17.1: Monty Hall Confusion

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Page ID: 48422

Eric Lehman, F. Thomson Leighton, & Alberty R. Meyer
Google and Massachusetts Institute of Technology via MIT OpenCourseWare

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Remember how we said that the Monty Hall problem confused even professional mathematicians? Based on the work we did with tree diagrams, this may seem surprising—the conclusion we reached followed routinely and logically. How could this problem be so confusing to so many people?

Well, one flawed argument goes as follows: let’s say the contestant picks door A. And suppose that Carol, Monty’s assistant, opens door B and shows us a goat. Let’s use the tree diagram 16.3 from Chapter 16 to capture this situation. There are exactly three outcomes where contestant chooses door \(A\), and there is a goat behind door \(B\):

\[\label{17.1.1} (A, A, B), (A, A, C), (C, A, B).\]

These outcomes have respective probabilities 1/18, 1/18, 1/9.

Among those outcomes, switching doors wins only on the last outcome, \((C, A, B)\). The other two outcomes together have the same 1/9 probability as the last one So in this situation, the probability that we win by switching is the same as the probability that we lose. In other words, in this situation, switching isn’t any better than sticking!

Something has gone wrong here, since we know that the actual probability of winning by switching in 2/3. The mistaken conclusion that sticking or switching are equally good strategies comes from a common blunder in reasoning about how probabilities change given some information about what happened. We have asked for the probability that one event, [win by switching], happens, given that another event, [pick A \text{AND} goat at B], happens. We use the notation

\[\nonumber \text{Pr}[[\text{[win by switching}] \mid [\text{[pick A AND goat at B}]]\]

for this probability which, by the reasoning above, equals 1/2.

Behind the Curtain

A “given” condition is essentially an instruction to focus on only some of the possible outcomes. Formally, we’re defining a new sample space consisting only of some of the outcomes. In this particular example, we’re given that the player chooses door A and that there is a goat behind B. Our new sample space therefore consists solely of the three outcomes listed in (\ref{17.1.1}). In the opening of Section 17.1, we calculated the conditional probability of winning by switching given that one of these outcome happened, by weighing the 1/9 probability of the win-by-switching outcome, \((C, A, B)\), against the \(1/18 + 1/18 + 1/9\) probability of the three outcomes in the new sample space.

\[\nonumber \text{Pr}[[\text{[win by switching}] \mid [\text{[pick A AND goat at B}]] = \text{Pr}[(C, A, B) \mid \{(C, A, B), (A, A, B), (A, A, C)\}] + \dfrac{\text{Pr}[(C,A,B)]}{\text{Pr}[\{(C, A, B), (A, A, B), (A, A, C)\}]} = \dfrac{1/9}{1/18 + 1/18 + 1/9} = \dfrac{1}{2}.\]

There is nothing wrong with this calculation. So how come it leads to an incorrect conclusion about whether to stick or switch? The answer is that this was the wrong thing to calculate, as we’ll explain in the next section.