Talk:Monty Hall problem/Archive 20

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Falk to the rescue

It is always interesting to actually read the sources claimed to criticise the simple solutions. Here is what Falk says (with my bolding and comments in square brackets) about the symmetry argument:

The symmetry condition holds in the original version of the problem, where P(T) = P(H) = l/3, and so does the invariance of Dick’s probability for pardon. It should be pointed out, however, that the equality P(T) = P(H) is not by itself a sufficient condition for securing no change in the target probability (nor is it a necessary condition, for that matter). The derivation of this condition has depended crucially on the specific triplet of conditional probabilities (likelihoods).

Shimojo and Ichikawa (1989) are right in claiming that the symmetry condition [car is originally placed with equal probability behind doors 2 and 3 in the MHP] is sufficient for invariance of Dick’s probability provided we assume the warden is indifferent, when Dick is to be freed, about naming Tom or Harry [host chooses uniformly between goat doors in the MHP]. If we drop that assumption, the symmetry condition is no longer sufficient. Suppose the situation is changed only in that the warden will always answer “Harry” when he has a choice between Tom and Harry. Because the warden does now discriminate between Tom and Harry, P(h D) changes from 112 to 1. 1 Replacing the old value by the new one in formula (1) yields P(D h) = l/2. 1 Dick’s survival probability has thus changed (from l/3 to l/2) in spite of the equality of P(T) and P(H), thereby refuting the generality of the symmetry heuristic [if the host does not choose goat doors equally].

If, however, we keep the initial equal probabilities and insist on assuming the warden is unbiased (i.e., P(h D) = l/2), then everything is symmetrical with respect to Tom and Harry, and naming either of them cannot affect Dick’s chances. The argument is now impeccable. That set of conditions, which can be labeled the complete symmetry, is surely sufficient (though not necessary) for obtaining no change in Dick’s survival probability. All the requirements of complete symmetry are satisfied in the classic three-prisoner story supplemented by the assumption of an unbiased warden.

Translated into the MHP equivalent the section I have bolded above would be:

If, however, we keep the initial equal probabilities with which the car was placed and insist on assuming the host is unbiased (i.e., P(H=2) = P(H=3) = l/2), then everything is symmetrical with respect to Doors 2 and 3, and opening either of them cannot affect the players chances of winning by switching. The argument is now impeccable.

So Boris' suggestion of using the powerful, well respected, and often used mathematical principle of symmetry to simplify the solution to the MHP turns out to be supported by a reliable source. No one seems to have noticed that before. Martin Hogbin (talk) 18:04, 11 July 2010 (UTC)

Um, what? This is exactly what Selvin says in his second letter. No one has ever said that if the host's choice is unbiased then the problem is not symmetrical. Quite the contrary. Specifying, or assuming, the host's choice is unbiased is precisely what makes the problem symmetrical. The problem we keep having is that you want to simply assume the problem is symmetrical - without any argument for why this is so. -- Rick Block (talk) 18:20, 11 July 2010 (UTC)
In the standard problem formulation the host opens a goat door uniformly at random, thus the problem is symmetrical thus the probability the car is behind the originally chosen door is unchanged thus simple solution is fine. Martin Hogbin (talk) 20:30, 11 July 2010 (UTC)
Which wording of the problem says "the host opens a goat door uniformly at random" - and, which published simple solution is based on this? -- Rick Block (talk) 21:09, 11 July 2010 (UTC)
The K&W statement in the article states that 'If both remaining doors have goats behind them, he chooses one [uniformly] at random'. All the simple solutions are based on the principle that opening a goat door does not affect the probability that the car is behind the originally chosen door. We spent a long time discussing this, remember? Martin Hogbin (talk) 21:47, 11 July 2010 (UTC)
??? - Rick, just read the first words of the article. The words:
"Suppose you're on a game show ..." are immediately followed by the statement:
"Although not explicitly stated in this version, solutions are almost always based on the additional assumptions that the car is initially equally likely to be behind each door and that the host must open a door showing a goat, must randomly choose which door to open if both hide goats, and must make the offer to switch..."
You didn't read the article? So, just read its first words, at least.
The original "Ask-Marilyn-question", not mentioning any presumption regarding the host's behavior, evidently was implying "no deviant host's behavior whatsoever at all", i.e. "No host's bias". This wasn't even into anyones mind. Obviously no need to mention this.
But, just that fact not having been expressly named, Morgan et al., rather crooked, felt free to allowing for a "Suppose he is biased", and a new deviating variant was born, applauded by (some) crowd.
In reporting the sources it is important to clearly distinguish those two (or more) variants, never presenting confusing mixtures. Granted: Not everyone is capable of doing it, as years of discussion show. More precisely: Even years after the starting of the attempt to weed out: Many (?) or few are still unable to. But just let's try it, yet. --Gerhardvalentin (talk) 22:22, 11 July 2010 (UTC)
OK. The K&W statement says this.
Yes, and the article says, 'According to Krauss and Wang (2003:10), even if these assumptions are not explicitly stated, people generally assume them to be the case'. We have all agreed that this is the standard version of the problem, supported by many reliable sources. Martin Hogbin (talk) 09:15, 12 July 2010 (UTC)
Which simple solution (please be specific) is based on this version? Rick Block
All the simple solutions, including the two (vos Savant's and 'combining doors') given in this article. These are all based on the assumption that revealing a goat does not change the probability that the car is behind the originally chosen door. Martin Hogbin (talk) 09:15, 12 July 2010 (UTC)
I'll make it easier. Which simple solution says anything remotely like "since the host chooses between two goats randomly, the problem must be symmetric", or, even easier, which simple solution says anything at all related to how the host chooses between two goats - random or otherwise?Rick Block
The problem statement in this article tells us that the host chooses evenly. Falk tells us that, in that case, this does not change the probability that the car is behind the originally chosen door. The simple solutions given in this article are based on that assumption, as you have pointed out many times. Martin Hogbin (talk) 09:15, 12 July 2010 (UTC)
On the flip side, there are plenty of simple solutions that say things like "because we already know the host can and will open one of the two unchosen doors revealing a goat, doing so does not change the probability of the player's initially selected door hiding the goat". Rick Block
Correct. An in case of the standard version of the problem, as agree by everyone here, Falk confirms that that statement is correct. Martin Hogbin (talk) 09:15, 12 July 2010 (UTC)
And, we know there is a reliable source (Falk) who says that this statement is confusing cause and effect. Saying "because A, then B" means A logically implies B. We all agree A is true. We also all agree if A is true and the host chooses between two goats randomly, then B is true. But this is entirely different from agreeing that A implies B. What Falk is saying is that the statement in these solutions claiming A implies B is wrong. She goes on to show using a slight variant A may be true but B not (demonstrating A by itself does not imply B). If you object to calling it "unsound" pick another word. False? Incomplete? -- Rick Block (talk) 23:14, 11 July 2010 (UTC)
If the host is known to choose goat doors non-randomly, that is correct but in the same paper Falk calls a the same argument 'impeccable' when applied to the case that the host (warden) chooses evenly. Martin Hogbin (talk) 09:15, 12 July 2010 (UTC)

