Talk:Monty Hall problem

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Are the odds ever 50-50 ?[edit]

It's been established that the contestant and Monty have produced a pair of doors containing a car and a goat that do not have the same odds of producing a goat. Yet, on Marilyn vos Savant's website where she discusses this teaser, she perhaps yields too much ground to the academic furor that confronted her. She made this concession:

"Suppose we pause at that point [when two doors remain], and a UFO settles down onto the stage. A little green woman emerges, and the host asks her to point to one of the two unopened doors. The chances that she’ll randomly choose the one with the prize are 1/2, all right. But that’s because she lacks the advantage the original contestant had—the help of the host." [1]

But are the chances 50%? Probabilities don't care much for who does the choosing, and neither the contestant nor the green woman has insider information about the whereabouts of the car behind the doors. The contestant can know that the odds of finding a car behind the switched door are 2/3, but the odds don't change just because someone doesn't know the odds. If I know that four sides of a die are red and two are blue before I paint over the sides, the odds of someone who has never seen the die rolling red are not 1/2 due to her unawareness.

If the odds of randomly selecting a car among the two blind choices are 50-50 for anyone other than the contestant, then they must be 50-50 for everyone. We know that they are not, so the green woman approaches a stacked deck, so to speak. Otherwise, we end up saying that all odds between situations with two outcomes are 50-50 so long as we know nothing about the underlying conditions. That's not correct.

Vos Savant refers to the help of the host as being a missing ingredient, but that help actually settled the odds prior to his final selection, and it must've done so for anyone choosing between the two doors. We are now convinced that this problem is a matter of frequency of appearance of the car behind particular doors after particular actions have settled the odds, so vos Savant needn't have made any concessions to the contrary.

Thoughts? — Preceding unsigned comment added by Summers999 (talkcontribs) 14:24, 25 April 2015 (UTC)

The question is what odds are you talking about? If you pick randomly between two choices the odds of your choice being the correct one are 50-50. Note how I worded this - the odds we're talking about are the odds of "your choice". If we roll your red/blue die the odds are 4/6 that it ends up red and 2/6 that it ends up blue. If you roll it 100 times and I randomly guess red or blue, about 50 times I'll guess red and 50 times I'll guess blue. Of the 50 times I guess red I'll be right about 4/6 of the time, so about 33 times. Of the 50 times I guess blue I'll be right about 2/6 of the time, so about 17 times. Altogether I'm right about 50 times out of my 100 guesses, i.e. 50-50. If I don't know 4 sides are red and 2 are blue, my odds of correctly guessing are 50-50 whatever the "actual" odds may be because I have no choice but to randomly guess. If I always guess red (because it's my favorite color), I'll be right about 4/6 of the time, but now we're not talking about the odds of a random guess but the odds that the die ends up red (which is a different question).
If you want to talk about this further let's move the discussion to the Talk:Monty Hall problem/Arguments page. -- Rick Block (talk) 16:40, 25 April 2015 (UTC)
I'm only referring to a single selection of unequal odds. Namely, if the contestant or a green alien chose one of the remaining doors, it would still be better to choose a particular door. And this is true whether or not someone asked to select knows of the contestant's original choice. Even two of three random selections will yield a car from the alternative door, just as blind rolls of the die I described will yield red twice as often as blue. Summers999 (talk) 21:32, 25 April 2015 (UTC) user:Summers999
Like I said - if you want to talk about this further let's move it to Talk:Monty Hall problem/Arguments. -- Rick Block (talk) 04:09, 26 April 2015 (UTC)
It entirely depends on your information, that's all odds are.

The car is behind a specific door. Stick or switch are each either 1 or 0 probability of getting the car. The host knows the car is fixed so for the host choosing stick or switch the odds are 0 or 1 The contestant doesn't know where the car is but he knows the host could have opened door 2 and didn't the odds are 0.33 0.67 stick and switch. The alien doesn't know which door the host chose not to open and so the odds with that information are 50/50 SPACKlick (talk) 09:42, 1 May 2015 (UTC)

Edit War over Game Theory[edit]

There have been several edits back and forth about whether Game Theory is an appropriate descriptor of the problem. I cannot find a reference to support the inclusion but I would not be surprised if such sources existed. Are there any sources to support the inclusion of the term in the initial description? SPACKlick (talk) 13:17, 10 July 2015 (UTC)

