The Monty Hall Problem

A probability puzzle.

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There are three doors. A car is behind one of them and a goat is behind each of the other two. You choose a door, but it stays shut.

Three doors, and the one you chose.

The host knows where the car is. He opens one of the two doors you did not choose, and he always opens a door with a goat behind it.

Then he offers you a choice: keep your door, or switch to the other one still shut.

The host opens a goat door. Two doors are still shut: yours, and one other.

Most people say it makes no difference — two doors are left, so it must be an even chance. It is not.

There are only three places the car can be, and they are equally likely. Work through all three and you get your answer, because switching wins in two of them.

stay wins 1 time in 3 · switch wins 2 times in 3

Every case. You chose door 1; the rows are the three places the car could be. Switching wins the two shaded rows.

Another way to see it: your first choice is right one time in three, and the host opening a goat door does not change that. So the car is behind one of the other doors two times in three — and the host has just told you which of those two it is.

Your first choice, and the two thirds it leaves behind.

Arguing about it settles nothing, so the visualisation plays the game over and over and counts. Both answers are tracked at once: what would have happened if you had kept your door, and what would have happened if you had switched.

Ten thousand games, played here. The two rates settle where the three cases say they should.

What we are doing

the objective

Play the game 10,000 times and count how often each answer wins.

Hide the car behind one of the three doors, at random.
Choose a door, at random. Choosing thoughtfully cannot help: the car was hidden before you chose, and you have nothing to go on.
Open a goat door. The host, who knows where the car is, opens one of the two doors you did not choose. If both hide goats he picks between them at random.
Count both answers. Whether keeping your door would have won, and whether switching would have won. Every game votes on both, so the two rates are measured on exactly the same games.

The claim we are checking: switching wins about twice as often as staying. And the advantage does not come from there being two doors left at the end — it comes from the host knowing where the car is. At the end the same games are replayed with a host who opens a door at random, and switching stops helping.

the settings

Games 10000 Doors 3 Games shown in full 6 Seed 7

the host

More than three doors. With ten doors the host still opens every goat door but one. Staying wins one time in ten; switching wins the other nine.

The first few games are dealt at reading speed. The rest are played at full speed.

The puzzle is from Steve Selvin (1975) and the American game show Let's Make a Deal, hosted by Monty Hall. Every number on this page is counted from games played in your browser — no library, and nothing precomputed.

one game—

Four beats to a game: the doors, then the door you picked (marked YOUR PICK), then the host opening a goat door, then the question — keep yours, or switch to the one still shut. Every door opens at the end, so you can see what each answer would have won.

every game so far—

The count comes first, then one mark per game underneath — green where switching would have won, coral where keeping your door would have won. The notch on the bar is where the three cases say it should land.