Rick Block, you've been arguing this nonsense for 6 years now. Every one of your worthless arguments has been refuted via reliable sources. It's about goddam time you knocked it off. Your perverted POV should no longer strangle the Wikipedia MHP article and talk pages. Glkanter (talk) 23:18, 11 July 2010 (UTC)

@Martin - Your responses above are truly absurd. We have all agreed the K&W statement of the problem is the standard interpretation, not the standard version (there's a big difference). As far as I know none of the simple solutions reference the K&W version or say anything at all about how the host chooses between two goats. The reasoning they all use ONLY applies if the host chooses randomly in this case, but they never say anything about it. Nobody has ever said it's impossible to justify these solutions, but you can't justify solutions after the fact by saying something they themselves do not (this is WP:OR). And, concerning the specific point we're talking about here (whether basing a solution on the assumption that the host's action can affect the probability that the player's initial choice of door hides the car is sound), you're completely twisting what Falk is saying. She indeed says a symmetry argument is impeccable, but only if this argument is justified by saying the host picks randomly. She is actually criticizing sources for NOT making this argument. -- Rick Block (talk) 13:57, 12 July 2010 (UTC)
Vos Savant did make clear that she had taken the host to choose goat doors uniformly at random thus her solution is completely correct for the assumptions that she made. Selvin also specified that the host chose uniformly at random.
Vos Savant did not explicitly make the symmetry argument in her solution but no source criticises her for that specific failing either. Morgan criticise her for failing to take the host goat door choice into account. So, if yo really want to be fussy the vos Savant solutions stands. Martin Hogbin (talk) 21:53, 12 July 2010 (UTC)
I suspect I know what you're talking about, but just to make sure we're on the same page here where exactly did vos Savant make it clear she had taken the host to choose between goat doors uniformly at random? I know the Selvin reference since I added it to the article in the first place. Is the vos Savant reference one of the references in the article? Better yet, where is her revised solution published that says anything like "because the host can always open a door showing a goat and he chooses between two goats uniformly at random, the probability that the player's initially selected door hides the car remains unchanged when the host opens a door"? Moreover, if this is what she was thinking don't you think it's a little odd that she didn't mention this in her otherwise explicit-in-every-detail experimental procedure described in her 3rd Parade column (copy here)?
From a sourcing perspective, if you want to present a simple solution that explicitly says it's based on the assumption that the host chooses between two goats randomly I think you need to actually find one that says this. Selvin's is a possibility, but I think it's obvious that most simple solutions say nothing about this. To be clear, no one is saying it's not true that the probability of the player's door doesn't change if the host chooses between two goats randomly, just that the sources presenting simple solutions don't (at least not typically) say this. In particular, many of them appeal to what Falk calls the no-news intuition without saying anything about an essential condition that makes this argument valid. We can certainly construct a valid argument from bits and pieces here and there, but that would be wp:synthesis. The fact that many sources overlook this is a notable omission. If your primary goal is to further our readers' understanding of the problem, surely you must agree that this should be mentioned - presumably the closer to where the omission occurs the better. -- Rick Block (talk) 01:40, 13 July 2010 (UTC)
  1. The contestant's State of Knowledge is the key factor here. It's a game show. Indifference makes that 50/50. vos Savant and the others have no obligation to spell this out, as it is so obvious.
  2. Carlton's (and others') simple solution and decision tree show that the contestant's SoK and/or how the host chooses between goats is immaterial. It is not restricted to just 50/50. It can be anything, and the contestant should still switch. Which is both Whitaker's and Selvin's actual questions. Why do you continue to ignore or argue these points? Glkanter (talk) 04:14, 13 July 2010 (UTC)
Falk's point is that the statement "because we already know the host can and will open one of the two unchosen doors revealing a goat, doing so does not change the probability of the player's initially selected door hiding the goat" (that appears in many simple solutions, including vos Savant's) is based on a very strongly held intuition but does not mention an essential condition required for this to be a true statement. What we're talking about is how to incorporate this point into the article again (since Martin deleted the paragraph making this point shortly before the article was protected). Are your points #1 and #2 supposed to be related to this discussion somehow? If not, can I suggest you start a new section rather than continue to try to wp:disrupt this discussion? -- Rick Block (talk) 04:49, 13 July 2010 (UTC)