There are a few sources that analyze the problem from a game theoretic perspective, as mentioned in the body of the article, but the vast majority do not. As a degenerate case for games of competition (cf. Seymann, R. G. (1991), "Comment on Let's make a deal: The player's dilemma", American Statistician 45: 287–288: "Simply put, and quite clear considering her [vos Savant's] suggestions for simulation procedures in her two later columns, the host is to be viewed as nothing more than an agent of chance.") MHP may be a useful example for explaining game theory, but using game theory to explain MHP is like shooting a mosquito with the proverbial elephant gun. I do not think it needs to be mentioned in the lede, and it should certainly not be used to state the nature of the problem like this in the opening sentence. ~ Ningauble (talk) 15:54, 10 July 2015 (UTC)

Has nobody noticed vos Savant's problem in the intro is NOT the same problem described under the Standard Restrictions section?[edit]

The standard restrictions section includes "The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999)." This is not part of vos Savant's problem. The host does in fact open a door the contestant didn't pick... she she didn't say he's REQUIRED to. This changes the truth table needed to solve the problem, and thus changes the probabilities, because it assumes that had she picked door #2 instead, he couldn't still have opened door #2. This In short, it's a different problem. Her stated answer to her stated problem was incorrect. Her stated answer to the problem with the additional restriction from the below section would have been correct... but that's not the problem she stated. In her stated problem, the odds that the prize is behind either of the remaining doors is 50%, and her answer was wrong. SteubenGlass (talk) 22:07, 10 August 2015 (UTC) SteubenGlass (talk) 22:07, 10 August 2015 (UTC)

A friend has pointed out that the host "must" open a door she didn't pick, otherwise offering her the choice to pick a different door makes no sense. But in vos Savant's version of the problem, it's never specified that the contestant MUST be offered that choice. vos Savant totally botched the setup of the problem, that's why there's so much confusion. SteubenGlass (talk) 22:25, 10 August 2015 (UTC)

Yes, it is well known that vos Savant did not describe the problem very well. Many sources have discussed what they believe the precise problem intended by vos Savant was and the consensus of these sources is that she intended to specify what we have called the 'standard problem'. I think this is all made clear in the article. Martin Hogbin (talk) 08:50, 11 August 2015 (UTC)
Yes, many people have remarked that her formulation of the problem was imprecise. She herself admitted, obliquely, in a follow-up column that the intended conditions of the problem are only implicit in the answer she gave, not in the question as originally stated. She wrote: "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. ... Anything else is a different question." [emphasis added]

I imagine she thought there was something implicit in saying "Suppose you’re on a game show", and in the general nature of this sort of riddle, that would make her original intent clear. It was evidently not clear enough, or was sufficiently open to nit picking, that later clarification was necessary.

This is a sorry state of affairs because so much attention has been given how sloppy her formulation of the problem was, rather than the much more interesting fact that the intended problem defies common sense: most people get it completely wrong even when it is stated in a completely precise form. Vos Savant surely had no idea of the importance this simple riddle would take on, or she would have been a little more careful about how she presented it. ~ Ningauble (talk) 15:00, 11 August 2015 (UTC)

I still think this needs to be addressed in the lead section. The fact that the answer depends so subtly on conditions of this type is actually one of the most interesting things about the whole paradox. I am not at all convinced that vos Savant had considered this issue in her original column; she may well have assumed that it didn't make any difference. Even at this late date I have run into people who cannot be convinced that it does make any difference. --Trovatore (talk) 03:49, 12 August 2015 (UTC)
The lead section already alludes to problems of interpreting the question when it says "under the standard assumptions". (At one time the article explained those assumptions in the very first section after the lead.) Perhaps slightly more could be said about it there, but it is not really so ambiguous.

"Suppose you're on a game show, and you're given the choice of three doors" and after you pick a door the host says "Ok, you win/lose. Thanks for playing." Not much of a game, is it? Suppose the host opens the door you picked and asks if you want to pick another. Kind of a no-brainer, isn't it? Suppose the host shows you where the car is, and asks whether you want it. What game show ever worked like that? No, let's suppose you're on a realistic game show.

I am quite convinced vos Savant did not consider such scenarios in her original column: the answer she gave in that column makes it clear what scenario she had in mind. The essence of the problem lies in being given a second chance with additional information that is still inconclusive. As vos Savant wrote: "Anything else is a different question."

I agree that quite a few people are convinced that this makes no difference. Many theories have been suggested to explain why people don't get it, and there are probably multiple factors in play. The two that are most interesting to me are confirmation bias (the goat behind door #3 taken as confirming evidence for the car being behind door #1), and the problem that common sense intuitive reasoning is ill equipped for interpreting selective evidence (as if it doesn't matter that Monty knows what's behind the doors).

