Understanding the Monty Hall Problem

Previously I talked about how our intuitions are not so great in the context of classical physics, but this is hardly the limit to what we haven’t the best intuitive understanding of. In the world of statistics and probability, our intuitions really are not great, and one classic example of this is the Monty Hall problem.

What is it? Suppose you are on a game show (hosted in this case by Monty Hall of Let’s Make a Deal), and there are three doors labeled 1, 2, and 3. Behind two of those doors is a goat, but behind one of them is a new car. You want to win the car, but you only have a 1 in 3 chance of guessing the right door.

Here is where things get interesting. You pick one door which remained closed, and Monty will show you one of the remaining doors has a goat. Monty then asks you if you want to choose the remaining door. The question is: do you stay, which the other door, or does it not matter at all?

When this was posed, most people have the intuition, included well educated people, that there is no benefit to switching. After all, if there are two doors left and one of them has the car, there is a 50/50 chance it is behind either the door you chose or the remaining one, right?

There are apparently a number of nuances about how the question is asked, but these pieces of nuance ultimately matter because of what other information it gives the game show contestant, and the way it is set up here has the surprising results: you ought to switch to the other door; your chances of winning double.

Why is this? Suppose you take Door 1 as your first pic (it doesn’t matter which so long as the car is equally-probable behind any of the 3 given doors), you can have two sets, doors picked and doors not. The set of doors picked has one member (Door 1), and the set of doors not picked has two members (Doors 2 & 3). That means the probability that the car is in the second set (doors not picked by you) is 2/3, while the probability of the first set (doors you did pick) is 1/3. Now, when Monty shows a door with a goat (say Door 2), it is from the second set of doors (the ones you did not pick), meaning he shows one member of that set is not the one with the car; this does not affect the set of doors you picked. The probability of either set having the car is still the same, but now you know one member of the second set is not the correct one, while only one member remains. Thus, that remaining member of the second set (Door 3) has a 2/3 probability of having the car, while your first choice of the door (Door 1) still has a 1/3 chance. Thus, you are twice as likely to get the car if you switch rather than stay put.

Crazy, it seems. It also means that if we did this with 100 doors, you picked one, and Monty revealed goats behind 98 of the 99 remaining doors, the chances of getting the car by switching is 99/100, way more likely than if you stayed. And all because Monty gave you more information about the set of doors you didn’t pick.

The Wikipedia page also gives other proofs, such as the use of the decision tree as well as more formal proofs, but the set format I used seems the most helpful, at least to me. If you still feel like there is some trickery, this has been also verified empirically:

You can see from this computer sim now the probabilities asymptotically reach 2/3 (66.7%) and 1/3 (33.3%) respectively. You can try out a simulation as well.

But as one last bit of fun, let’s see what we get using a Bayesian approach. Bayes’ theorem is as follows:

P(A|B) is the probability A is true given B evidence, P(A) is the initial probability A is true, P(-A) is the probability not-A is true (or probability A is false), P(B|A) is the probability of seeing evidence B given A is true (same applies for P(B|-A)). We have the rules that P(A) + P(-A) = 1, and all probabilities must be between 0 and 1. So, what are the numbers? (Another treatment is given here, but I think my method may be a bit easier to understand for this example).
Well, P(A), the prior probability before we have Monty open any goat doors, is 1/3, and the probability of P(-A) is necessarily 2/3. Now, evidence B is the opening of a door with a goat. So, P(B|A) should be 1/2 because the goat would have been behind either one of the remaining doors given we picked the right door from the start (that is, A is true). P(B|-A), on the other hand, is 1. Why? Because if the car is behind one of the doors you did not choose, then the unopened door must have the car (remember, P(B|-A) is the probability of having a goat behind a door if in fact the car is behind one of the doors not chosen). 

Put that all in, and you find the probability of staying with the initial door is 1/3 (P(A|B) = 1/3), and the probability of winning by switching is 2/3 (P(-A|B) = 2/3). So now we have two proofs, and one using Bayesian methods.

And I push Bayes’ theorem because it is so powerful and will need to be used more and more in the sciences to make sure we draw proper conclusions. But the debate between Bayesians and Frequentists must wait for another post.


One thought on “Understanding the Monty Hall Problem

  1. Pingback: Using Statistics to Find Fiddled Statistics | Fleeing Nergal, Seeking Stars

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