The Monty Hall Problem
Three doors, one car, two goats. The host opens a door with a goat. Should you switch?
Pick a door: click it, or press 1, 2 or 3.
What’s going on
You choose one of three doors. Behind one is a car; behind the other two are goats. The host knows where the car is. He always opens one of the doors you did not choose, always shows a goat, and then lets you change your mind.
Two doors are left, so it feels like a coin toss. It is not. Switching wins two times out of three.
The key is your first choice. It is right one time in three, and that does not change when the host opens a door, because he can always show you a goat, whatever you picked. So:
- One time in three you picked the car. Switching loses.
- Two times in three you picked a goat. The host must open the other goat door, so the one door he leaves closed is the car. Switching wins.
The host is not opening a random door. He is giving you information: he steers around the car. When you stick, you ignore that information.
Make it obvious with more doors
Change the scene above to ten doors. You pick one; the host opens eight doors, all goats, and leaves one other door closed. Would you stay with your first pick, which had a one-in-ten chance, or take the single door the host carefully avoided? With ten doors, switching wins nine times in ten.
Try it yourself
1. The host forgets where the car is
This time the host opens one of the other two doors at random, and it happens to show a goat. Should you switch now?
Answer
Now it really is fifty-fifty. Out of six equally likely situations (where the car is × which door he opens), the host reveals the car in two of them; those games are discarded. In the remaining four, sticking wins in two and switching wins in two. What made switching good was that the host knew and avoided the car.2. Three prisoners
Three prisoners, A, B and C. One of them, chosen at random, will be pardoned. A asks the guard: “Tell me the name of one of the others who will not be pardoned.” The guard says: “B.” A is pleased: “Now it is me or C, so my chance went up to one half.” Is A right?
Answer
No. This is the Monty Hall problem in disguise. The guard can always name someone other than A, so A's chance stays one in three. C's chance is now two in three — C is the “switch” door.3. Play smarter than the formula
Play twenty games in the scene above by hand, always switching. Then press “Play 1,000”. Why do your twenty games rarely show exactly two thirds?
Answer
Twenty games is a small sample. Two thirds of twenty is about 13, but 11 or 15 wins are perfectly normal. The share settles near 66.7% only over many games — that is the law of large numbers.Go deeper
- The puzzle became famous in 1990, when Marilyn vos Savant answered it in Parade magazine and received thousands of letters, many from mathematicians, insisting she was wrong. She was right.
- The original game show, Let’s Make a Deal, was hosted by Monty Hall. In the real show the host did not always offer a switch, which changes the maths.
Inspired by the article «Мерседес за тремя дверями» by S. Dorichenko in Kvantik magazine, July 2012. The text, code and pictures on this page are our own.
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