Question

How to solve Monty hall problem by using conditional probability?

Answer 1

Monty hall problem is
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

There are essentially two intuitive (i.e., not based on conditional probability) attempts at solutions. If Door #1 is chosen, and Door #3 is opened:

1. The car was equally likely to start behind any door, and that has to remain true after removing Door #3 as a possibility. Since Door #3 now has probability zero and Door #1 and Door #2 have the same probability, it is 1/2 for either.
2. The car was equally likely to start behind any door, and that has to remain true after removing Door #3 as a possibility. Door #1 started with a 1/3 probability, and that remains the same. So the set of all other possibilities - now only Door #2 - has a 2/3 chance.

Both of these explanations are based on the fallacy that some probabilities have to remain the same. When you learn information about a random outcome, conditional probability applies, and you have to allow for anything to change. So everything past “has to remain true,” in both explanations, is based on an error.

1. Explanation #1 uses correct mathematics after that point. As a consequence, it gets the wrong answer.
2. Explanation #2 makes two more errors - that there is something that can change when it assumed nothing can, and that it is only the probability for Door #2 that can change. By sheer luck (a fortuitous assumption that the explanation does not recognize is implicit), this combination of three errors ends up getting the correct answer. People accept it as a correct explanation because it gets the correct answer.

The reason most explanations don’t use conditional probability, is because they either are naive and don’t consider it, or they see only the wrong condition that “Door #3 has a goat.” Using this condition leads to explanation #1.

The correct condition is “Monty Hall choose #3 because he had to choose a door, other than #1, that has a goat.” Using “C” to represent where the car is, and “M” the door that Monty Hall opens, the correct solution is found by Bayes Law:

Pr(C=2|M=3) = Pr(M=3|C=2)*Pr(C=2)/Pr(M=3).

Now, Pr(C=2)=1/3 because the car was, indeed, equally likely to start behind any door. Pr(M=3|C=2)=1 because Monty Hall had no choice if the car was behind Door #2. But if we don’t consider where the car is, then it is just as likely that he’d open Door #2; that is, Pr(M=3)=Pr(M=2)=1/2. So:

Pr(C=2|M=3) = (1)*(1/3)/(1/2) = 2/3.

Note that this is based on the assumption that Monty Hall chooses randomly. Since we don’t know how he chooses, “randomly” is a good assumption. But Pr(M=3|C=2) can be anything between 1/3 (if he never opens #3 unless it is his only choice) and 2/3 (if he always opens #3 unless it has the car). This leads to answers anywhere between 1/2 and 1, proving that all of the probabilities can change.