After a goat is revealed, should you stay or switch?

You pick one of three doors. One hides a car, the other two hide goats. The host, who knows where the car is, opens a different door to show a goat, then lets you stay or switch. It feels like 50–50.

The Monty Hall problem

If you stay
1 in 3
you win only when your first pick was right
If you switch
2 in 3
you win whenever your first pick was wrong
What most people guess
50–50
two doors left, so it feels like a coin flip

Every way the game can go

You pick Door 1 (outlined). The host always opens a goat door you didn't pick.

Three equally likely cases. If the car is behind Door 1, staying wins and switching loses. If the car is behind Door 2 or Door 3, the host opens the other goat door, so staying loses and switching wins. Staying wins 1 of 3 cases; switching wins 2 of 3. Where the car is You pick Door 1; the host opens a goat door Stay Switch Door 1 1 time in 3 Car Goat 3 Win Lose Door 2 1 time in 3 1 Car Goat Lose Win Door 3 1 time in 3 1 Goat Car Lose Win Total 1 of 3 2 of 3

10,000 simulated games

Random car and first pick each game; the host always reveals a goat

Bar chart of 10,000 simulated games: always staying won 3,316 games, about 1 in 3; always switching won 6,684 games, about 2 in 3. Always stay 3,316 wins Always switch 6,684 wins 1 in 3 2 in 3

Switching wins about twice as often. Your first pick starts with a 1-in-3 chance, and the host's reveal doesn't reset the two remaining doors to 50–50. The host knows where the car is and always opens a goat door you didn't pick, so the 2-in-3 chance that you were wrong gets concentrated onto the one other unopened door.

The same trap shows up whenever new information comes from someone who knows the answer: a reveal that was chosen isn't the same as one that happened by chance. Before updating your odds, ask how the information you just got was selected.

What did your gut say the odds were, and why?

Why does switching work?

Your original choice is correct only 1 time out of 3. The other two doors together begin with a 2-in-3 chance. Because the host knows where the car is and always opens a goat door you did not choose, that 2-in-3 chance is concentrated on the one other unopened door. If your original pick was wrong, which happens 2 times out of 3, switching wins.

It's easier to see with 100 doors. You pick one, and the host, who knows where the car is, opens 98 of the others to show goats. Would you stick with your first pick, or take the one door the host carefully avoided?

Explore other counterintuitive probability results

It sits in the same family as the birthday paradox: a probability result that keeps feeling wrong even after a simulation proves it, because intuition evaluates the choice in front of it and quietly ignores the information locked up in everything that did not happen.

Related exhibits

Sources: first rigorously stated by Steve Selvin in a 1975 letter to The American Statistician; popularized, and initially disputed by many mathematicians, after Marilyn vos Savant answered a reader's question about it in Parade magazine in 1990. The simulation is one run of 10,000 games with a fixed random seed.