The Monty Hall Problem Explained

The Monty Hall Problem is this: you're on a game show and you're given the choice of three doors. Behind one door is a car and behind the other are goats. You choose, say, door A and the host, who knows where the car is, opens another door, B, behind which is a goat. He now gives you the choice of sticking with door A or switching to door C. Which should you do?

This problem appears almost whenever people are talking about counter-intuitive examples in probability. I got it wrong, the only reason you haven't is because you already knew the answer, but don't worry, we're all in good company: even Paul Erdos didn't get it first time, didn't like the answer when he did hear it and couldn't come up with a decent proof when he had.

What you and I and everyone except Marilyn vos Savant said was something like this: now you know there is a goat behind door B, there is a car and a goat left behind two doors, so there is a 50% chance the car is behind door A and a 50% chance it is behind door C, so it makes no difference if you switch doors. Two outcomes, two doors, it must be 50:50. Right?

Wrong. Computer simulation and complicated decision-trees shows that switching to door C is your best choice as it doubles your chances of winning. So far no-one has come up with a neat argument, let alone one “from the Book”, as Erdos would say, to explain why. Here's my attempt.

Your choose door A, with 1/3 probability of getting the car, leaving the doors (B or C) with 2/3 probability of getting the car. If you could somehow say “I'll choose (B or C)” you would have a 2/3 chance of getting the car. You cannot do that directly – but the host does something that makes it possible. He opens the losing door (B) in (B or C). This is guaranteed by the rules of the game. So by switching and choosing C you have effectively picked '(B or C)' with one choice and so benefited from the 2/3 probability.

If that sounds a little odd, look at what happens with four doors. You choose A and the host opens D. The car has probability ¾ of being in the set (B or C or D) but this time you have only a 50:50 chance of choosing the winning door for that set, because the odds are ¼ each that the car is behind B or behind C. So if you switch, your chances are 1/2x ¾ = 3/8 and you still have a 50% better chance of winning if you switch. The general formula is pretty obvious: with N doors, the probability of winning if you switch is: p = (1-1/N) / (N-2).

For N =3 this 2/3, N = 4 it is 3/8 and for 10 about 1/9. The benefit from switching never disappears but does get small fairly quickly.

We should put in an argument from Bayes' Theorem, just to prove we can. You open door A, with a 1/3 probability of getting the car. We want to know the posterior probability of P(Car behind B) after the host opens door C. We have: P(Car behind B given host opened C) = P(Host opened C given Car behind B)P(Car behind B) / P(Host opens C).

The prior probability of P(Car behind B) = 1/3 and P(Host opened C given Car behind B) = 1 by the rules of the game. What's not obvious is the prior probability of P(Host opens C). Since the host can only open B or C and since you don't know either way, the indifference principle says you should put P(Host opens C) = ½. Hence: P(Car behind B given host opened C) = (1/3) / (½) = 2/3.

So why did those professors (and me) fall into what trap? Our argument runs something like this: there are three doors and there's a goat behind C, so the car must be behind A or B. There's an equal chance it's either so there's no benefit in switching. Why is it an equal chance? Well, because there's one goat left and one car: two doors, two objects. What's wrong with that?

Nothing, because it's a correct calculation of the conditional subjective probability of the probability of P(Car behind A given Goat behind C).

Everything, because that's not the probability you need to make the decision. Why not? Because it's not what you know. You know much more than that and you haven't put it into your model. What you know is that the host either chose C at random (probability ½) or had to choose it because the car was behind B. The probability of that is 1/3*1/2 (at random) + 1*1/3 (car behind B) = ½ . So P(Car behind B given host opened C and you chose A) = 1/3*2/1 = 2/3. Voila! The 50:50 answer is the result of arguing correctly from the wrong model of the probabilities.

One final thought. If it's this hard to model a simple game show correctly, how confident are you about any other argument or mathematical model using probabilities?

(I wrote this back in 2003 or so and lately I've seen one or two other arguments using this approach.)