate any "closer" probability to win by switching than to be forever within the fixed range of "at least 1/2 (but never less!), to max. 1/1, but never more :-)", all of that forever will remain "bootless and useless closer information" without any relevance as to the decision to switch or to stay, and so conditionalists are indeed unable to give you better advice than always to switch, also. Because staying never can be better (see my "Question" below). Gerhardvalentin (talk) 22:49, 11 April 2011 (UTC)
Thank you for your explanation. However it does not explain why the referred reasoning should be correct. It is not about the answer of 2/3, but about the way it is derived. You say the chance for the originally chosen door "has remained" 1/3. What do you mean by that. Of course do all the chances remain the same. May be you mean that the conditional probability after the initial choice and the action of the host also has the value 1/3. Am I right? Still the way Devlin reasons seems not correct to us. Handy2000 (talk) 08:51, 12 April 2011 (UTC)
On my talk page Gerardvalentin explains that Devlin is merely trying to give a way of understanding rather than a complete solution. He admits that the resulting probabilities of 2/3 and 0 for the remaining and the opened door are not the original probabilities, but conditional probabilities. The question remains whether Devlin also considers these probabilities to be conditional. At least it looks somehow a little misleading by giving an way of understanding and presenting this as a solution, without referring to the conditional nature of the resulting probabilities. Agree? Handy2000 (talk) 12:33, 13 April 2011 (UTC)
Oh dear Rick, we are back your misunderstanding of what probability means. I am not sure if we are allowed to discus this any more, perhaps it is OK on this page.
Rick, perhaps you could tell me what you understand the term 'probability' to mean.
Richard confirms above that there are two interpretations of probability, Bayesian and frequentist. Although they both should give the same answer to any exact question it is very confusing to flip interpretation in the middle of an argument. I am happy to discuss the MHP using either model but let us start with the Bayesian perspective first.
To a Bayesian, probability is a state of knowledge. That is the definition of what Bayesian probability means. In the Bayesian model, what is not known does not exist. That is not something that I have just made up, I am sure Richard will confirm it or you can confirm it using WP or any good text book. So, if the player does not know where the car is placed, or the host's door opening strategy, the probability of winning by switching is exactly 2/3. It may be that the car is always placed behind door 2 and the host always opens door 3 but, if we do not know that, it makes no difference to our (Bayesian) probability calculation, which depends by definition only on the information that we have, . The concept of average probability makes no sense, there is just 'the probability' which is based only on our state of knowledge. Martin Hogbin (talk) 11:26, 8 April 2011 (UTC)
Re: [quote deleted], No, that is not allowed here. I have placed a warning on your user page. I strongly advise against anyone else who may be tempted to respond in a nnon-civil manner to resist the temptation. (issue resolved) Guy Macon (talk) 06:19, 9 April 2011 (UTC)
I apologise and have struck out the offending comment and have replaced it with a more appropriate intro. I hope Rick will now respond to the content of my post, which explains much of the pointless disagreement here. Martin Hogbin (talk) 08:21, 9 April 2011 (UTC)
@Martin - we have discussed this to death and back numerous times before. In the spirit of a fresh start, I'll try once (only once) more.
Point 1: We're not talking about interpretations of probability here, but rather what probability is of interest. Is it P(win by switching), or P(win by switching|player picks door 1), or P(win by switching|player picks door 1 and host opens door 3)? These are each distinct probabilities whether we're using frequentist or Bayesian interpretations of probability, i.e. this difference has absolutely no relevance to this question. P(win by switching) is what I'm calling the "average" probability. It is what Carlton, and Grinstead and Snell (and others) call the success of a "switching strategy" (i.e. decide before even picking a door that you're going to switch). In a Bayesian sense (as you seem to be thinking about it) it is the probability of winning by switching knowing the rules of the game, before going on the show. In a frequentist sense, it is the limit of (number of players who switch and win, plus number of players who stay and lose) / (total number of players) as the number of players grows large. It is what the simplest solution (you have a 2/3 chance of initially selecting a goat, if your initial choice is a goat and you switch you end up with the car, so if you switch you have a 2/3 chance of winning the car) computes - and this probability is 2/3 (assuming the player's initial choice is independent of the initial car location, and the host must open a door showing a goat and must make the offer to switch) whether you're interpreting probability as a frequentist or a Bayesian.
Point 2: vos Savant's solution (which enumerates three equally likely cases assuming the player picks door 1, with switching losing in one and winning in 2) computes P(win by switching|player picks door 1) - and this probability is 2/3 (assuming the car is uniformly randomly located and the host must open a door showing a goat and the host must make the offer to switch) whether you're a frequentist or a Bayesian. As a Bayesian, this means you're deciding to switch after picking door 1 but before seeing which door the host opens. As a frequentist, it is the limit of (players who pick door 1 and switch and win, plus players who pick door 1 and stay and lose) / (players who pick door 1).
Point 3: If you're putting the point of the player's choice after the host opens a door, the state of knowledge of the contestant includes the BOTH the number of the door she originally chose and the number of the door the host opens. This means, for a frequentist or a Bayesian, the probability of interest (assuming the player has picked door 1 and the host has opened door 3) is P(win by switching|player picks door 1 and host opens door 3), NOT simply P(win by switching) and NOT P(win by switching|player picks door 1).
Point 4: Per Puza et al. (Teaching Statistics, vol 27 number 1, Spring 2007 - quite a nice paper BTW), the (clearly Bayesian) probability of winning by switching given the player picks door 1 and the host opens door 3 is (their notation):
which varies from 0 to 1 depending on what assumptions you make. Once again, this is the probability regardless of whether you're a Bayesian or a frequentist. As a Bayesian it means you don't know for sure your probability of winning by switching if you've picked door 1 and have seen the host open door 3 - unless you know or make some assumption about all of the probabilities involved. As a frequentist, it means (players who pick door 1 and see the host open door 3 and switch and win, plus players who pick door 1 and see the host open door 3 and stay and lose) / (players who pick door 1 and see the host open door 3) will approach something between 0 and 1, but you could do a number of experiments to determine the individual probabilities involved.
Point 5: (also per Puza et al.) Assuming what they (referring to Morgan et al.) call the "vos Savant" scenario (i.e. the car is uniformly randomly placed, the host must open a door showing a goat, and the host must make the offer to switch) they show this reduces to the 1/(1+q) answer, with q=P(H3|C1S1) (i.e. the probability the host opens door 3 if the car is behind door 1 and the player initially chose door 1). Once again, this is the probability regardless of whether you're a Bayesian or a frequentist. As a Bayesian it means you don't know for sure your probability of winning by switching if you've picked door 1 and have seen the host open door 3 (even with these assumptions), but you do know it's between 1/2 and 1. As a frequentist, it means with these assumptions that (players who pick door 1 and see the host open door 3 and switch and win, plus players who pick door 1 and see the host open door 3 and stay and lose) / (players who pick door 1 and see the host open door 3) will approach something between 1/2 and 1.