It's very simple, Rick. As an editor, you are giving greater merit to Falk's statement. Obviously, the reliable sources that published the various simple solutions she is criticizing don't agree. To include her criticism in the Sources of Confusion section (no matter what the wording) keeps her POV, and by extension, your POV, that the simple solutions are somehow inadequate in the article, in violation of the Wikipedia NPOV policy. The same with that paragraph that begins the Solution section. They belong only in a Controversies section that enumerates the various MHP reliably published POVs without editorializing. That section would include Morgan's paper of 1991, as well as their letter of 2010, of course. Glkanter (talk) 12:05, 13 July 2010 (UTC)

I see. You're suggesting anything that doesn't agree with your POV should go in a section called "Controversies". This is a common enough POV structure that it's explicitly mentioned in WP:NPOV (see wp:structure). What you're doing here is called disruption. Please stop. -- Rick Block (talk) 14:06, 13 July 2010 (UTC)

No, Rick. I'm saying rather than the editors act as referees, anything that the reliable sources disagree on should end up in a a Controversies section. Quit making Straw Man arguments, please. Glkanter (talk) 20:21, 13 July 2010 (UTC)

One for the pedants

Any complete solution of the MHP must show all possible doors that the player might have picked.

Why? If the MHP is the version in which door 1 has been chosen and door 3 opened, it is sufficient to consider this situation. (And we know it stands model for all the other combinations as well.) Nijdam (talk) 17:09, 11 July 2010 (UTC)
Please read the paragraph below.

It is no good saying that the player has, in fact, picked a specific door. How do we know that the result might not have been different had the player picked another door. If there is one thing that the MHP teaches us it is that in probability problems, we must not only take account of what did happen but of what might have happened. (The difference between a host who only opens goat doors and a host who happens to reveal a goat by chance nicely illustrates this) There are, no doubt, possible variations of the MHP in which the door picked by the player does matter.

Do you still claim that we need not consider the player's choice?
How do you prove this?
Are you saying that the player's door choice can never make any difference to the answer (probability of winning by switching)?

Martin Hogbin (talk) 17:29, 11 July 2010 (UTC)

Nijdam, do you have answers for these questions? Martin Hogbin (talk) 12:49, 14 July 2010 (UTC)

So for the pedantic amongst us, no solution of the MHP is complete unless it covers all the doors. For some reason this is not considered necessary in the article. Maybe the editors who propose the 'complete' solutions shown here have appealed to a simple and obvious symmetry between the doors. If the car is placed uniformly at random and the player chooses uniformly at random,

[AND the host chooses uniformly at random (Gill110951 (talk) 13:01, 3 July 2010 (UTC))]

the problem is clearly symmetrical with respect to the originally-chosen door.

The distribution of the choice of the player is unimportant. Nijdam (talk) 17:09, 11 July 2010 (UTC)
[NO, since it is not obvious how the host would have determined his choices in other situations (Gill110951 (talk) 13:01, 3 July 2010 (UTC))].

We can, without loss of generality, pick any door and solve the problem on the basis that that door was chosen.

Similarly, if the host chooses a goat door uniformly at random, the problem is obviously symmetrical with respect to the door opened by the host and we can, without loss of generality, choose either door or even an unspecified door, making the problem unconditional.

The simple solution is therefore as well justified as the so called 'complete solutions' presented here. Martin Hogbin (talk) 09:30, 3 July 2010 (UTC)