The real difficulty of the Monty Hall problem lies in the nature of the problem itself, not in the vagueness of saying "suppose you're on a game show". ~ Ningauble (talk) 14:30, 12 August 2015 (UTC)

I don't know how old you are; you may never have watched the actual show. Americans of my generation or older mostly have watched it at some point, and their analysis is likely to have been influenced by that.
My recollection is that Monty appeared to make these offers (to switch) more or less ad lib. You didn't actually know whether it was scripted or not, could have been, or he could have decided on the spot whether or not to make the offer. It was entirely obscure whether his decision to make the offer or not depended on whether you had picked the door with the car.
So I disagree with your assertion that "it is not really so ambiguous" for a "realistic game show", because it is completely ambiguous for the actual game show. --Trovatore (talk) 16:59, 12 August 2015 (UTC)
Well, it is completely different than the actual game show (which I do indeed remember from my youth), where you never got a second shot at a choice among the same three options – only an offer to trade for something else.

Have you read Krauss & Wang (2003), cited in the article? In the section "Are There Possible Effects of Incomplete Information?" at pp. 9–10, they make a good case that it is not ambiguity of the rules that leads people astray, but an inability to appropriately model the scenario for solution.

Some people do get hung up on guessing at Monty's motivations, but Krauss & Wang observe that such guesses would not logically lead to the common answer that the odds are 50:50! (Any lack of impartiality in the host is really not in the spirit of game shows from his era, for those who remember, but harkens back to a much earlier era of crooked games on the radio, which were before my time.)

I am not aware of any reliable source that demonstrates or claims that people get it wrong due to trying to apply rules from the actual game show. ~ Ningauble (talk) 19:14, 12 August 2015 (UTC)

In fact there's a letter from Monty Hall to Steve Selvin, cited in the article, in which he clarifies that on the show the rules were "no trading boxes after selection" (this was 15 years before vos Savant's column) - so applying rules from the game show to the problem simply doesn't work. What was offered was cash in lieu of whatever was behind the selected door, and sometimes a non-selected door was opened (making the contestant think her odds of winning had gone up). This is discussed in the NY Times interview with Monty (also referenced in the article) in which Monty Hall agrees that you should switch if the host must open a door and make the offer to switch doors [there's an implicit assumption that the host knows where the car is and never reveals it], and then goes on to demonstrate (!) that if the host is not compelled to do this you could lose every time by switching. -- Rick Block (talk) 15:43, 13 August 2015 (UTC)
The majority of the people in general may get it wrong even among the standard assumption (according to Krauss/Wang). However much/most of academic discussion of the subject deals exactly with the ambiguity of the original problem, the justification of those standard assumptions and what is happening when they're dropped. So having that in mind it is indeed justified to mention something to that regard in the lead.
Having said that however this however, this is also reopening old (and apparently neverending bitter) edit disputes over the most appropiate descrtiption of the problem and which aspects are supposed to be played down or up and which aspect is the "really important" one and the "true reading" of the problem. So in doubt be prepared for yet another round.--Kmhkmh (talk) 20:02, 13 August 2015 (UTC)

Paragraph which explains why chance doubles by switching[edit]

Any random choice of 3 things has a 1/3 chance of finding a certain 1. If you are then shown 1 of the other 2 its not, that changes (in your knowledge) the total chance of the remaining 2 to 100%, but it cant change the 1/3 chance that the 1 you already chose is the winner because the 2 could be eliminated cant be that. Since your first choice has 1/3 chance and total remaining chance is 2/3, you have a 2/3 chance of winning by switching. (talk) 16:17, 14 September 2015 (UTC)

This was removed for being an "unsourced addition", but math is factual and needs no source. (talk) 16:17, 14 September 2015 (UTC)

This is a little way beyond the "Basic arithmetic, such as adding numbers, converting units, or calculating a person's age" permissible under WP:CALC. It also seems to be a not particularly clear reframing of the existing "Monty is saying in effect: you can keep your one door or you can have the other two doors" explanation given in the "Simple solutions" section. --McGeddon (talk) 16:23, 14 September 2015 (UTC)

Beyond basic arithmetic? It doesnt use any numbers bigger than 3 (counting 100% as 1/1). — Preceding unsigned comment added by (talk) 20:02, 14 September 2015 (UTC)

There would need to be a source for the methodology presented. The emphasis or otherwise of elements of the puzzle would be OR without it. SPACKlick (talk) 12:53, 15 September 2015 (UTC)