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Probability

The Monty Hall Problem: Why You Should Switch Doors

Published November 4, 2026 · 2 min read

Three doors, one car, two goats. After the host reveals a goat, switching wins two times out of three — an answer that famously angered thousands of readers.

The puzzle

You are on a game show facing three closed doors. Behind one is a car; behind the other two are goats. You pick a door — say, Door 1. The host, who knows where the car is, opens another door, say Door 3, to reveal a goat. He then offers you a choice: stay with Door 1 or switch to Door 2. Should you switch?

Yes. Switching wins the car two-thirds of the time. Staying wins only one-third of the time.

The controversy

The puzzle is named after Monty Hall, host of the TV show Let’s Make a Deal. Statistician Steve Selvin posed it in The American Statistician in 1975. It became famous in 1990, when Marilyn vos Savant answered it in her Parade magazine column. She received thousands of letters, many from readers with doctorates, insisting the answer had to be 50/50. She was right; they were not.

Why switching works

When you first pick, you have a 1-in-3 chance of being right and a 2-in-3 chance of being wrong. The host’s reveal does not change that first pick — he will always be able to show a goat, whichever door you chose. So your original door still has a 1/3 chance. The remaining 2/3 has to go somewhere, and it all lands on the one door left closed.

Every possible starting position when you pick Door 1
Car is behindHost opensStay wins?Switch wins?
Door 1Door 2 or 3YesNo
Door 2Door 3NoYes
Door 3Door 2NoYes

Another helpful version: imagine 100 doors. You pick one, and the host opens 98 goat doors, leaving one other closed. Would you keep your original 1-in-100 guess?

The assumptions matter

The 2/3 answer relies on specific rules: the host always opens a door, always reveals a goat, never opens your door, and always offers the switch. If the host opened a door at random and happened to reveal a goat, the odds really would be 50/50. If the host offered a switch only when you had picked the car, switching would always lose. Many arguments about the puzzle are really disagreements about these rules.

Try it yourself

Use three playing cards — one ace for the car, two others for goats. Have a friend play host, run 30 rounds always switching, and count your wins. Divide wins by 30 and you should land somewhere near 67%. Small samples wobble, so more rounds bring you closer.

Sources & notes

  • Selvin, S. (1975). A Problem in Probability (letter to the editor). The American Statistician, 29(1).
  • vos Savant, M. (1990). Ask Marilyn. Parade Magazine, September 9, 1990, and follow-up columns.

This article is for general education and is not financial, tax, or legal advice.

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