Monty Hall Problem: Why Switching Wins

simulator beginner ~8 min
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Switch ≈ 66.7% — always switch!

With 3 doors, switching wins approximately 66.7% of the time versus 33.3% for staying. The host's reveal transfers probability to the remaining door.

Formula

P(win|switch) = (N-1)/N for N doors (host opens N-2 losing doors)
P(win|stay) = 1/N
Advantage = P(switch) - P(stay) = (N-2)/N

The Most Controversial Probability Puzzle

In 1990, Marilyn vos Savant published the correct answer to the Monty Hall problem in Parade Magazine: you should always switch doors. Nearly 10,000 readers wrote in to disagree — including hundreds of PhD mathematicians. The problem is named after Monty Hall, the host of the TV game show Let's Make a Deal, where contestants chose between three doors.

How the Game Works

You pick one of three doors. Behind one is a car; behind the other two are goats. The host, who knows what's behind each door, opens one of the remaining doors to reveal a goat. You're then offered the chance to switch to the other unopened door. Should you switch? The simulation above runs thousands of games to show that switching wins approximately 66.7% of the time.

The Bayesian Explanation

When you pick a door, you have a 1/3 chance of choosing the car. That means there's a 2/3 chance the car is behind one of the other doors. When the host reveals a goat, they don't change these probabilities — they just concentrate the 2/3 probability onto a single door. Switching captures this 2/3 probability. This is a beautiful application of Bayes' theorem in action.

Beyond Three Doors

The generalized Monty Hall problem with N doors makes the logic even clearer. Imagine 100 doors: you pick one (1% chance of being right), then the host opens 98 doors with goats. Would you switch to the one remaining door? Of course — it has a 99% chance of hiding the car. The 3-door version follows exactly the same logic, just less dramatically.

FAQ

Why does switching win more often in the Monty Hall problem?

When you first pick a door, you have a 1/3 chance of being right. The host then reveals a goat behind another door, but this doesn't change your original 1/3 probability. The remaining door inherits the 2/3 probability that your first choice was wrong.

Does the Monty Hall problem work with more than 3 doors?

Yes, and the advantage of switching increases dramatically. With N doors, staying wins 1/N of the time while switching wins (N-1)/N × 1/(N-2) after the host opens N-2 doors. With 100 doors, switching wins about 99% of the time.

What assumptions does the Monty Hall problem require?

The problem assumes the host always knows where the car is, always opens a door with a goat, and always offers you the chance to switch. If the host opens doors randomly, the advantage disappears.

Why do people get the Monty Hall problem wrong?

People intuitively think that after a door is opened, the remaining two doors each have 50/50 odds. This ignores the crucial information that the host's choice was constrained — they could only open a door with a goat, which is a non-random action that shifts probabilities.

Sources

Embed

<iframe src="https://homo-deus.com/lab/probability/monty-hall/embed" width="100%" height="400" frameborder="0"></iframe>
View source on GitHub