Point 6: Further assuming an uninformative prior (all values of this q are equally likely) Puza et al. end up with 2/3. This is the only thing in this whole discussion where Bayesian and frequentists might conceivably disagree - although even here there is a distinct equivalence. As a frequentist, this would mean uniformly randomly selecting q per trial that (players who pick door 1 and see the host open door 3 and switch and win, plus players who pick door 1 and see the host open door 3 and stay and lose) / (players who pick door 1 and see the host open door 3) will approach 2/3.
Rick. this is all fine, but not terribly important, since for most readers it is already exciting enough to learn that you can win with unconditional probabiity 2/3 rather than unconditional probability 1/3. You are talking about ways to prove rigorously that you cannot do better than 2/3. Some people might find this interesting, others not. One way to do this is through a derivation and analysis of the formula which you just showed us. Most readers of the wikipedia page will have no use for that whatsoever. Fortunately, another way is simply to remark that by symmetry the specific door numbers on the door chosen by the player and the door opened by the goat are irrelevant. You can't use the knowledge of the numbers on the doors to improve your strategy since they have no bearing whatsoever on the question whether the car is behind your door or the other closed door!
Here I am taking the point of view of almost all ordinary readers who will instinctively use probability in the subjective sense - which is perhaps the only way possible, given Vos Savant's question, since she doesn't tell us anything about the procedure whereby the door hiding the car is determined nor the door opened by the host. Her solution is perfectly adequate to her question, and the world-wide school-children's experiments confirmed the correctness in precisely the same way. Moreover there are academic sources which do it Vos Savant's way (Georgii, Gill, and no doubt others too). The fact that certain formulas are duplicated again and again by writers of elementary texts in probability and statistics in the chapter where they prove and illustrate Bayes theorem does not give these formulas some kind of academic superiority. Monty Hall problem is treated by textbook after textbook on game theory and optimization and there you will see completely different formulas. Fortunately, formulas are not necessary at all. Richard Gill (talk) 11:29, 10 April 2011 (UTC)
Intrusion: Rick, that was no "reply to Martin", even if you call it so. Martin's question was:
"Rick, perhaps you could tell me what you understand the term 'probability' to mean." – And he added "there are two interpretations of probability, Bayesian and frequentist. Although they both should give the same answer to any exact question it is very confusing to flip interpretation in the middle of an argument. I am happy to discuss the MHP using either model but let us start with the Bayesian perspective first. [...] In the Bayesian model, what is not known does not exist. [...] (Bayesian) probability calculation, which depends by definition only on the information that we have [...] there is just 'the probability' which is based only on our state of knowledge."
May I assume that your (pointless) "citing of sources" says that you do not even think of answering Martin's question? Not paying regard to Martin's distinction between "Bayesian" vs. "frequentist's" – "state of knowledge" vs. "all kinds of possibilities might exist in the world out there and could be assumed"?
I think Martin showed us a way that really could be of help for the article. If we just "knew" what we (contradictory) are talking about, and how to distinguish the different perspectives of the sources. Regards, Gerhardvalentin (talk) 18:42, 9 April 2011 (UTC)
Gerhard - I believe my points above are very clear. The overriding point is that the distinction between frequentist and Bayesian has nothing whatsoever to do with what we keep arguing about. -- Rick Block (talk) 23:48, 9 April 2011 (UTC)
Rick, in defining the different meanings of the term "probability", Martin just said to you "In the Bayesian model, what is not known does not exist", and he added "there is just 'the probability' which is based only on our state of knowledge." – Sticking on that denotation means that "Before and After" gives no "additional information". Whereas your accounting from the outset for "doors no. 1 and no. 3" denotes that you just converted the 'actual denotation' of the term "probability", and transformed it to include a lot of things that might be given in the world out there, converting the 'actual denotation' of the term probability to "what we actually do not know, but what could be assumed as any additional information that might exist in the world out there." I guess you do not like to stick for a moment to the denotation of "probability" that Martin actually was talking about. Please try. Gerhardvalentin (talk) 01:19, 10 April 2011 (UTC)
Gerhard - Perhaps there's a language barrier here, please try to understand what I'm saying. A player, Bayesian or frequentist, that knows she has picked door 1 and has seen the host open door 3 knows the identity of the two doors. For this player, Bayesian or frequentist, the probability of interest is the conditional probability P(win by switching|player picks door 1 and host opens door 3), NOT P(win by switching). Whether you're interested in the conditional probability or not has NOTHING to do with whether you're a Bayesian or a frequentist. In both interpretations there is the concept of conditional probability, and in both it follows Bayes Rule. The Puza et al. paper presents a completely general, Bayesian, analysis. The conditional probability of winning, to a Bayesian, is the expression above (from the Puza et al. paper). You don't need to take my word for this - read the Puza et al. paper. -- Rick Block (talk) 02:38, 10 April 2011 (UTC)
Rick, I do no see any language barrier in what Gerhard says but I am puzzled by your reluctance to address my point. I am not proposing a long discussion on philosophy, just that you make clear what definition of probability you are using in your arguments. If you do not do this is is impossible to discus the issues logically. As Gerhard says above, based only on a Bayesian interpretation of Whitaker's question the difference between the 'conditional' and 'unconditional' formulations becomes extremely pedantic. We cannot revise our probability estimate on seeing a specific door opened because we are given no additional information by the number number of that door.
