Monty Hall and Bayes

In the nineteen seventies there was a game show on TV called “Let’s Make a Deal”.  The host was Monty Hall and for the game  he would present three doors to a contestant.  Behind one door was the grand prize (e.g. a living room set) and behind another door was a goat.  The contestant would choose a door.  Then Monty Hall would open one of the two doors not selected to show that the prize was not there.    The contestant could then choose to keep their original door or switch doors.  The problem became famous in 1990 because Marilyn Vos Savant, who reportedly had/has the world’s highest IQ at 228, wrote in her column in Parade magazine that you should switch doors because the probability of winning increased from 1/3 to 2/3.  The column prompted thousands of letters, including many from those claiming to have statistics and mathematics PhDs, saying that she was wrong and that it didn’t matter since the probability of either door having the prize was 1/2. I must confess that when I first heard the problem (I didn’t know about Bayesian inference) I also believed that the probability of winning was 1/2 for either door so it didn’t matter and it took me awhile to understand the correct answer.  Now, this has been written about hundreds of times in books and articles (just google Monty Hall Problem), so there is nothing more that I can really add.  However,  just in case you didn’t know about this problem, it is a very nice example of how Bayes theorem can come in handy in real life.

Now the trivial way to solve the problem is that there is a 1/3 chance of  picking the right door, so that means there is a 2/3 chance of being wrong.  When Monty Hall opens a door, the door you didn’t select now has a 2/3 chance of being right, so you should always switch.  The faulty reasoning is that there are two doors and one door has the grand prize so it doesn’t matter if you switch. The difference is the history behind the situation and probability can change depending on history.  This is completely natural in the Bayesian framework but may give fits to a frequentist.

Here’s the Bayes calculation. Suppose there are three doors, a, b, c, you choose door a, and Monty Hall opens up door c.  The question then is what is the probability that the prize is behind door a given that Monty Hall opened door c versus the probability that the prize is behind door b, given that he opened door c.

Let  P(a | C) be the probability that the prize is behind door a and he opened door C.  Using Bayes theorem

P(a | C) = \frac{ P (C | a) P(a)}{P(C)}

where P(C | a) is the probability he opens door C given that the prize is behind a, P(a) is the prior probability that the prize is behind door a and P(C) is the total probability he opens C.  Initially, you have no idea where the prize is so P(a) = 1/3.  Monty Hall can then open one of two doors so P(C) = 1/2.  If the prize is behind door a then the conditional probability that he opens c is also 1/2 so P( C | a)  = 1/2.  Putting this together gives

P(a | C) = \frac{ (1/2) 1/3}{1/2}=\frac{1}{3}

So, if the prize is behind door a then the probability that it is behind door a if he opens door c is 1/3.  Now let’s calculate the probability that the prize is behind door b.  Again, Bayes theorem gives

P(b | C) = \frac{ P (C | b) P(b)}{P(C)}

and P(b) = 1/3 and P(C)=1/2.  The only difference now is that if the prize is behind door b, then Monty Hall can only open door c so P(C | b) =1.  This then gives

P(b | C) = \frac{ 1 (1/3)}{1/2}=\frac{2}{3}

which gives the correct answer.  There are three lessons from this problem and episode.  The first is that Marilyn Vos Savant is pretty smart (shouldn’t she be out trying to cure cancer?), the second is that having a PhD doesn’t innoculate one from being wrong and making a fool of oneself, and the third is that Bayesian inference is not just an academic exercise but can really make an impact on your life (like getting a brand new sofa).

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3 thoughts on “Monty Hall and Bayes

  1. The Bayes answer (or any other calculation I’ve seen) still doesn’t help my intuition.

    The way I convinced myself that it is at least plausible that you should switch is to take the “thermodynamic limit”.

    Consider 100 doors, with only one of them containing the prize. Of course, the odds of you picking the right door is the worst, 1/100. But if the host selectively opens a large number of wrong doors (say, 98), it’s clear that you’re better off switching.

    -A.

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  2. Hi Artur,

    You are absolutely correct. That is what convinced me as well. I think Bayes is useful as a way to compute the answer without having any intuition. I’m always concerned about intuition leading me astray so I like Bayes because I can compute probabilities without requiring insight.

    best,
    carson

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  3. If Monty invariably opens a door that does not contain the grand prize, and the contestant knows that Monty always behaves in this way, then it is correct that the contestant should change his/her selection because Monty has revealed the winning door for those 2/3 of cases in which the contestant will have guessed wrong initially. Thus the 2/3 chance that changing will result in the correct pick.

    If Monty doesn’t know which door hides the grand prize and opens a door at random, sometimes revealing the grand prize, then no new information has been revealed to the contestant. If the contestant has reason to totally trust that Monty’s door choice is an uninformed random one , the contestant has gained no new information and gains no advantage by changing his/her choice. On the other hand, his/her odds are not lessened by changing.

    If the contestant is uncertain whether Monty always will intentionally open a door not containing the grand prize, the contestant should ordinarily change his/her selection because even the chance that Monty intentionally opens a door not containing the grand prize increases the odds that changing is a better bet.

    If the contestant thinks that Monty knows the correct door and is playing some sort of bluffing game, then it all becomes a head game and the odds of random chance outcomes are irrelevant.

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