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)