In your response to Handy2000 above you talk of running an experiment. I therefore assume that you are talking about a frequentist interpretation of probability. If, purely for the sake of clarity of discussion, you would confirm this, I would like to explain where I think the weaknesses in your argument are. If you are not willing to do this (and I can see no logical reason for not wanting to) we are doomed to go on forever failing to understand the other's arguments.Martin Hogbin (talk) 21:00, 10 April 2011 (UTC)
If you're doing an experiment, you're either thinking of probability from a frequentist viewpoint or validating your Bayesian result. In my response to Handy2000 the interpretation of probability you're using makes no difference. Whether you're a Bayesian or a frequentist, the probability before the host opens a door is (typically) 1/3 per door. If you're a Bayesian or a frequentist and have chosen door 1 and have seen the host open door 3, after the host opens this door the probability of door 3 is definitely 0. If you're a Bayesian or a frequentist, the probabilities of the other two doors may or may not be the same as they were before the host opened door 3 - and you can express the after probabilities as conditional probabilities. The knowledge you've gained as a Bayesian is the identity of the open door (it is this knowledge that allows you to update its probability to 0). This same knowledge allows you (requires you) to update the probabilities of the other two doors. To avoid confusion, and so we can understand each other's arguments, I think we should use different names for these before and after probabilities. Both Bayesian and frequentist interpretations of probability use the language of conditional probability for this. I can see no logical reason for not using this terminology here. If you'd like to point out weaknesses and think the specific interpretation of probability makes a difference, please feel free to clarify what interpretation your comment pertains to. My assumption is it makes no difference, so unless it's clarified I'm assuming we're always talking about both. -- Rick Block (talk) 13:59, 11 April 2011 (UTC)
The term "conditional probability" is technical. It belongs to mathematics and it has a precise mathematical definition, namely as that definition which makes the chain rule true (prob of A and B equals prob of B times prob of A given B). Ordinary language does know about the intuitive concept of "the probability of A given B". Ordinary logic acknowledges the chain rule. Ordinary language and ordinary logic knows the concept of independence. Ordinary folk can agree that the identity (car or goat) of the object behind door 1 (the door chosen by the player) is independent of the identity (door 2 or 3) of the door opened by the host. This is intuitively true when we are using intuitive of subjective probability.
The difference between a subjectivist and a frequentist analysis of MHP is that for a subjectivist, the host *is* equally likely to open either door if the host has a choice (because we are subjectivist), while a freqentists doesn't know anything about the probabilities of his choices. Hence the frequentist either needs to leave them unknown (Morgan et al.) or has to be told "out of nowhere" that the host chooses randomly. The frequentist also has to be told that the door hiding the car is chosen at random. Alternatively, the frequentist finesses all discussion of unknowable host-side randomness by choosing his own door at random and switching. The frequentist chooses at random and switches and gets the car with probability 2/3. He doesn't know and doesn't care what is the conditional probability. The subjectivist chooses door 1 since that is their favourite number, and because he knows nothing about how the car is hidden, so it's equally likely behind any of the three doors. He'll switch and he knows he now has 2/3 chance of getting the car. He knows this probability, because he knows nothing at all. The frequentist doesn't know this probability, because he knows nothing at all! (I find that rather amusing). The frequentist who chose his door at random has got a strong guarantee that he'll get the car with (unconditional) probability 2/3 according to anyone's interpretation of probability. He's not interested in the conditional probability. The subjectivist has no guarantees whatsoever. Once off, he'll probably be OK, but how well he would do in many games is anybody's guess. And my subjective probability that't he'll get the car is less than 2/3 since I think Monty is smarter than he is.
Of course, subjective and objective probabilities satisfy the same rules, so you can check the results of calculations of subjective probabilities by actually doing many repetitions using "real" randomness. Richard Gill (talk) 16:43, 24 April 2011 (UTC)
In abstract mathematics, the conditional probability of A given B is *defined* by demanding that P(A and B)=P(A|B) P(B). In the real world, ordinary people who are frequentists define the probability of A given B by imagining many repetitions of the probability experiment in question, and imagining how often A occurs within that subsequence of repetitions in which B happened to occur. In the real world, ordinary people who are subjectivists define the probability of A given B by imaging a betting situation in which they are to bet on A versus not A, where the bet is only called if B happens to happen. So if you would be prepared to bet at most 5 Euros against my 2 Euro on A happening against it not happening where the bet is settled if and only if B happens to occur (otherwise we just keep our own money), then for you the odds on A (against not A) given B is 5 to 2, and your conditional probability is 5/7. It's a nice exercise to check that the frequentist's conditional probability satisfies the chain rule, and that the subjectivist conditional probability satisfies the chain rule. The different kinds of probability follow the same calculus. The same formal rules. When teaching probability to mathematicians we emphasize the formal rules and down-play the interpretation (since this has been a field of controversy for at least 300 years). Richard Gill (talk) 07:22, 25 April 2011 (UTC)
Devlin and others
As I wrote under "Course on probability", Gerhardvalentin tried to explain to me the reasoning of Devlin and others. This goes along the lines that the chance the contestant initially picks the car is 1/3 and as this chance is not changed when the door is opened by the host, the remaining 2/3 chance is on the two other doors. As the opened door has chance 0, the other must have chance 2/3. Gerhardvalentin calls it a way of understanding, not a full solution. I wonder why this yet is presented as a solution, rather than a way of understanding. Even then, as a way of understanding, it seems not a correct way of reasoning. Strangely also Rchard Gill - from his user page I discovered he is a professor in statistics - also reasons along these lines on the talk page of this article. I would say, that after door 3 has been opened by the host a new (conditional) probability law is governing the situation. Hence you can't say that the chance for the chosen door 1 does not change. In fact no chance whatsoever ever changes. You should argue that the new probability has the same value for the first door as the initial probability. And hence, as the new probability for door 3 is 0, door 2 must have a new probability of 2/3. I think this is a serious mistake in the arguing of Devlin and others. Handy2000 (talk) 09:23, 14 April 2011 (UTC)
Handy2000, I have to reject sharply your untrue and false statement about my words. I never said the thing you foist on me to have said. Stop immediately to spread your misinterpretation of what I have "told". Gerhardvalentin (talk) 10:32, 14 April 2011 (UTC)
Handy2000, I hope your probability teacher will also teach you Bayes' rule. Suppose we are interested in two possible scenarios or hypotheses (eg: car is behind door 1, car is not behind door 1). Our relative belief in the truth of those two scenarios can be measured by the odds of the one scenario relative to the other. For instance, initially the odds are 2 to 1 that the car is not behind Door 1. Bayes rule says that every time some new piece of info comes into your possession, you can find the conditional odds, ie the ratio of the conditional probabilities of the scenarios given the information, by just multiplying the prior odds by the so-called Bayes factor: the ratio of the probability of the information under scenario 1, to the probability of the information under scenario 2. You can do this in many steps. At each step, already incorporated information is included alongside the two scenarios, so the Bayes factor is always: ratio of probability of new information under scenario 1 and old info, to probability of new info under scenario 2 and old info.
