You are on a game show. Three closed doors: behind one is a car, behind the other two are goats. You pick door 1. The host, who knows where the car is, opens door 3 and shows you a goat. Then he asks: do you want to switch to door 2?
Most people say it makes no difference. Two doors left, one car, so it must be 50/50. That feels airtight, and it is wrong. Switching wins twice as often as staying.
Pick a door, watch the host open another, then choose. Play a dozen rounds, trying both strategies, and watch the tallies.
Pick a door to start.
Small samples are noisy. That is the point of the next panel.
Your first pick is right one time in three. That never changes, because the host has not told you anything about your door yet. The other two doors together hold the car two times in three.
Now the host opens one of those two doors. He is never allowed to open yours, and he never reveals the car. So he is not choosing at random: he is handing you the full two-thirds chance, concentrated on one door. If your first pick was a goat (2 in 3), the host is forced to open the only other goat, and the remaining door has the car. If your first pick was the car (1 in 3), switching loses.
So switching wins exactly when you started on a goat. A goat is what you start on two times out of three.
Here a simulation plays thousands of games with each strategy. The checkbox changes one rule: if the host does not know where the car is and just opens a random other door, something surprising happens.
Press the button to run the experiment.
With a knowing host, the results land close to 67% and 33%. With a host who opens a random door, about a third of games end early because he reveals the car by accident. In the games where he happens to show a goat, switching and staying each win about half the time. The same visible event, a goat behind the opened door, carries different information depending on why the host opened it.
The puzzle only works under these assumptions: the car is placed at random, the host knows where it is, he always opens a door you did not pick, he always shows a goat, and he always makes the offer. If he chose to offer a switch only when you had picked the car, staying would be the right move. The probability lives in the host's rules, not in the number of doors left.
The problem was first posed by Steve Selvin in a 1975 letter to The American Statistician. A second letter gave it the name it still carries: Monty Hall was the long-time host of the TV game show Let's Make a Deal. It became famous in 1990 when Marilyn vos Savant answered it in her Parade magazine column, saying the player should switch. She later estimated that around 10,000 letters came in, including close to 1,000 from people with PhDs, and that most of them insisted she was wrong. Even Paul Erdős, one of the most prolific mathematicians of the twentieth century, is said to have stayed unconvinced until he was shown a computer simulation of the result.
If it still feels wrong, imagine 100 doors. You pick one. The host, who knows where the car is, opens 98 other doors, all goats, leaving yours and one more. Would you switch? Almost everyone says yes. Your door had a 1 in 100 chance and still does; the other 99 doors had a combined 99 in 100, and that has all collapsed onto one door. Three doors is the same story with smaller numbers.