Switching Doors Wins 2/3 of the Time, and the Host's Rule Is the Whole Reason

Three doors, a car behind one and a goat behind each of the others. You point at door 1. The host, who knows where the car is, opens one of the two doors you did not pick and shows you a goat. He offers you the swap. Take it or keep what you have?

Almost everyone says it makes no difference. Two doors are left, one car, so the odds must be even. That answer is wrong, and it stays wrong no matter how confidently it is delivered. Switching wins two times in three.

Two doors left is not the same as two equal doors

The reflex treats the reveal as if it reset the problem. It did not. Your door was chosen when all three were equally likely, and nothing that happened afterwards could touch it. Door 1 held the car with probability $1/3$ before the host moved, and it still does after.

The other two doors held it with probability $2/3$ between them. That mass did not evaporate when one of them was opened. It piled up on the single door left standing.

You can settle it by writing out all three cases, which takes less time than arguing about it. Put the car behind door 1, then door 2, then door 3, and keep picking door 1 every time.

Read the figure as a bet on your first guess. Switching loses exactly when your first pick was right, and that happens a third of the time. Staying loses the other two thirds. The whole puzzle is that inversion.

The host’s rule is part of the problem

Here is the part that decides interviews, and it is not the number. The answer $2/3$ depends on two things the story never states out loud: the host knows where the car is, and he always opens a goat door and always offers the swap.

Those are assumptions about the process that generated what you saw, not about the doors. Change them and the number changes with them. A candidate who says “two thirds” and stops has answered a question about doors. A candidate who says “two thirds, assuming the host is required to reveal a goat” has answered a question about information, and that is the one being asked.

The same answer in one line

Write $C_i$ for the car being behind door $i$, and let the evidence be $E$, the host opening door 3 after you picked door 1. The priors are $1/3$ each. The likelihoods are where the work is:

If the car is behind your own door the host has a free choice between doors 2 and 3, so he opens door 3 half the time. If the car is behind door 2 he has no choice at all: door 3 is the only goat he is allowed to show. He never opens the car. Averaging gives $P(E) = \tfrac13 \cdot \tfrac12 + \tfrac13 \cdot 1 = \tfrac12$, and Bayes finishes it:

The factor of two between $P(E \mid C_2) = 1$ and $P(E \mid C_1) = 1/2$ is the entire puzzle. Everything else in the two calculations is identical.

Change the host, and two thirds becomes one half

Now suppose the host has no idea where the car is. He opens one of doors 2 and 3 at random, and this time it happens to be a goat. You are looking at exactly the same picture: your door closed, one goat revealed, one door left. Should you still switch?

No. It is now a coin flip. The likelihood that changed is the one that mattered:

An ignorant host opens door 3 half the time regardless of where the car is, so the evidence no longer discriminates between $C_1$ and $C_2$. Conditioning on the reveal being a goat removes the $C_3$ branch and leaves the remaining two equally weighted:

Two situations, identical to look at, worth $2/3$ and $1/2$. Nothing about the doors distinguishes them. Only the rule that produced the reveal does, which is the same machinery that separates $1/3$ from $1/2$ in the two children problem: a fact that arrives because someone was obliged to send it carries more than the same fact arriving by accident.

On a desk this stops being a puzzle. A quote you receive because a counterparty chose to show it to you is not a quote sampled at random from the market, and pricing it as though it were is how people get picked off. Selection is part of the likelihood.

A hundred doors, and the intuition finally lands

If the three-door version still feels like a trick, scale it. A hundred doors, one car. You pick door 1. The host, who knows, opens ninety-eight goat doors and leaves door 73 closed alongside yours.

Nobody hesitates here. Your first guess was worth $1/100$, and watching someone step around door 73 ninety-eight times in a row is not a coincidence you want to bet against. In general, with $n$ doors:

which returns $2/3$ at $n = 3$ and $0.99$ at $n = 100$. Three doors is simply the smallest case, where the effect is real but too small to feel. The mechanism does not change on the way up.

Asked this cold, the answer that scores is $2/3$, said quickly, followed by the sentence that earns the rest of the marks: it is $2/3$ because the host is constrained, and it drops to $1/2$ the moment he is not. State the protocol you are pricing and there is nothing left to trap you with.