Now think of the host first opening a door revealing a goat, and only afterwards telling the player (he was looking the other way) which door it was that the host opened (2 or 3). Two scenarios: car behind Door 1, car not behind Door 1. Initial odds: one to two. First info: a goat is revealed behind one of the two other doors, we don't know yet which. Bayes factor: one to one, since the host is certain to do this under either scenario. Odds that car is behind door 1 after opening of another door: 2 to 1 against. Now we get the info it was door 3. What is the chance it was door 3 when the car was actually behind door 1? 50-50 (host can choose, either choice equally likely in the eyes of the player). What is the chance it was door 3 when the car was actually not behind door 1? 50-50 (car is equally likely behind door 2 or 3, host opens 3 or 2). The Bayes factor is therefore 0.50 to 0.50 or 1:1. The final odds are still 2 to 1 against. Richard Gill (talk) 16:30, 14 April 2011 (UTC)
PS, here's the proof of Bayes rule. Let D stand for the info (data), let H and K stand for the two scenariors (hypotheses). We know P(H|D)=P(H & D)/P(D) = P(H)P(D|H)/P(D) by applying the chain rule (=definition of conditional probability) twice. Similarly P(K|D) = P(K)P(D|K)/P(D). Divide the one equality by the other. P(H|D):P(K|D) = P(H):P(K) * P(D|H):P(D|K). Posterior odds equals prior odds times likelihood ratio. Whoever thought of this was a genius. I have no idea who it was. Richard Gill (talk) 16:37, 14 April 2011 (UTC)
Richard - the confusion here is clearly the conditional sounding reasoning applied at the step where you don't yet know which door is open. Saying the probability of the open (but still unknown) door is 0 and reasoning that the probability of the remaining door is 2/3 is completely and entirely different from saying that the probability of door 3 is 0 and reasoning that the probability of door 2 must be 2/3. The first of these is valid, since the probability of the host opening "a door" is 1, so the "original" probability of door 1 is not changed. However, the second (applying this same reasoning to door 3 and door 2) is not valid, since knowing which door the host opens at least potentially affects the probabilities of all the doors (so you cannot assert without some sort of argument that the original probability of door 1 is unchanged).
Handy2000 - Delvin published a followup column [4] where he says "it may be easier to find the relevant mathematical formula and simply plug in the appropriate values without worrying what it all means". Smashing advice in this case. -- Rick Block (talk) 17:36, 14 April 2011 (UTC)
*My* argument only considered the question whether or not the door you first chose hides the car. I don't think there is any confusion there. Devlin in his second article admits that his first article jumped over one issue, namely the question whether or not *which* door is opened has any relevance. He was scared by his mistake didn't know how to get out of it and advised to fall back on insight-less accountancy. Pity he didn't see that Bayes' rule would help him over the difficulty. Not a difficulty at all if you actually know probability. Moroever it solves the problem of "what it all means" since it shows explicitly exactly what it all means. Richard Gill (talk) 21:07, 14 April 2011 (UTC)
Gerhardvalentin: Sorry I misunderstand what you've written on my talk page. May be be you can clarify what you meant.
RickBlock: Thank you for pointing to the Devlin article. Although he gives a correct solution in the latter part, he still maintains his wrong argumentation in the first part. Somehow it seems to me he tries to justify this error in the last part of his article.
Richard Gill: We learn Bayes' theorem if that's what you mean, and with this theorem we calculate the conditional probability. We do not speak off odds, but it looks as if you compare the probability on chosen door 1, opened door 3 and car behind 2, with the probability on chosen door 1, opened door 3 and car behind 1. This is, as far as I can see, equivalent to the calculation of the conditional probabilities.
My conclusion: Indeed is the Devlin solution, as well as Cecil Adam's, mentioned in the solution section under "Simple solutions" incorrect. And I'm surprised that this wrong way of reasoning is still presented as a solution. Handy2000 (talk) 23:26, 14 April 2011 (UTC)
I agree Handy2000, solutions whose reasoning is actually wrong ought not to be included! We had a long discussion about Devlin some months ago. Part of the reason he screwed up was because he apparently doesn't know about Bayes rule. I think that the other part of the reason Devlin stumbled was because if you want to give a "full solution" you have to take account of the chances that the host opens either door when he has a choice. And this forces you to think about what probability means. If you are a subjectivist then this is 50-50 because you don't know anything. If you are a frequentist you are stuck.
Regarding Bayes' rule and Bayes' theorem: of course both of these are trivialities and they are completely equivalent. Bayes' theorem is an application twice of the definition of conditional probability (the chain rule). Bayes' rule is obtained by dividing the formula for Bayes' theorem for two different events both conditioned on the same "given". You can derive Bayes theorem from Bayes rule by applying Bayes' rule to a collection of events which are mutually exclusive and exhaustive. So you can think of them as equivalent, both are trivial.
The difference is that Bayes' theorem is a formula, while Bayes' rule is expressed in words: posterior odds equals prior odds times likelihood ratio (aka Bayes factor). You have to remember Bayes theorem (a formula) by remembering how to prove it, while you remember Bayes' rule by remembering the words. And the words contains a collection of truly important concepts. You need the concept of odds and you have to introduce the concept of likelihood ratio or Bayes factor. It's a very simple thing and easy to remember. Bayes' rule shows how your uncertainty is changed by getting new information, showing precisely how this depends both on the prior knowledge and new information. And this dependence is the most simple form you can possibly imagine. Bayes' rule is a gift from the Gods.
My experience in talking about probability and statistics to lawyers and medics and journalists - very intelligent but mathematics-challenged people - is that you can explain Bayes' rule but not Bayes' theorem to them. You can explain the concepts. You can go on to do numerical examples. I must say this is a little easier for native English speakers, since the concept of odds is native, it's part of ordinary language. The English love gambling. On mainland Europe, there is not a word for odds in any of the languages I know. The Dutch for instance preferred to make money by offering bets to English sailors, and the concept of a Dutch book is a collection of bets which a better foolishly accepts and loses money whatever the outcome. The Dutch think that betting is a sin but making money from foolish people is a virtue.
Rosenthal in his article and book uses Bayes' rule. His experience too is that this is the way to explain conditional probability to ordinary people. It's such a pity Devlin didn't know Bayes rule. I repeat, he's not from probability and statistics.
