The Monty Hall problem

You are on a game show. There are three doors. Behind one is a car; behind the other two are goats. You want the car.

You point at a door, say door 1. The host, who knows where the car is, does not open your door. Instead he opens one of the other two, always revealing a goat. Now two doors are still closed: yours, and one other. He asks if you would like to switch. Should you?

Yes. Switching gives you the car two times out of three. Staying only wins one time in three. It feels like it should be fifty-fifty with two doors left, but it is not, and you can prove it to yourself by playing a lot of rounds. The simulator above does exactly that.

The quick way to feel it: when you first picked, you almost certainly picked a goat, because two of the three doors are goats. Most of the time your first door is wrong. So most of the time, switching away from it is right.

The key is that the host knows what is behind the doors, and he never opens the car. That turns his choice into information.

When you first pick, the chance you landed on the car is \(1/3\), and the chance the car is behind one of the other two doors is \(2/3\). That does not change just because a door gets opened. What the host does is take that whole \(2/3\) and, by law, funnel it onto the single unopened door he left alone. So the door you can switch to carries a \(2/3\) chance, while your original door still carries only its first \(1/3\).

Another way to see it is to list the cases. Suppose the car is behind door 1 and you always start by picking door 1. If you stay you win. But if the car is behind door 2, the host must open door 3, and switching lands you on the car. If the car is behind door 3, the host must open door 2, and switching wins again. So out of the three equally likely places for the car, switching wins in two of them and loses in one.

The stayer wins only when the first guess was already right, which happens \(1/3\) of the time. The switcher wins whenever the first guess was wrong, which is the other \(2/3\). The two strategies split the outcomes perfectly, and switching gets the larger share.

If you find it slippery, make it a hundred doors. You pick one, the host throws open ninety-eight goats, and leaves one other door shut. Would you really keep your first one-in-a-hundred guess? Switching to the door the host so carefully avoided wins ninety-nine times in a hundred.

The Bayesian bookkeeping. Say you pick door 1 and the host opens door 3. Write \(C_i\) for "the car is behind door \(i\)" and \(H_3\) for "the host opens door 3." The priors are \(P(C_i) = 1/3\). The host's behaviour sets the likelihoods: if the car is behind 1 he may open 2 or 3, so \(P(H_3\mid C_1) = 1/2\); if the car is behind 2 he is forced to open 3, so \(P(H_3\mid C_2) = 1\); if the car is behind 3 he cannot open it, so \(P(H_3\mid C_3) = 0\). By Bayes' theorem \(P(C_2\mid H_3) = 2/3\) and \(P(C_1\mid H_3) = 1/3\). The maths agrees with the intuition: switch.

Why the host's rule matters. Every step of that depends on the host knowing the layout and deliberately revealing a goat. Change the rule and the answer changes. In the "Monty Fall" variant the host stumbles and opens a random door that happens to show a goat; now the reveal carries no intention, the two remaining doors really are \(1/2\) each, and switching gains nothing. Same picture on screen, different information behind it. The probability lives in the process, not in the final tableau of two closed doors.

The general case. With \(n\) doors, one car, you pick one and the host opens \(k\) goat doors, staying keeps probability \(1/n\) while switching (spread over the \(n-1-k\) doors left) does progressively better as \(k\) grows. The classic show is \(n=3\), \(k=1\): stay \(1/3\), switch \(2/3\). The lesson is the same one that runs through Bayes' theorem and the base-rate fallacy. Our instinct treats the two survivors as symmetric and ignores how the situation was set up, and that is precisely where the extra two-thirds hides.