I really can't see what you want to prove? Nijdam (talk) 17:09, 11 July 2010 (UTC)
For a subjective Bayesian you don't have to consider "what would have happened if...". Your probabilities are measured by your own personal betting odds. All doors equally likely initially means you would bet at even odds between any pair of doors; the quizmaster's choices equally likely means you would be happy to bet at even odds either way. If you are a subjectivist and you are not stupid your betting odds will reflect probabilities which satisfy the probability calculus. You will therefore be prepared to bet at odds 2:1 that the other door has the car, or at odds 1:2 that your initial choice has the car. Subjective probabilities in, subjective probabilities out.
If you are a frequentist then you are thinking of many many repetitions and you imagine that the host would in the long run open one of the two doors with a particular relative frequency when he has a particular choice, and so on.
I don't see why the MHP has to be solved using subjectivist probability or by frequentist probability. All mathematical models are just that ... models. Garbage in, garbage out. The more sound common sense and relevant real world knowledge you put in, the more sound and relevant the conclusion will be.
I do agree that the simple solution can be as well justified (and as well mathematized) as a "more complete" solution, but that is for different reasons than @Martin gives. Moreover I think there are even more complete solutions (game theoretic is more complete than conditional probabilistic, like it or not). I like to distinguish between problems and solutions. And I like to distnguish between problems at different levels. I like to distinguish between mathematical problems (problems inside mathematics), and problems of mathematical modelling (problems of converting real world problems into mathematical problems, and problems of converting solutions of mathematical problems back to the real world again). The MHP is meta-problem from the point of view of mathematics, a problem of mathematical modelling. An important part of this meta-problem is to convert an ambiguous common language question into a well-formulated mathematical problem. Different people published different sequences of English language words. Different people converted such sequences into different mathematical problems. One could say "Selvin invented the MHP as a statistical problem" or one could say "Nalebuff made the MHP famous in economics and decision theory, where it belongs" or one could say "vos Savant made the MHP famous to the man in the street" or one could say "Whitaker proposed the MHP to vos Savant" ... and so on. The story is very rich.
Defining "THE MHP" by the authority of ONE reliable source does not actually reflect the reality of the MHP at all - unless you are a Catholic and the Pope has pronounced on it, a fundamentalist Christian and you can find it in the Bible, or a Muslim and Mohammed wrote down Allah's words on the matter in the Holy Koran. The average wikipedia reader met the problem in a discussion about an imaginary quiz show situation with a friend at a party, and got into an argument about whether or not one should switch. You can't resolve those people's argument by saying "Mormon et al. told us what the MHP is and what it's solution is".
We can't solve the meta-problem by choosing 1) exactly one reliable authority's choice of English language question about a quiz show, 2) one particular (reliabably sourced) applied mathematician's mathematization thereof, 3) one particular (reliabably sourced) pure mathematician's solution thereof, 4) one particular (reliabably sourced) applied mathematician's translation of the maths answer back to a real world answer.
Here are three sound real-world advices:
Level 0: "Switch, because (ignoring door numbers) you'll win 2/3 of the time - a least, provided initially that all three doors were equally likely to hide the car"
Level 1: "Switch - at least provided initially that all three doors were equally likely to hide the car - there's then nothing better that you can do"
Level 2: "I guess your question is hypothetical and you'ld really like to know what to do whatever door numbers are involved: do yourself a favour then and when you get on this show you talk about choose your own initial door completely at random and thereafter switch - you'll win with probability 2/3 and you can't do better"
Note that level 0 and level 1 make an assumption which might be reasonable if we talk about probability as a way to reflect our own (lack) of knowledge, but otherwise is hard to justify. Note that level 2 finesses this issue by telling you to choose your own door at the start by grouping the six faces of a fair die into 3 pairs of 2, one pair corresponding to each door - choose your door by tossing your die. The validity of the answer relies only on your dice throwing being a good way to generate uniform random digits between 1 and 6. The probability it mentions is valid under all probability interpretations I know of.
Important addition: The reason Level 2 goes further than 0 and 1 is because the host might know that the player is a subjective Bayesian for whom subjectively in advance all doors are equally likely and therefore he obstinately always chooses Door 1. Monty also knows that the player has read wikipedia, so knows he is going to switch. Monty therefore rather often hides the car in advance behind Door 1 and always gets to keep the car and have a good laugh. The level 2 solution gives you protection from assuming that your personal ignorance means that your initial choice doesn't matter and hence you can may as well start by taking Door 1.
Pigeons are not fools. But they have very small brains. They have evolved to be excellent at on-line learning. Initially they use a lot of randomization to learn good solutions, and later they keep injecting a bit of random noise into their solution so as to keep ahead of the game and keep on having an opportunity to learn if the environment changes. That way they protect themselves from getting into a rut. Humans fool themselves into thinking they know the answer and hence they get themselves fooled by other humans. Gill110951 (talk) 05:17, 5 July 2010 (UTC)
Gill110951 (talk) 13:43, 3 July 2010 (UTC)

"Ignorant Monty" Simulation

This discussion page seems like a pretty hot place, so I'm a little timid to ask. But, can anyone describe a simulation of the Ignorant Monty that would converge to P("win by switching"|"ignorant host opened a door that happened to reveal a goat")=0.5? As a casual reader with some background in probability this article really helped my understanding, and has clarified most of my questions except for this detail. Maybe some small improvement in the article might come out of raising this question. —Preceding unsigned comment added by 69.134.3.102 (talk) 17:48, 11 July 2010 (UTC)