I'm glad the explanation was useful. It could surely have been shorter, but it's difficult to predict how much detail, and where, is needed by the intended reader. Prize: a bottle of good wine, or equivalent value as Amazon.com gift token, or Paypal cash, or donation to your favorite charity... Richard Gill (talk) 17:57, 15 April 2011 (UTC)
Bayes' rule, not Bayes theorem. Who first said and used posterior odds equals prior odds times likelihood ratio, or if you prefer posterior is proportional to prior times likelihood. Stephen Stigler told me that he doesn't know, but he's sure it's only well into the 20th century. Richard Gill (talk) 18:22, 25 April 2011 (UTC)
I noticed that the way of arguing of Devlin is in fact the way the pictures alongside the simple solution section explain the solution. The pictures are wrong, and so is Devlin's arguing. Why is this still kept as a kind of solution, instead of an example of false reasoning? Sorry, forgot to login. Handy2000 (talk) 14:34, 5 May 2011 (UTC)
This discussion has been taken from the main talk page
I would be very interested to see the sources that support your statements above. Let us start with, 'in either a frequentist or subjectivist view the player's probability is a function of the host's preference between two "goat doors", and if the player doesn't know this preference the best we can say is the probability of winning by switching is an unknown value between 1/2 and 1' and consider the subjectivist (Bayesian) view. Martin Hogbin (talk) 16:01, 5 May 2011 (UTC)
As you are well aware Morgan et al. (which is based on a Bayesian view) says this virtually verbatim, as does Gillman (specific references in the article). Puza et al. [5] and Eisenhauer [6] as well. -- Rick Block (talk) 18:47, 5 May 2011 (UTC)
I found Probability to be incredibly helpful in understanding this. Is says,
"The word probability does not have a consistent direct definition. In fact, there are two broad categories of Probability interpretations, whose adherents possess different (and sometimes conflicting) views about the fundamental nature of probability:
Frequentists talk about probabilities only when dealing with experiments that are random and well-defined. The probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment. Frequentists consider probability to be the relative frequency "in the long run" of outcomes.
Bayesians, however, assign probabilities to any statement whatsoever, even when no random process is involved. Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, or an objective degree of rational belief, given the evidence."
An interesting aside: in my engineering work, we pretty much follow the Propensity probability interpretation. It's extremely useful for getting work done, but is it the way the real world works? Hard to tell. --Guy Macon (talk) 19:26, 5 May 2011 (UTC)
Thanks Guy for reminding us all what the Bayesian interpretation means. It means the probabilities are based on (or defined to be) a degree of belief given the evidence. If there is no evidence at all we must take all outcomes as equally likely. Thus, as Whitaker's statement does not tell us how the car was placed, how the player chooses, and how the host chooses, we must take all possibilities to be equally likely. That is pretty well the meaning of the Bayesian definition. Thus using a Bayesian definition, based strictly on Whitaker's statement alone, the answer is obviously exactly 2/3. It can be nothing else.
Where exactly in Morgan does it say that they are using a Bayesian interpretation in the main body of their paper?
In fact in their conclusion they say [my italics], 'In general, we cannot answer the question "What is the probability of winning if I switch, given that I have been shown a goat behind door 3?" unless we either know our host's strategy or are Bayesians with a specified prior '. Or to put this more clearly If we are Bayesians we can answer the question in general The appropriate prior, as they point out elsewhere, is a noninformative one. Martin Hogbin (talk) 21:42, 5 May 2011 (UTC)
In moving some comments to here they have become separated from other comments which may have been intended as responses to me. I did not move them because I was not sure if they were intended as suggestion on how to improve the article or responses to me. Please feel free to move them her if they were intended to be responses to me. Otherwise, I await response from Rick or indeeed anyone else. Martin Hogbin (talk) 13:00, 8 May 2011 (UTC)
Martin - what is your source for your claim that "if there is no evidence we must take all outcomes as equally likely" or that this is the meaning of the Bayesian definition? Regarding where Morgan et al. say they are using a Bayesian interpretation they (of course) don't specifically say this - however they're not talking about measurements taken from repeated experiments so given the dichotomy you seem to think exists between frequentist and subjectivist they are rather obviously on the subjectivist side. They make very clear "the probability" (without distinguishing frequentist or subjectivist - which might reasonably lead us to believe they think this distinction is not important) of winning by switching, given the clarifications vos Savant made explicit in her later columns (initial car placement is uniformly random, host must open a goat door and must make the offer to switch) is 1/(1+q). Bayesians can in addition assume a specific prior to arrive at a single numeric answer - but assuming a prior is an additional assumption just as much as assuming the host chooses uniformly between two goats. -- Rick Block (talk) 17:06, 8 May 2011 (UTC)
As the question gives us no information as to the way the host may choose a legal door door we naturally want an uninformative prior. The simplest and most obvious of the possibilities is a uniform distribution. This is the same prior that Morgan use on their Bayesian calculation earlier in the paper to get a (erroneous) single numeric answer. Martin Hogbin (talk) 17:57, 8 May 2011 (UTC)
Are we disagreeing about anything here? Given all of vos Savant's clarifications, the Bayesian answer is 1/(1+q). Additionally assuming an uninformative prior this works out to 2/3. -- Rick Block (talk) 18:20, 8 May 2011 (UTC)
Strongly disagree. Quite the contrary. You wrote "Additionally assuming an uninformative prior of q=1/2". This is an incorrect and misleading formulation / assertion.
"Additionally assuming an uninformative prior of q=1/2" in effect is no additional assumption at all. It's just a given fact (!) based on your "lack of knowledge". It never can be any "additional assumption". Such assertion is misleading and not correct.
I repeat: No "additional assumption". – As q is totally unknown, there is NO USE "to use" any superfluous "assumption". You don't need to assume. Just honestly cut down on what you really know.
As you actually don't know anything about the value of "q", nor will ever know but in "using q", though, you are strictly forced to take q as 1/2. That's a fact. Once more: In case you superfluously should "like" to use "q", you are forced to attribute the value of 1/2 to q. Period.
And if you should "like" to attribute any other value to "your q" for the actual game show in question, then you will just get what you "like" to get. Quite outside the MHP. Never addressing the famous "MHP-question" anymore, but just "testing" Bayes' capability to come up with any "assumptions and illusions" you like to assume. Unnecessarily for the MHP. Quite outside the famous MHP-question.
