A car is behind one of three doors. You pick one, the host opens a different door showing a goat, and offers you a swap. Most people say it is 50/50. Play it yourself, then watch ten thousand games settle the argument.
Your own record, kept in this page only. Play ten or twenty rounds, trying both strategies. With few games the luck is noisy, which is exactly why the simulator below exists.
Same rules, no human. Each game puts the car behind a random door, the player picks at random, and the host always opens a goat door that the player did not pick.
| Strategy | Wins | Rate |
|---|---|---|
| Always stay | - | - |
| Always switch | - | - |
It gets obvious with more doors. Pick the number of doors, play one round where the host opens every other goat door but one, and see where the car tends to be.
The two fractions are computed from the number of doors N: staying wins 1 out of N, switching wins N minus 1 out of N. At N = 3 that is 1/3 against 2/3, so switching doubles your chances.
The result depends on one rule: the host knows where the car is and always opens a goat door. If he opened a random door and it happened to show a goat, switching would be no better than staying. Test it.
| Clueless host (random door) | Games kept | Rate |
|---|---|---|
| Stay | - | - |
| Switch | - | - |
Games where the random door revealed the car are thrown away, since the question only applies when a goat was shown. About a third of games are discarded.
All numbers on this page are produced by the code running in your browser (random games) or follow directly from the arithmetic above. I did not consult an external source for the puzzle history, so none is quoted. The puzzle is described in the Wikipedia article "Monty Hall problem" (secondary source, not opened for this page).