To simulate the "Monty forgets" variant (where the host forgets where the car is, opens a door, and expresses a sigh of relief when a goat is, in fact, revealed) similar to the vos Savant cup/penny experiment [1] that answers the question of the probability faced by a player who picks door #1 and sees the host open door #3
  1. label three cups #1, #2, #3
  2. the contestant looks away and the host rolls a die until it comes up 1, 2, or 3 and hides a penny under the indicated cup
  3. the contestant rolls a die until it comes up 1,2, or 3 and chooses the indicated cup
  4. the host rolls a die until it comes up with one of the two remaining cup numbers (not the one the contestant "chose") and lifts up this cup
  5. START OVER AND DO NOT COUNT THIS TRIAL (either as a success for switching or staying) unless the contestant picked cup #1, the host lifted up cup #3, and the penny was not under the cup the host lifted up
  6. count this trial (excluding the ones above) as a success for staying if the penny is under the player's cup and as a success for switching if the penny is under the other cup
-- Rick Block (talk) 20:23, 11 July 2010 (UTC)
Thanks for your reply. I programmed it just as outlined, counted the trial only if the contestant randomly picked cup 1, the host randomly revealed cup 3, and the penny was not under the cup randomly revealed by the host. But, I still got that 1/3 of the time the contestant wins by staying, 2/3 of the time the contestant wins by switching after a large number of trials. I'm don't know much about Wikipedia rules and protocol. Is it appropriate/useful for me to share matlab code so others could review/critique/figure out what's going on here?69.134.3.102 (talk) 22:48, 11 July 2010 (UTC)
Programmed in matlab? You could try posting the source here. Just out of curiosity, are you counting total trials (including the ones you're throwing away) and also wins by switching and staying? If so, if the total is (say) 900, you should be ending up with about 50 wins each by switching and staying. In about 600 cases the player won't pick cup #1, of the remaining 300 the host should lift cup #2 half the time leaving 150. Of these, the penny should be under cup #3 about 1/3 of the time leaving 100. These should be split roughly 50/50 win/lose. -- Rick Block (talk) 23:23, 11 July 2010 (UTC)
Ok sorry, got it now just as stated above. Example run: out of 900 trials (actually rounding from 900,000 trials), the contestant doesn't pick cup one 600 times, the host lifts cup two 150 times, the penny was under cup three 50 times, leaving 100 valid trials. Out of these 50 win by staying, 50 win by switching (50%/50%). Cool, now that we're on the same page--if we change that simulation to be this:
  1. label three cups #1, #2, #3
  2. the contestant looks away and the host rolls a die until it comes up 1, 2, or 3 and hides a penny under the indicated cup
  3. the contestant rolls a die until it comes up 1,2, or 3 and chooses the indicated cup
  4. the host rolls a die until it comes up with one of the two remaining cup numbers (not the one the contestant "chose") and lifts up this cup
  5. START OVER AND DO NOT COUNT THIS TRIAL (either as a success for switching or staying) if the penny was under the cup the host revealed
  6. [EDIT: but we DO allow the contestant to pick cup #1, #2, or #3]
  7. [EDIT: and we DO count the trial regardless of which "unchosen" cup he revealed, as long as he didn't reveal the penny]
  8. count this trial (excluding the ones above) as a success for staying if the penny is under the player's cup and as a success for switching if the penny is under the other cup
Here, out of 900 trials, the ignorant host revealed the penny 300 times. Out of the surviving 600 trials, 300 win by staying, 300 win by switching (50%/50%).
Now, back to standard assumptions, if we have the host (randomly) reveal an "unchosen" cup that doesn't contain a penny and don't need to discard any trials, out of 900 trials, 300 win by staying, 600 win by switching (33%/66%).
I hope it is helpful and not annoying for me to post all this, but I hope maybe posting my learning process can help to illuminate how to guide other readers. Thanks for your help, Rick 69.134.3.102 (talk) 03:46, 12 July 2010 (UTC)
Restricting it to player picks #1 and host reveals #3 is to literally match the question. This actually relates to the interminable argument going on elsewhere on this page. Counting all player picks and any cup the host reveals inherently assumes the answer to the specific case of player picks #1 and host reveals #3 is the same as any other specific case, which it will be if the penny is randomly distributed to start with and the host is picking randomly. If you're absolutely sure this is the case - or if the question is slightly different and requires the player to decide whether to switch before seeing what the host does - then doing this is OK, but you should realize you're not actually simulating any specific case. To see that this matters, if you change your simulation so the host always reveals the highest numbered cup and keep track of each of the six possible pairs player pick and host reveal separately, you'll see quite different results. In the "host forgets" case you'll only get results for 3 of the 6 possible pairs (about 200 each if you're running 900 trials) and switching will win about 50% of the time in each of these 3 pairs. In the "standard assumptions but host always reveals the highest numbered goat" scenario (i.e. the host opens the highest numbered door the player didn't pick that is a goat) out of 900 trials you should see about 300 for each initial player pick, but if the player picks door #1 there won't be the same number of trials where the host reveals #2 or #3 (should be about 100 where the host reveals #2 and about 200 where the host reveals #3), and the probability of winning by switching will be different in these cases. Out of all players who pick door #1 (combining those who see the host open #2 and #3) 2/3 will win, but a different percentage will win depending on whether they see the host open #2 (100%) or #3 (50%). -- Rick Block (talk) 05:44, 12 July 2010 (UTC)
Yep, I see. Systematic host behavior affects probabilities for a particular case. I guess if it really were a game show and you were trying to "beat" it you could analyze historical game data and look for bias in which goat the host tends to reveal (and also nonrandom initial placement of the car) and create a door-specific strategy to maximize the probability of winning. Another point is, if this were a game that I were playing for the first time and the host unexpectedly revealed a goat and offered a switch (and I had no indication that the offer to switch was a permanent "rule" of the game), I would find myself trying to weigh the probability of "Monty from Hell" behavior and adjust the random probability of winning-by-switching accordingly.69.134.3.102 (talk) 04:48, 13 July 2010 (UTC)
Precisely. In the case where the host is not required to make the offer to switch, if the host is actually trying to keep as many contestants as possible from winning then the game theoretic solution is you don't switch. This is something that should go in the article eventually - Mueser and Granberg discuss this in Appendix A of their paper. "Monty from Hell" is one equilibrium of a continuum where the overall chance of winning by staying is 1/3 and by switching is anything from 0 to 1/3 with corresponding conditional chances of winning by staying of anywhere from 100% to 50%. This is something else you could determine by analyzing historical game data if it were a real show. If 2/3 of the time the host says "you've won a goat" without offering the chance to switch (and all players who switch lose), you're in the "Monty from Hell" scenario and absolutely should not switch. Since this would be a rather dull game, you might more realistically observe that 1/3 of the time the host says "you've won a goat" without offering the chance to switch and of the remaining players those who switch win 1/2 of the time (but still only 1/3 of the time overall). If the host is not required to open a door and make the offer to switch you're never worse off staying, which is completely the reverse of the situation where the host must make the offer to switch. If I were running a show like this, I would find it highly amusing to have the host make the offer to switch 1/2 the time, resulting in a 2/3 chance of winning by staying and a 1/3 chance of winning by switching. -- Rick Block (talk) 14:56, 15 July 2010 (UTC)