Because, as long as you do not know the exact "known" host's bias and its direction, it doesn't matter whether the host has some bias or not. So it's superfluous to speculate about that. – On the other hand, if you like to use Bayes and any "q", you are free to speculate and to do what you "like". But it's an illusion to think that you could get something reasonable to answer the famous question on the actual game show in question. On no account. You will just always get just the illusion of "what you like to get", and that will be and forever will remain within the given strict bonds. Not addressing the famous question about the actual game show in question anymore, but just addressing Bayes' theorem, and just showing that conditional probability can be applied wherever and whenever you want to, whatever unproven "assumptions" you should like to include. Outside the famous question about the actual game show in question. But in any case obviously remaining within the strict bonds shown and given by just simply asking "what if he always" and "what if he never": Probability to win by switching will remain within the range of at least 1/2 to 1. Read the actual sources. Gerhardvalentin (talk) 19:13, 8 May 2011 (UTC)
We're talking here specifically about what Morgan et al. say, which I believe is exactly what I've said above. If you'd like to bring up what other sources say feel free, but please name them. -- Rick Block (talk) 19:20, 8 May 2011 (UTC)
Rick, I think you are talking about a different bit from me. I am referring to the bit at the bottom of the left column on page 286 which starts, 'This provides an excellent opportunity to bring in the Bayesian perspective...', it then mentions a noninformative (uniform) prior, and ends with a single figure answer of 0.693. This was the figure which Nijdam and I showed should have been 2/3. This is the bit that Morgan refer to in their conclusion. You will not that it gives a single value answer to the question. Martin Hogbin (talk) 21:57, 8 May 2011 (UTC)
We're talking about the same thing. The 1/(1+q) answer is a perfectly Bayesian answer. What they're saying is that Bayesians can additionally obtain a single numeric answer by assuming a particular prior. With a noninformative prior their answer is .693 (which Puza et al., or your recent letter, correctly show is actually 2/3). Per the top of the next column, with different priors Bayesians can arrive at other answers - but never less than 1/2. Again, are we disagreeing about something here? -- Rick Block (talk) 22:29, 8 May 2011 (UTC)
We are still talking about Morgan et al., right? Are you asking me to make a conjecture about something that this paper doesn't say, or are you claiming this paper says anything other than "the probability" (even the Bayesian probability) is 1/(1+q) 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)? -- Rick Block (talk) 01:57, 9 May 2011 (UTC)
I am asking you what other prior there could possibly be other than a noninformativce one, considering we are given no information. 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)? -- Rick Block (talk) 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. ISBN1602063281) and has remained so ever since.
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. Martin Hogbin (talk) 08:45, 10 May 2011 (UTC)
Please note I am not a wordsmith and I am just a layman fascinated by the mathematics quoted in the above article which seems 100% correct in the context used.
However to me with a basic understanding of mathematics the original choice given you have a 1 out of 3 chance of "winning" is not correct.
The actual chance you have of "winning" are greater at the outset of 1 out of three because if 1 of the 3 choices of "winning" is already arranged to be removed prior to your choice of a "pick" of 1 out of 3 (if you are incorrect in picking a winner at the beginning) then the mathematic formula’s quoted surely is incorrect
Tony Gray
New Zealand —Preceding unsigned comment added by 114.134.7.115 (talk) 22:40, 6 May 2011 (UTC)
The statement means that right at the start, before any doors have been opened, you have a 1 in 3 chance of picking the car. It does not refer to your probability of going on to win the game. If you are smart, this is 2/3. Martin Hogbin (talk) 22:54, 6 May 2011 (UTC)
Lundy's Solution
Mathematician Nicholas Lundy suggests numbering your goats or making them different colors to reveal that there are 6 possibilities.
C -|- G1 -|- G2
C -|- G2 -|- G1
G1 -|- C -|- G2
G1 -|- G2 -|- C
G2 -|- C -|- G1
G2 -|- G1 -|- C
Show a goat in any room and you remove two possibilities (the ones with cars) and therefore always have 1/2 chance of winning by switching or staying. — Preceding unsigned comment added by Camtin (talk • contribs) 04:05, 25 June 2011 (UTC)
This explanation appears to assume that the host may sometimes open a door with a car and/or that he may sometimes open the already-selected door. A significant amount of confusion about the Monty Hall problem arises due to the ambiguity of the problem statement. I don't know what 141.161.133.130's "wrong goat in the revealed door" comment means, but he/she very likely has the "standard" version of the problem in mind, which instead assumes that the host always opens one of the non-selected doors, and never opens the door with the car. I believe that this illustrates why it is important to make such assumptions explicit. 139.99.16.28 (talk) 19:17, 18 October 2011 (UTC)
Simple explanation! - Really
Ok so it seems to my lay self that there is a lot of over complication of a pretty basic problem.
I will now try to explain how the it works in the simplest possible manner.
1. Pick a door (chances of getting a goat = 2/3)
2. A goat gets revealed, you know that the door you chose to begin with most likely holds a goat, and that one door has the car
3. Switch, since you're pretty sure you have a goat (2/3 chance) you can be equally sure that the other door is a car.
4. Therefore after switching you have 2/3 chance that you have the car.
If that seems confusing, think about the problem with 1,000,000 doors and 999,999 goats and 1 car.
1. Pick a door, you almost definitely pick a goat
2. All goats but one are removed, leaving one door with the car and one with the goat.
3. You basically know that you chose a goat in the first round, so you obviously change your choice, making it almost guaranteed that you pick the car.
Gerhardvalentin, indeed assuming that you have a small insider's advantage, AND assuming almost symmetric host the conditional probability is close to 2/3. Einverstanden?RocksAndStones (talk) 18:45, 6 August 2011 (UTC)
Nein, RocksAndStones, not agreed, for that's the dodgy matter of dispute. Maths teachers like sundry "assumptions" as waste conditions and, according to their input forever will get then the adequate results. So that point is moot. For the "standard problem" all of this is not relevant to get the "only correct answer". You are right, assuming to have insider's advantage will result in probability to win by switching for the most extreme case of at least 1/2 (factor 2/3) and at the same time(!) likewise to 1 (factor 1/3). So always varying around 2/3.
What can conditional probability and superfluous assumption contribute to getting the only correct solution, the only correct answer, the only correct decision? Nothing but to confirm once more that nothing can be better than to switch here and now, in this one special game the question is about. That's no news at all.
Superfluous additional assumptions and their handling in conditional probability theory belong to the realm of probability theory. Solving the paradox of the MHP is easy, as anyone can clearly find out that two doors have double chance. And if you like you can add "have exactly double chance, because probability to win by switching will be at least 1/2 (factor 2/3) to even 1 (factor 1/3) in this one special game the question is about". That's all what conditional probability theory can contribute. So no forum here to teach and to learn conditional probability theory and to teach and to learn application of Bayes' theorem. That belongs to the maths' forum. Regards, Gerhardvalentin (talk) 11:14, 7 August 2011 (UTC)
Gerhardvalentin, Perhaps, V R speaking about the same thing, so a "scheinbar" disagreement results from wording, instead of writing formulas. What I meant to say was: if Connie is a bit clairvoyant and can guess the worst door with chance say 0.35, her win probability is close to 2/3 no matter what Monte does. If we understand standard assumption as the uniform distribution, then always-switching gives 2/3 (no matter what Monte as door-opener does), although you really need an additional argument why Connie cannot do better (no matter what Monte does). That an argument is needed, can be seen with 4 doors, when Monte can help win with 100% probability by signaling through sequence of 2 doors to reveal. Now, as far as I know, Olle Haggstrom and Richard were first to focus on this Holy Grail argument. Combinatorially, the Holy Grail result means that if Connie's decision policy is not allowed to exploit the actual location of prize (which formally means she "does not know" where it was hidden) then she cannot win in all three cases out of three under any circumstances. I agree, that the site is not a forum for teaching the conditional probability. However, from the viewpoint of Markov decision processes (and other standpoints), the conditional probability is a natural quantification of the advantage of switching. Although the existence of cond. prob. is unnecessary assumption and its computation adds nothing to the right decision, it is a valuable thing. In particular, if the distribution of the prize is not uniform, and the first quess was not done in optimal way, you may decide for sticking on base of conditional probability. I do not object the cond. prob. approach, it only needs to be given a proper place as a tool to quantify the advantage.RocksAndStones (talk) 12:14, 7 August 2011 (UTC)
Rocks&Stones, R U interested in my view? Again, I M not fully einverstanden. Let us stick to the standard version, where the host not just offers to switch if you should have chosen the prize, etc... Let us stick to the standard version, where unproven "assumptions" of clairvoyance and opaque doors never will influence the decision asked for.
