You may have heard of the so-called Monty Hall problem: you’re on a game show, there are three doors, and there’s a car behind one door. You choose door 1. The host, Monty, opens a door which (1) is different than the door you chose and (2) has no car behind it. So let’s say he reveals that door 2 is empty. Now he offers you a choice: Should you switch to door 3?

The “standard” explanation goes something like this: When you first chose, the probability that the car was behind door 1 was 1/3, and the probability that it was behind another door was 2/3. Monty’s opening door 2 doesn’t change that, so the whole 2/3 not-door-1 probability “falls on” door 3. So the probability of door 1 stays 1/3, and the probability of door 3 becomes 2/3, ergo you should switch doors.

I say this is the “standard” explanation not because I’ve done any rigorous empirical canvassing of explanations, but because it’s an explanation I’ve heard a lot, and most of the explanations on the problem’s Wikipedia page have that flavor. (Admittedly, the Wikipedia page was kind of tl;dr, but at least many of the explanations toward the top had that flavor.)

But this explanation, though it comes up with the right answer, is totally false. (The later discussion on the Wikipedia page does explain the true solution, but I’m not sure that the criticism of the intuitive solutions really shows clearly enough why the intuitive solutions of the type I described above are false.)

To get some of the true intuition, consider the effect of various assumptions about Monty’s behavior. If the car is behind door 3, Monty must show you door 2. If the car is behind door 2, Monty must show you door 3. But if your choice of door 1 was correct in the first place, then Monty has a choice of whether to show door 2 or door 3. Call p the probability that Monty shows you door 2 when he has a choice. p can vary between 0 (Monty never shows you door 2 when he has a choice) and 1 (Monty always shows you door 2 when he has a choice). Now let’s work through some assumptions:

• Suppose p=0: Monty never shows you door 2 when he has a choice. Now suppose you see Monty open door 2. That means he didn’t have a choice, so the car is behind door 3. The fact that Monty chose door 2 was informative to you. Even though the old probability that the car was behind door 3 (the “prior”) was 1/3, the new probability after you’ve gotten this new information (the “posterior”) is 1, so you should definitely switch to door 3.
• Suppose p=1: Monty always shows you door 2 when he has a choice. Now suppose you see Monty open door 2. If the car is behind door 1, he always shows you door 2; and if the car is behind door 3, he likewise always shows you door 2. So his “choice” to show you door 2 was completely uninformative: the probability that the car is behind door 3 is just 1/2, same as the probability that it’s behind door 1. You could switch, but there’s no affirmative reason for you to do so.

First, it’s clear that any explanation that says something like “the probability of door 1 was 1/3, and nothing can change that…” is automatically fishy: probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance. Thus, maybe if we knew all the variables, we would be able to exactly predict the outcome of a coin toss; saying “heads and tails each have a probability of 1/2” is just a fancy way of saying “I don’t know what those variables are, and heads and tails are similarly situated on a coin (or perhaps I’ve seen the results of many other coin tosses), so there’s no reason for me to assume these results aren’t equally probable in this case.”

Second, it’s clear that the probability that you’ll win by switching all depends on the value of p — in other words, it depends on how Monty chooses door 2 when he has a choice. In general, whenever p<1, we can say the following: If the car is behind door 3, he must show you door 2. But if the car is behind door 1, he may show you door 2. So showing you door 2 is more likely under the door-3 scenario than under the door-1 scenario, so his choice of door 2 makes the door-3 scenario more likely. That’s the Bayesian method: whenever the evidence is “more consistent” with world A than with world B (meaning, the evidence would be more likely if A is true than if B is true), you revise your probabilities to believe A is more likely than you thought it was before.

The second bullet above shows you how Monty’s “choice” was uninformative when p=1. You can imagine other uninformative scenarios, where the new information you see is equally likely under either the door-1 scenario or the door-3 scenario so they don’t alter your view of their relative probabilities:

• A meteorite happens to come through the studio and destroys door 2. You happen to see that there’s nothing behind door 2. They quickly replace the door (thinking that you didn’t see anything) and continue with the game. The meteorite was uninformative, so the probabilities of doors 1 and 3 are both 1/2.
• You magically get a quick burst of X-ray vision. You happened to be looking at door 2 at the time and can see that there’s nothing there. You look to door 3 but, alas, your X-ray vision was highly temporary and has now worn out. This, too, was uninformative (you didn’t know the car is, so there was no particular reason you were looking at door 2 at the time), so again the probabilities of doors 1 and 3 are 1/2.

It’s not hard to see with notation: let’s call C1, C2, and C3 the events “car behind door 1,” “car behind door 2,” and “car behind door 3,” and M1, M2, M3 the events “Monty shows you door 1,” etc. Our priors are Pr(C1) = Pr(C2) = Pr(C3) = 1/3. We want Pr (C3|M2): the conditional probability that the car is behind door 3 given that Monty has shown door 2. By the rules of conditional probability:

Pr(C3|M2) = Pr(C3M2)/Pr(M2)

= Pr(C3M2) / [Pr(M2C1) + Pr(M2C2) + Pr(M2C3)]

= Pr(C3M2) / [Pr(M2|C1) Pr(C1) + Pr(M2|C2) Pr(C2) + Pr(M2|C3) Pr (C3)]

= Pr(C3) / [p Pr(C1) + 0 Pr(C2) + 1 Pr(C3)]

= 1/3 / [p/3 + 0 + 1/3]

= 1 / (p+1).

So the chance that you’d win by switching is 1/(p+1). When p=0, you win with a probability of 1 if you switch, and when p=1, you win with a probability of 1/2 if you switch — those were the two scenarios in the bulleted list above. When p=1/2, you win with a probability of 1/1.5 = 2/3 if you switch. (In any event, the probability that you win by switching, 1/(p+1), is always at least 1/2, so you can never do worse by switching; might as well switch even if you don’t know p.)

So the standard explanation gets you the right result, but for a somewhat random reason. Yes, this 2/3 happens to be equal to the prior probability of not-door-1, but that seems like just coincidence. The true explanation is that Monty must show door 2 if the car is behind door 3, but he may show door 2 if the car is behind door 1, so his choice to show door 2 gives you a moderate amount of information in favor of the door-3 scenario; his choice is more informative as p approaches 0.