The Monty Hall problem

Three doors, one prize. You pick a door. The host, who knows where the prize is, opens a different door and shows you it is empty, then offers a swap. Swapping wins two times in three. Staying wins one time in three.

Why it is famous

Almost everyone's first answer is fifty-fifty: two doors left, one prize, so it cannot matter. That answer is wrong, and it stays wrong after you have been told it is wrong. The problem is a clean demonstration that "how many options remain" is not the same as "how likely each option is," and that what the host did not do carries information.

The mechanism

Your first pick is right one time in three. Nothing the host does afterwards can change where the prize already is, so your door stays at 1/3. The other 2/3 of the probability sits behind the two doors you did not choose.

Now the host acts, and the rule he follows is the whole point: he must open a door you did not pick, and he must open one that is empty. When your door is wrong — two times in three — one of the other doors has the prize and the host is forced to open the other one. The unopened door then holds the prize for certain. When your door is right — one time in three — the host opens either remaining door and the swap loses.

So the swap wins exactly when your first pick was wrong, which is 2/3 of the time. The host's rule funnels all of the 2/3 that was spread over two doors onto the single door he left shut.

Change the rule and the answer changes. If the host opened a random door without knowing where the prize was, and it happened to be empty, then the two remaining doors really would be fifty-fifty. The advantage comes from the host being forced.

In one breath

Your first pick is wrong 2/3 of the time. When it is wrong, the host has no choice about which door to open, and the closed one he leaves is the prize. Swapping converts that 2/3 into wins.