Yes, Richard's papers confirm that indeed there never will be any better decision than to switch here and now in this very special game the question is about, and moreover that there never will be any better decision than to switch in every one of these games, if this special game should ever be on stage in real life. I would like to show the chances for switching of 2/3 in "odds form" just at the beginning, also. Just to help convincing the reader. And later that the chances forever will remain within the very strict range of (at least!) 1/2 to even 1, and without given better knowledge forever exactly 2/3. But I M strictly against the totally unproven brazen assertion that the MHP is incomplete without the conditional calculus showing a variable representing the totally unknown "host's special behaviour" in this special game. Such variant should be treated where it belongs, in some maths forum, for it is not needed within the MHP article, and it does not belong to the MHP article. It is an interesting maths aspect, but is without any relevance to the question asked for and to the correct decision to switch. And the fulminant nonsensical history can be shown in the "history" section. Regards, Gerhardvalentin (talk) 13:57, 7 August 2011 (UTC)
Gerhardvalentin, sure I am interested in your viewpoint and appreciate it. Let us review your statements word-by-word, to avoid ambiguity. Richard's papers confirm that indeed there never will be any better decision than to switch here I agree, vorausgesetzt uniform distribution, this was shown by Olle Haeggstroem (Lehrbuch Streifzuege der W-Theorie) and Richard by somewhat different arguments. I would like to show the chances for switching of 2/3 in "odds form" just at the beginning, also. If you mean unconditional odds - I agree this is easy and can be done at the very beginning, but if you mean conditional odds you need the assumption of symmetric host for the case when he has a freedom of choice. And later that the chances forever will remain within the very strict range of (at least!) 1/2 to even 1, and without given better knowledge forever exactly 2/3. I agree, under the assumption that host tosses a perhaps biased coin when he has the fredom of choice.But I M strictly against the totally unproven brazen assertion that the MHP is incomplete without the conditional calculus showing a variable representing the totally unknown "host's special behaviour" in this special game. I agree with you at this point completely.Such variant should be treated where it belongs, in some maths forum, for it is not needed within the MHP article, and it does not belong to the MHP article. It is an interesting maths aspect, but is without any relevance to the question asked for and to the correct decision to switch. And the fulminant nonsensical history can be shown in the "history" section. Well, the subject belongs to the MHP site, and the "history section" is indeed to where it belongs. Now, I have a straightforward question to you: have you ever taken 15 minutes to understand the dominance argument, explained since March at least half a dozen of times on these discussion pages?RocksAndStones (talk) 18:01, 7 August 2011 (UTC)
Thanks for your questions, RocksAndStones, I try to reply as follows
Uniform distribution of the three objects at the beginning is only affordable if the guest does not choose her door randomly. If she has no information on the distribution of the three objects and chooses her door uniformly at random, then a uniform distribution is not even necessary.
Richard's papers confirm that nothing can beat switching in each and every single game, no matter (!) whether "uniform distribution" or "biased host", and even more of that: The more biased, the better! In the end it never matters whether the host's method of showing one goat can tell you sth to revise the odds on the door first selected, or not! (By the way see also the "Total Solution" in "The Three Doors" by Sasha Gnedin May 31, 2011, e.g.: You can answer the question without needing probability at all). We should not stick on dated sources that do not even address the solution of that veridical paradox, but just use the "MHP" to present their competence and expertise in conditional probability theory. And textbooks written to teach conditional probability theory, using the MHP as a welcome example. No room here for such sources, showing correct calculi and explaining probability theory, but not the famous paradox.
The prior and the conditional probability can easily be shown in odds form, using Bayes' rule:
Prior odds:
Prob(C=1) : Prob(C=2) : Prob(C=3) = 1 : 1 : 1 (as long as no other information is available)
Likelihood of location of car C=1,2,3, given data Q=3:
The posterior probabilities are therefore 1/3, 2/3, 0 since they must add to 1, and must be in the just mentioned ratios.
This computation is easy to generatlise to the situation where the host may cause asymmetry. As long as the initial distribution of the location of the car remains uniform, the posterior odds are
As Prob(Q=3|C=1) can conceivably be anything between 0 and 1, the posterior odds on the location of the car being door 1 to door 2 is anything between 0 : 1 to 1 : 1. All we can say is "switching will never hurt".
As you can see, the assumption of a symmetric host – for the case that he has all the freedom of choice in the world – is not necessary at all, as switching will give you the car on average in 2 out of 3, and will never hurt.
The host could be "assumed" to be (extremely) biased to open only the door with the lowest number e.g., whenever he can, i.e. if a goat is behind his preferred door. He can open his preferred door in two of three cases:
if in 1/3 he has got two goats to show and switching will LOOSE the car
if in 1/3 he has got the car and one goat, and the goat being behind his preferred door, and switching will WIN the car.
Whenever he opens his preferred door in 2 out of 3 cases, you know that the chance to win by switching is 1/2 (and never ever less).
but if in the last 1/3 he has got one goat and the car, but the car being behind his preferred door, he has to open his avoided door with the higher number, and in this 1 out of 3 -case switching is very likely to win for sure.
Thank you, had a look there, and that's exactly the principle and affirmation to where any vain, though prominent, shiftless failed attempt imho irrevocably belongs to. Gerhardvalentin (talk) 16:43, 8 August 2011 (UTC)
Conditional probability and the MHP
The MHP is a good example in teaching and learning conditional probability theory. But what is weight of conditional probability theory for solving the world famous paradox called "Monty Hall paradox"?
What "variant" of the MHP? It's about a "fair variant". Special presuppositions: at best "none".
The distribution of the three objects behind the three doors is of no relevance as long as it is unknown to the candidate who must assume 1/3:1/3:1/3.