Another proposal for Falk's point

I don't think Martin's proposed text and the text I proposed above are irresolvably different. Martin's main objection seems to be the last sentence. Can we agree the paragraph to be added back can start with this?

In addition to the "equal probability" intuition there is a competing deeply rooted intuition that is the basis for some solutions that lead to the correct answer for the standard interpretation of the problem but an incorrect answer for slight variants. The intuition is that revealing information that is already known does not affect probabilities. Although this is a true statement, it is not true that knowing the host can open one of the two unchosen doors to show a goat necessarily means that opening a specific door does not affect the probability that the car is behind the initially-chosen door. If the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation) this probability indeed remains unchanged, but if the host can choose non-randomly between such doors then the specific door that the host opens does reveal information about this probability.

What I proposed as the last sentence of the paragraph (that Martin objects to) is

Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially-chosen door, although very persuasive, are therefore not mathematically sound (Falk 1992:207).

How about this as the last sentence?

Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially-chosen door are very persuasive, but lead to the correct answer only if the host chooses randomly between two goats (Falk 1992:207).
I think that is fine. Maybe we could talk about the rest of your version. I thought mine was clearer. Martin Hogbin (talk) 18:06, 12 July 2010 (UTC)

Would this address the concern? -- Rick Block (talk) 14:43, 12 July 2010 (UTC)

Put your nonsense and opinions that are 'loosely', if at all, based on published sources in a 'Controversies' section. Not before any 'Solutions'. Not after any 'Solutions'. Not in 'Causes of Confusion'. Not in 'Aids To Understanding'. Simply enumerate the sources, and what they say. Period. No 'helpful explanations'. Glkanter (talk) 15:24, 12 July 2010 (UTC)
@Glkanter - What we're talking about here is restoring a clarified version of a paragraph that was deleted without consensus (just before the article was protected). This paragraph describes a confusion, not a controversy. We can talk about moving it, but I think it's helpful to focus on one thing at a time.
@Martin - your version (here) focuses on the answer rather than the reasoning whereas what Falk is actually talking about is the seductiveness of the (not always correct) reasoning. Is this going to be another "it must be the way Martin wants it and there will be no compromises" situation, or are you willing to bend a bit? -- Rick Block (talk) 18:59, 12 July 2010 (UTC)
All I said was, 'Maybe we could talk about the rest of your version. I thought mine was clearer'. Martin Hogbin (talk) 21:30, 12 July 2010 (UTC)
So, fine, lets talk. I obviously prefer my version for the reasons I've stated. What about it do you find unclear? -- Rick Block (talk) 21:41, 12 July 2010 (UTC)

Wikipedia policy does not 'protect' article text from change. Not even FA articles. Martin's change to the article was wholly within Wikipedia policy and spirit. There is nothing 'inferior' or 'untoward' with Martin's edit that requires any extraordinary remedy. Glkanter (talk) 21:13, 12 July 2010 (UTC)

I'm well aware of Wikipedia's policies, thank you. -- Rick Block (talk) 21:38, 12 July 2010 (UTC)

@Martin and anyone else - Is there anything unclear about the following? How about if we treat the text in the collapsible box as if it were in the article (3rd paragraph in the "Sources of confusion" section), directly editing it as a draft paragraph? -- Rick Block (talk) 14:38, 13 July 2010 (UTC)

+++ Thank you, please goon like this. --Gerhardvalentin (talk) 20:57, 13 July 2010 (UTC)
paragraph about Falk's no-news intuition

In addition to the "equal probability" intuition, a competing and deeply rooted intuition is that revealing information that is already known does not affect probabilities. Although this is a true statement, it is not true that just knowing the host can open one of the two unchosen doors to show a goat necessarily means that opening a specific door cannot affect the probability that the car is behind the initially-chosen door. If the car is initially placed behind the doors with equal probability and the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation) this probability indeed remains unchanged, but if the host can choose non-randomly between such doors then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat and (in the standard interpretation of the problem) the probability that the car is behind the initially-chosen door does not change, but it is not because of the former that the latter is true. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially-chosen door are very persuasive, but lead to the correct answer only if the problem is completely symmetrical with respect to both the initial car placement and how the host chooses between two goats (Falk 1992:207,213).

