Pattern · probability
The Door You Didn't Pick
Three doors. A car behind one, a goat behind each of the others. You choose; the host opens a different door to show a goat; then he offers you the swap. Keep your door, or take the last one? The answer feels like a coin-flip and isn't — and the reason it isn't hides in a detail almost every telling leaves out.
Play it — the three-door game
Pick a door to begin.
One game tells you nothing — luck drowns the signal. So play a few, watch your own two tallies drift apart, and then let the machine do what no contestant ever could: play the same game ten thousand times.
Skip the suspense — simulate
Uses the host behaviour you selected above: Host knows.
Why two-thirds, not one-half
Your first pick is right one time in three — that never changes, because nothing about your door is revealed. So a strategy of staying wins exactly 1/3. The car is behind one of the other two doors the remaining 2/3 of the time — and the host, by opening the goat among them, quietly sweeps that entire two-thirds onto the single door he leaves closed. Switching doesn't gamble on one door; it collects both of the doors you didn't pick, minus the goat the host kindly removed. That is the whole trick: the host's open door is not new randomness — it's a concentrated clue.
The detail the internet leaves out: the host has to know
Flip the toggle to “Host forgets” and run ten thousand games. Switching drops to a coin-flip — 1/2. Nothing about your choice changed. What changed is the host. The two-thirds edge was never really yours; it was information the host injected by deliberately avoiding the car. A host opening a door at random injects nothing: when he happens to miss the car (two times in three), the car is now equally likely to be behind your door or the other one, and switching gains you nothing. The famous answer “always switch” is not a fact about doors. It is a fact about a host who knows where the car is and is bound to reveal a goat. Change that rule and the answer changes with it — which is exactly why the puzzle has fooled so many careful people.
Turn up the doors (a knowing host) — and the intuition arrives
You pick one. The host throws open every goat door but one, then offers the swap. Switching wins 2 / 3 = 66.7%.
At 100 doors the host all but points at the car — and staying loyal to your one-in-a-hundred guess suddenly looks as foolish as it always was.
The check — show the working
Every probability here is computed two independent ways and they agree: by exact enumeration of the whole game tree, and by the same seeded Monte-Carlo you can run above. The certificate is research/monty-hall/verify.mjs — all checks pass:
- Knowing host: staying wins
1/3, switching wins2/3(enumerated exactly; 400,000-game simulation lands at 0.333 / 0.668). - Forgetful host: among games where a goat is shown, staying and switching each win
1/2— and the host reveals the car outright in1/3of all games. - N doors, knowing host opening all but one other: switching wins
(N−1)/N—2/3at N=3,0.99at N=100 (simulation matches each). - Price of the misconception: a switcher wins twice as often as a stayer in the classic game —
2/3vs1/3.
The smartest people in the world got this wrong
When Marilyn vos Savant gave the correct answer — switch — in her Parade column in 1990, she estimates the magazine received some 10,000 letters insisting she was wrong, close to 1,000 of them from PhD holders, many on university mathematics-department letterhead.1 Paul Erdős, one of the most prolific mathematicians who ever lived, refused to accept the result until a colleague showed him a computer simulation of it1 — the very thing you just ran. The problem had been posed and correctly solved fifteen years earlier, in 1975, by the statistician Steve Selvin in a pair of letters to The American Statistician; his second letter is the first known use of the name “Monty Hall problem,” after the host of the game show Let's Make a Deal.1
Honest apparatus
The classic answer rests on stated rules, not vibes. “Always switch”
requires that the host (a) always opens a door, (b) always opens one with a goat, never
the car or your door, and (c) always offers the switch regardless of your first pick. These
are the standard assumptions of the puzzle; vos Savant later clarified them after the
controversy, because a host with other habits gives other answers. The “Host forgets”
toggle drops assumption (b) and the answer moves to 1/2 — the page makes the
dependence operable rather than burying it. A host who only offers the switch when you've
picked the car could make switching lose every time; that adversarial case is
named here but not simulated.
What's proven vs. asserted. The probabilities are proven for the stated rules — enumerated exactly and matched by simulation (see the check). The historical figures (≈10,000 letters, ≈1,000 PhDs; Erdős persuaded only by simulation; Selvin's 1975 priority) are reported from the sources below, not independently audited; vos Savant's letter counts are her own estimate.
- Wikipedia, “Monty Hall problem” — the standard assumptions, vos Savant's 1990 Parade column and the ≈10,000-letter response, Paul Erdős's simulation, and Steve Selvin's 1975 American Statistician letters. en.wikipedia.org/wiki/Monty_Hall_problem. Origin: S. Selvin, “A Problem in Probability,” The American Statistician 29(1), 1975.
Verifier: research/monty-hall/verify.mjs (exact enumeration + seeded Monte-Carlo for the knowing host, the forgetful host, and the N-door version).