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
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
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
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
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).