I have made some changes. What do you think Martin Hogbin (talk) 21:40, 13 July 2010 (UTC)

why do you reject the specification "though"? My English is not so good, but I think it's pointing the "however"-aspect. --Gerhardvalentin (talk) 22:18, 13 July 2010 (UTC)
Not that it matters much given the current version, but a long sentence starting with "although" and ending with ", though" does not exactly suggest clarity of thought. glopk (talk) 05:29, 14 July 2010 (UTC)

'restricted or assumed to' vs 'taken to'

The meaning that I wanted to express by using 'taken to to choose randomly between two goats' was this: in some problem formulations the host is defined to choose a goat door at random but, in others, the host goat door policy is not defined. In these cases, because we choose to answer the question from the SoK of the player, and as the player is not likely to know the host's goat door policy, we may quite reasonably take the policy to be uniform at random. This is an appropriate uninformative prior in this case. It is not an 'assumption', that is to say something taken to be the case, through carelessness or inatention, it is a valid decision based on the way that we have chosen to address the problem. Martin Hogbin (talk) 12:47, 14 July 2010 (UTC)

Identifying the two cases you mention is the exact reason I changed it to "restricted or assumed". Does "assumed" necessarily imply "carelessly assumed" to you? It seems quite neutral to me about the intent. It may be, as you say, a considered, deliberate choice (based on SoK, or one's personal understanding of the rules of game shows, or one's conviction that the doors described in the problem are indistinguishable) in which case I agree "taken" would be appropriate - but unless there's an explicit statement saying why this assumption was made all we know is that it was an (implicit!) assumption, i.e. it may have been made with good reason or it may have been made due to carelessness or inattention. Unlike "taken" which distinctly implies deliberation, I think "assumed" covers both of these, without taking any stance at all on the reasoning behind the assumption. -- Rick Block (talk) 14:00, 14 July 2010 (UTC)
I do think 'assume' has negative connotations, people are often warned not to assume things. I think it is normal in serious mathematical answers to treat the reader as someone who knows what they are doing, even if this may not actually be the case. Thus in presenting an argument we should treat the reader as someone who does know about probability and thus has decided, after proper consideration of of the facts presented in the problem statement and the of approach that they have decided to take, treat the host's choice as goat door choice random. On other words it is normal to give the benefit of any doubt to the reader, even though we know that most people will not have even considered the possibility that the door chosen by the host might make a difference. Martin Hogbin (talk) 16:34, 14 July 2010 (UTC)
How about "restricted or explicitly assumed" - or go back to what you originally said was acceptable, i.e. "only if the host chooses"? -- Rick Block (talk) 17:43, 14 July 2010 (UTC)
On reflction, I think that 'taken to be' is exactly right, on the basis that it neutrally covers every possibility. To give an example from everyday speech, to say 'I took him to be British' could mean everything from, 'for no particular reason and with no reasonable evidence I assumed this man to be British, possibly mistakenly' to 'I had checked his passport and therefore made a considered judgment that he was British'. To use 'assumed' in that example strongly suggests the former case.
I do not understand your objection to 'taken'. What POV do you think it pushes? Martin Hogbin (talk) 09:10, 15 July 2010 (UTC)
If you've checked his passport, you're not "taking" him to be British - he IS British. "Taken", by itself, loses the distinction between the two cases you identify above - i.e. the one where the host's choice is specified in the problem definition (in this case the host is "restricted" to choose randomly) and the one where this behavior is not specified. From a POV standpoint, "taken" distinctly implies (to me) a deliberate decision whereas "assumed" can be explicit or implicit, and if implicit deliberate or not. Since we're sourcing this to Falk, perhaps we should rephrase this to be more directly what she says, i.e. change the last sentence to something like:
The host can always open a door revealing a goat and (in the standard interpretation of the problem) the probability that the car is behind the initially-chosen door does not change, but it is not because of the former that the latter is true.
This is nearly a quote. -- Rick Block (talk) 15:17, 15 July 2010 (UTC)
No, the passport might turn out to be a forgery. The point is that 'taken' does not assert either a strong or a weak degree of confidence that the fact is true. In Falk context it just refers to the approach actually taken by the person answering the question, it does not reflect on the decision-making process that led them to take that approach. Martin Hogbin (talk) 09:16, 16 July 2010 (UTC)
The problem with just giving that quote out of context is that it gives the impression that the fact that the host can always open a door revealing a goat is not one of the reasons that the probability that the car is behind the initially-chosen door does not change. As we all know, and as Falk makes quite clear in her paper, the fact that he host can always open a door revealing a goat and the fact that the host chooses a goat door uniformly (if this is the case) taken together do mean that the probability that the car is behind the initially-chosen door does not change. Without both being true the probability can change. Martin Hogbin (talk) 08:55, 16 July 2010 (UTC)
Rather than continue to argue about "taken" my suggestion is that we resolve this by staying much closer to what Falk actually says. I think we're providing full context for the near quote. If you don't like essentially quoting her is there some other wording you'd like? Falk's point, in context, is that knowing the host can and will open a door showing a goat is not sufficient to conclude the probability that the car is behind the initially-chosen door does not change.
Agreed. But Falk also states the circumstances when it is true that the probability does not change. We should say that as well.
BTW - Falk says even knowing the host can reveal a goat and chooses between two goats uniformly at random is not sufficient - it also requires equal priors as well. The easiest way to specify what is required in addition to knowing the host can reveal a goat is to say that the doors are indistinguishable (to both the contestant and the host). This forces the problem to be symmetrical and makes it equivalent to the urn problem we've repeatedly discussed - but puts the problem squarely in the realm of unrealistic imagination since everyone knows the doors are numbered and even if they're not they have a persistent positional identity. -- Rick Block (talk) 15:16, 16 July 2010 (UTC)
If by 'equal priors' you mean that the car was originally placed behind each of the three doors with equal probability that is fine and is a point that I have made many times before. On that basis all that is required to conclude that the probability that the car is behind the initially-chosen door does not change is that host can always open a door revealing a goat and the host chooses a goat door uniformly. Are you disagreeing with this now? Martin Hogbin (talk) 17:11, 16 July 2010 (UTC)
No, I'm not disagreeing about this. I'm saying Falk is saying that there's even more required than knowing 1) the host can and will open a door showing a goat, and 2) the host opens a random goat door - we also need 3) the car is initially uniformly distributed. The point is, again, that the intuition that knowing the host can and will open a door showing a goat is sufficient to mean opening a door does not change the probability of the player's initially selected door is wrong. There's more to it than this. The fact that anything else is required is enough to make the point. Precisely what is required can be determined by looking at the parameters involved in the Bayesian expansion - and people's intuition about what may or may not be important are not in the least reliable.
Why do you think it's important to enumerate the circumstances required for the probability to not be changed by the host opening a door? She doesn't include this in the section of the paper we're talking about. She's specifically addressing the no-news intuition and essentially saying it leads to convincing, but unsound, solutions. -- Rick Block (talk) 21:16, 16 July 2010 (UTC)
I think it is important for two reasons. Firstly, this is an article on the MHP, that is exactly the kind of thing the article should talk about. Secondly, Falk discusses those reasons in her paper. Why do you want to suppress the facts? Martin Hogbin (talk) 13:07, 17 July 2010 (UTC)
How about the changes as above? I've added the uniform prior in the (existing) sentence that describes what is required for the host opening a door not to change the probability of the player's initially selected door, added the near quote, and changed the end of the last sentence to "but lead to the correct answer only if the problem is completely symmetrical with respect to both the initial car placement and how the host chooses between two goats" (and added the appropriate page number for the symmetry bit to the citation). -- Rick Block (talk) 16:17, 17 July 2010 (UTC)
OK, except I think we should change 'but it is not because of the former that the latter is true' to something that makes it clearer that the condition in question is necessary but not sufficient. Martin Hogbin (talk) 18:58, 17 July 2010 (UTC)
Well, the point she's making is that it's not sufficient (even though the intuition says it is). Isn't what is necessary amply described by the rest of the paragraph? -- Rick Block (talk) 04:18, 18 July 2010 (UTC)
The problem is that your proposed wording might be seen to conflict with the rest of the paragraph. Can we not make the 'necessary but not sufficient' condition clearer? Martin Hogbin (talk) 10:37, 18 July 2010 (UTC)
The "not because of the former that the latter is true" wording (including the italics) is directly from Falk - and is the entire point of this paragraph (and the cited section of her paper). She's not saying it's "necessary but not sufficient", but rather that knowing the host can and will open a door showing a goat is not what makes the probability of the player's door remain unchanged, even though this connects to a strongly held intuition. If this sentence seems to conflict with the rest of the paragraph, it's the rest of the paragraph that should be changed. -- Rick Block (talk) 17:03, 18 July 2010 (UTC)