Special preferences of the candidate in choosing one door are not relevant, nor the special behavior of the host in opening a door, as long as he observes the rules that say he has to open one door showing a goat and he has to offer his still closed door as an alternative. If he should dispose of two goats it's completely up to him which door to open. Because even if he should be extremely biased, to switch never can be of disadvantage to the candidate.
Whether the host should act extremely biased, or biased to a smaller extent, or not biased at all is without importance for this famous paradox. Morgan et al. however succeeded in luring almost everyone interested in this paradox on the wrong track. Conditional probability is OK as long as it is not misused to lure in the wrong direction in fixing the view on insignificance. The current scientific literature has helped here to provide clarity.
What is "known"? The first known fact is the overall chance of winning by switching of 2/3. The chances of the "door first selected" to the "alternative still locked door" = "1/3 : 2/3". Agreed? And we can visualize an extremely biased host who, in opening his favored door, signals that switching is of no disadvantage for the candidate ("1/2 : 1/2" in two out of three) but who - in the last third of cases, in exceptionally opening his unfavored door, sigals that switching is very likely to win the car ("0/3 : 3/3" or "0 : 1").
Once more: We know the "hypothetical" assumed worst case of "1/2 : 1/2" and the assumed best case of "0 : 1". Just assumptions, but a valid demarcation of the intermediate range. Not a wide range, but a fixed, closely limited range that categorically excludes that staying could ever be of advantage. And, besides assumptions, the only thing we really "know" is the exact overall probability of "1/3 : 2/3". So we can see that it is absolutely pointless to try to define hypothetical variants within this fixed, closely limited range, that never can advise that staying could be of advantage. Trying to define hypothetical variants by using conditional probability theory is absolutely meaningless. We know from the outset that this fixed, closely limited range forever discards staying, because, in the MHP, staying never can be the better decision.
Conditional probability theory can never leave this fixed, closely limited range that – at the outset – categorically excludes that staying could ever be of advantage, and so never can be an option within the MHP.
It should be shown to the reader that in the MHP it is fruitless to use Bayes in search of variants within this narrow fixed range that, just from the outset, forever imperatively discards staying. Gerhardvalentin (talk) 22:07, 13 September 2011 (UTC)
Game Theory
Gnedin has shown in a solution using Game Theory that all strategies that consider "staying" as a possible option, and not imperatively decide "unexceptional switch" have to be discarded just from the outset. His solution shows that conditional probability, given the door numebers, or any assumptions regarding some special behavior of the host is completely irrelevant for the MHP. See Gnedin: The Doors. So I just said on Nijdam's talk page about the host's strategy in his section "Game Theory":
And yes Nijdam, the host indeed is able to determine his own strategy just as he should like. That's completely up to him.
Moreover, it is completely irrelevant whether the candidate knows about such strategy, or not.
Please note that any host's strategy is completely irrelevant for the decision asked for.
All sources agree that staying never can be of advantage for the candidate and never can be the "better" decision.
So any thought that considers "staying" is dominated and can be discarded just from the outset, as game theory can show and already did show.
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.
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. 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? 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. -- Rick Block (talk) 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. 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. -- Rick Block (talk) 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.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!! 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. -- Rick Block (talk) 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.
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. -- Rick Block (talk) 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.
"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.
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)? -- Rick Block (talk) 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.
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? -- Rick Block (talk) 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.
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. -- Rick Block (talk) 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?
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? -- Rick Block (talk) 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.
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. -- Rick Block (talk) 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.
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). -- Rick Block (talk) 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.
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.
Please consider carefully that basic interpretation of the 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, 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. Gerhardvalentin (talk) 14:12, 17 December 2011 (UTC)
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.
“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,
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. “
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.
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 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 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. -- Rick Block (talk) 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.
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.
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).
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.
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. Richard Gill (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.
Are you interested in talking about this, or are you absolutely convinced you're right and nothing anyone says can change your mind? -- Rick Block (talk) 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. — Preceding unsigned comment added by Kwinzenried (talk • contribs) 03:31, 7 January 2012 (UTC)
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? -- Rick Block (talk) 04:34, 7 January 2012 (UTC)
I do not agree that vos Savant's solution is flawed, but that's another matter.
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?
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).
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.
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.
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.
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!
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!
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", "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. Richard Gill (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.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.
One does not have to do conditional probability calculations in order to solve MHP adequately. Careful considerations of strategy do the job, too. Richard Gill (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. -- Rick Block (talk) 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.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). -- Rick Block (talk) 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).
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. -- Rick Block (talk) 20:23, 26 January 2012 (UTC)
The Monty Hall Problem — solved visually.
Legend:
[W]
Win (a car) if you choose this option.
[L]
Lose (get a goat) if you choose this option.
[X]
Option chosen.
[ ]
Option not chosen.
[\]
Not an option, because host of game show excludes it in stage 2.
:W
Result of scenario is a Win for contestant.
:L
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:
[W][L][L]
[W][L][L]
[W][L][L]
:Result
[X][ ][ ]
[X][\][ ]
[X][\][ ]
:W
[X][ ][ ]
[X][ ][\]
[X][ ][\]
:W
[ ][X][ ]
[ ][X][\]
[ ][X][\]
:L
[ ][ ][X]
[ ][\][X]
[ ][\][X]
:L
[L][W][L]
[L][W][L]
[L][W][L]
:Result
[X][ ][ ]
[X][ ][\]
[X][ ][\]
:L
[ ][X][ ]
[\][X][ ]
[\][X][ ]
:W
[ ][X][ ]
[ ][X][\]
[ ][X][\]
:W
[ ][ ][X]
[\][ ][X]
[\][ ][X]
:L
[L][L][W]
[L][L][W]
[L][L][W]
: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:
[W][L][L]
[W][L][L]
[W][L][L]
:Result
[X][ ][ ]
[X][\][ ]
[ ][\][X]
:L
[X][ ][ ]
[X][ ][\]
[ ][X][\]
:L
[ ][X][ ]
[ ][X][\]
[X][ ][\]
:W
[ ][ ][X]
[ ][\][X]
[X][\][ ]
:W
[L][W][L]
[L][W][L]
[L][W][L]
:Result
[X][ ][ ]
[X][ ][\]
[ ][X][\]
:W
[ ][X][ ]
[ ][X][\]
[X][ ][\]
:L
[ ][X][ ]
[\][X][ ]
[\][ ][X]
:L
[ ][ ][X]
[\][ ][X]
[\][X][ ]
:W
[L][L][W]
[L][L][W]
[L][L][W]
: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 [W] and the [\] align do not exist because the host would never reveal nor exclude the winning door as an option in stage 2.
Scenarios where the [\] and [X] 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.
