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SWITCH.1/3 → 2/3
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THE MONTY HALL PROBLEM

Your first pick wins one time in three. Switching wins two times in three. The host knows where the car is, and never opens it.

$34
SMLXLXXL
Printed on demand in the US · Sizes and shipping shown in the listing
OBJECT
009
DROP
01
RELEASED
2026
BODY
100% cotton, black
PRINT
White, chest block + sleeve index
WHY 2/3?
STAY: 1/3
SWITCH: 2/3

Three doors, one car, two goats. You pick a door. The host — who knows what is behind all three — opens a different door, always revealing a goat, and offers you the swap. Switching wins two thirds of the time. Staying wins one third. Both numbers are exact, and the result has been confirmed by simulation, by formal proof, and by a large volume of angry correspondence.

The clean way to see it is to notice that your first pick is made in ignorance and never improves. It had a one-in-three chance of being the car when you made it, and nothing the host does afterwards can change a probability that was already fixed. So the door you chose holds the car one third of the time, and the other two doors together hold it two thirds of the time. The host then removes a goat from that pair — for free, because he was never going to open the car. The remaining unopened door now carries the whole two thirds by itself.

The enumeration is short enough to check by hand. Say the car is behind door 1 and you pick door 1: the host opens 2 or 3, you switch, you lose. Car behind 2, you pick 1: the host must open 3, you switch to 2, you win. Car behind 3, you pick 1: the host must open 2, you switch to 3, you win. Two wins, one loss, from three equally likely arrangements.

Everything hinges on the host's constraint, and this is where most disagreements actually live. He knows the layout, he always opens a door, and he always shows a goat. Change any of those and the answer changes. If he opens a door at random and it happens to reveal a goat, the two remaining doors are each one half and switching gains nothing. If he only offers the swap when you picked the car, switching is fatal. The famous 2/3 is a fact about that protocol, not about doors.

The problem entered the literature as a letter from Steve Selvin in The American Statistician in 1975, and became notorious in 1990 when Marilyn vos Savant gave the correct answer in Parade. She received thousands of letters telling her she was wrong, many from people with doctorates. Paul Erdős reportedly refused to accept it until he was shown a simulation. The result is not hard; the intuition against it is unusually strong.

What that intuition gets wrong is a single thing: it treats the host as a source of randomness rather than as a source of information. He is not flipping a coin. He is telling you something about the two doors you did not pick, and switching is how you collect it.

P(win | stay) = 1/3 P(win | switch) = 2/3 (host knows the layout and always reveals a goat)
SOURCES —
· S. Selvin, "A Problem in Probability" (letter), The American Statistician 29(1), 1975
· M. vos Savant, "Ask Marilyn", Parade, 9 September 1990 and the 1990–91 follow-ups
· J. Rosenhouse, The Monty Hall Problem (Oxford University Press, 2009)
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