There is no 'learning from events' on a game show.

I guess that's been my point all along. Maybe in some other setting the puzzle-solver's State of Knowledge can change (improve) over time. But not on a game show, and neither Selvin's or Whitaker/vos Savant's statement of the MHP has ever indicated that there would be.

"David Blackwell, a statistician and mathematician who wrote groundbreaking papers on probability and game theory and was the first black scholar to be admitted to the National Academy of Sciences, died July 8 in Berkeley, Calif. He was 91."

http://www.nytimes.com/2010/07/17/education/17blackwell.html?_r=1&hpw


Glkanter (talk) 19:58, 18 July 2010 (UTC)

Source for the Mathematical Formulation in odds form

Courtesy of Google Books, I was able to fish out an English reference for the odds form of the Math. Form. This one is also interesting because it shows in a very simple way why the host's bias does not affect the rational player's decision problem (though it obviously affects their risk). Would the other editors care to comment about using it to replace or augment the current Mathematical Formulation section? The relevant page is here. glopk (talk) 15:14, 14 July 2010 (UTC)

Looks like Morgan's argument again with even more pointless maths. Martin Hogbin (talk) 16:25, 14 July 2010 (UTC)

Care for elaborating? Are you suggesting that the solution (there's no "argument" there) is incorrect, or that it is "pointless" to write math in the article, or ...? glopk (talk) 18:53, 14 July 2010 (UTC)
Morgan show, and we all agree, that whatever the host policy it is always better or equal to switch. How is your reference an improvement? Martin Hogbin (talk) 22:21, 14 July 2010 (UTC)
This reference is an improvement because the current one for the Mathematical Formulation (Henze) is written in German. Further, it is an improvement because, by not requiring the explicit computation of the stick and switch posterior probabilities, but only of their ratio, it does not require the marginalization with respect to the position of the car (i.e. the C): one step less. This benefit was pointed out by Gill weeks ago, and I agree. glopk (talk) 07:22, 19 July 2010 (UTC)
Since the basic Bayes expansion (with marginalization wrt the car's position) is so common I think the current formulation should remain in the article. Presenting the ratio approach as an alternate would be fine. -- Rick Block (talk) 14:31, 19 July 2010 (UTC)
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