You pick one of three doors. The host, who knows where the car is, opens a different door with a goat, then asks: stick or switch? Almost everyone says it makes no difference. It does. Switching wins twice as often, and you can prove it here by playing two hundred games in two seconds.
Most people think that after a goat is revealed the last two doors are 50:50. In fact your first pick stays right only one-third of the time, and the host's knowing reveal funnels the other two-thirds onto the single unopened door, so switching wins 2/3.
What's actually happening
The setup comes from the old game show Let's Make a Deal, and the storm it caused is half the fun. In 1990, Marilyn vos Savant answered it correctly in Parade magazine — switch, it doubles your chances, and received some ten thousand letters insisting she was wrong, around a thousand of them from people with PhDs. The greatest mathematicians weren't immune: Paul Erdős reportedly refused to accept the answer until he was shown a computer simulation. The tally counter above is exactly that simulation.
Here's the way to feel it rather than fight it. Your first pick is right one time in three, and nothing the host does afterwards can reach back and improve a guess you already made. So two-thirds of the time, the car is behind one of the doors you didn't pick. Now the host does you an enormous, easily-missed favour: from those two doors, he removes a guaranteed goat. He hasn't shuffled anything; he's taken the two-thirds share and funnelled all of it onto a single door. Sticking keeps your original ⅓. Switching inherits the ⅔.
The detail everything hinges on: the host knows. He always opens a goat door, never the car, never yours. If instead a clueless host opened a random door (sometimes revealing the car and spoiling the game), the leftover odds really would be 50:50, and intuition would be right. The puzzle isn't about doors at all; it's about what information someone's deliberate behaviour leaks. That instinct, formalised, is Bayesian reasoning, and it runs spam filters and medical-test interpretation alike.
Switching wins the car twice as often, because the puzzle is really about what a host's deliberate, informed choice leaks — the seed of Bayesian reasoning.
- 1Hide a coin under one of three cups while a friend looks away. They pick a cup; you (knowing where the coin is) lift an empty one from the other two, and they decide: stick or switch.
- 2Play twenty rounds with them always sticking, twenty always switching, tallying wins.
- 3Sticking lands near 7/20; switching near 13/20. Watching their own tally beat their own intuition is the moment it clicks — it never works by argument alone.
Common questions
Your first pick stays right one-third of the time no matter what the host does. So two-thirds of the time the car is behind a door you did not pick, and the host removes the goat from those two, concentrating all that probability onto the one he leaves.
He always opens a goat, never the car. If a clueless host opened a random door, the leftover odds really would be 50:50. The puzzle is about what someone's deliberate behaviour leaks — the seed of Bayesian reasoning.
Yes. After Marilyn vos Savant gave the correct answer in 1990, about ten thousand letters insisted she was wrong, roughly a thousand from people with PhDs. Paul Erdős only accepted it after seeing a simulation.
Pick one of a hundred, then the host opens ninety-eight goats, leaving your door and one other. Now switching obviously wins: your first guess was right only 1 time in 100, so the remaining 99% piles onto that single other door. The three-door game is the identical trick, just small enough that intuition fights it.