Monty Hall problem

Why does switching your choice of door improve the odds when only two doors are left?

Three doors, one car, two goats. You point at a door. The host opens one of the others to show a goat and offers you the swap. Two doors remain, one prize between them, and the symmetry is almost impossible to argue with: the standard reply, offered in 1990 to Marilyn vos Savant by thousands of correspondents including many with doctorates, was that it must now be even money.

The usual correction — your door was one in three and stays one in three, so the other must be two in three — gives the right number, and is also the least interesting sentence in the problem, because it does not say what did the work. Two doors look identical; something makes them unequal. Until we can name it, we cannot tell when the answer holds.

Start with a rule that governs all of this: an observation changes your view of a hypothesis only to the extent that it was more likely under that hypothesis than under its rival. If a thing would have happened either way, it is not evidence.

So ask the awkward question about your own door. Given that you picked the car, what is the chance the host then opens some door showing a goat? One — both remaining doors hide goats, and he takes his pick. Given that you picked a goat, what is the chance he opens a door showing a goat? Also one — exactly one of the other two is a goat, and he opens it. The ratio is one to one. His action was guaranteed whatever you had done, so it carries no information about your door at all, and one in three it remains.

But look what has happened on the other side of the table. When you picked a goat — which is two times out of three — the host had no choice; the door he left shut is the car, every time. When you picked the car, he had a free choice, and the door he leaves shut is a goat. So the surviving door is a car in exactly the cases where you were wrong. Its two-in-three is not left over from anywhere; it is the frequency with which the host was cornered.

That reframing earns its keep, because it tells us what to vary. Suppose the host does not know where the car is and opens one of the other two doors blindly, and it happens to be a goat. Now redo the ratio. If you picked the car, a blind opening shows a goat with probability one. If you picked a goat, a blind opening shows a goat only half the time — the other half he exposes the car and the game is void. The likelihood ratio is now two to one in favour of your door, which lifts it from one in three to one half, and switching genuinely gains nothing. Same two doors, same goat, same view from your chair — and a different answer, because the rule generating the reveal changed.

Push it once more. Suppose the host offers the swap only when you have already picked the car. Switching now loses every single time, and the offer itself is the evidence.

So the mechanism is not the doors and not the counting. It is that the host's behaviour depends on where the car is, and the strength of that dependence is exactly how much you learn. A useful confirmation for the intuition: run it with a hundred doors. You point at one, the host opens ninety-eight goats, and the single door he skipped past ninety-eight times is suddenly very conspicuous.

Three refinements hold the argument together.

First, the two-thirds answer is a conclusion about a protocol, not about a scene. It requires that the host always opens a door, always shows a goat, always offers the switch, and never opens your door. Those conditions are stipulations of the puzzle, not observations from the studio — and the real television programme did not reliably satisfy them, which is why a good deal of the original argument was people reasoning correctly from different unstated rules.

Second, there is a residual and genuinely arbitrary choice: when you have picked the car, which of the two goats does the host reveal? The standard answer assumes he picks between them evenly. If he had a known bias — always the leftmost, say — then seeing the rightmost opened would tell you more, and the odds on the two remaining doors would no longer be a flat two-thirds and one-third.

Third, the common gloss that "your first choice was probably wrong" is true but incomplete on its own. Being probably wrong is not enough; you also need the surviving door to have been selected against being a goat. Take the same three doors and let a gust of wind blow one open on a goat, and you are back at even money, still probably wrong about your original pick.

• Why a randomised host and an informed host produce different odds from an identical-looking outcome, and what that says about conditioning on what you might have seen.
• The three-prisoners problem and Bertrand's box, which are the same mechanism wearing different costumes.

Every term the collection defines is gathered in the glossary.