The Monty Hall Problem Explained

You're on a game show. There are three doors. Behind one is a car; behind the other two, goats. You pick Door 1. The host, who knows where the car is, opens Door 3 to reveal a goat. Then he asks: do you want to switch to Door 2?

Almost everyone says it doesn't matter. Two doors left, so it's 50-50. But that's wrong. Switching wins two-thirds of the time. Staying wins only one-third.

When this answer appeared in a popular magazine column in 1990, it triggered one of the most famous public math arguments in history.

The Host's Choice Gives You Information

The 50-50 intuition treats the two remaining doors as equal. They aren't, because the host doesn't open a door at random. He always opens a door that:

1. You didn't pick, and
2. Has a goat behind it

His choice is constrained by where the car is. That constraint leaks information, and it all flows to the door you didn't pick.

Explanation 1: Your First Pick Is Probably Wrong

When you first choose, you have a 1/3 chance of picking the car and a 2/3 chance of picking a goat.

• If you picked the car (1/3 chance): switching loses.
• If you picked a goat (2/3 chance): the host must open the other goat door, so the remaining closed door must hide the car. Switching wins.

Switching wins exactly when your first pick was wrong, which happens 2/3 of the time. Nothing the host does changes the fact that your original door had a 1/3 chance.

Explanation 2: List Every Case

Suppose you always pick Door 1. The car is equally likely to be behind any door:

Staying wins in 1 of 3 equally likely cases. Switching wins in 2 of 3.

Explanation 3: Imagine 100 Doors

Still not convinced? Scale it up. There are 100 doors and one car. You pick Door 1. The host, who knows where the car is, opens 98 other doors, all goats, leaving only Door 1 and, say, Door 57.

Do you really think your first random pick out of 100 is as likely as the one door the host carefully avoided opening? Your door has a 1/100 chance. Door 57 has 99/100.

Explanation 4: Bayes' Theorem

For a formal proof, let C₂ mean "car behind Door 2" and H₃ mean "host opens Door 3." You picked Door 1.

• If the car is behind Door 1, the host picks Door 2 or 3 at random: P(H₃ | C₁) = 1/2
• If the car is behind Door 2, the host must open Door 3: P(H₃ | C₂) = 1
• If the car is behind Door 3, he can't open it: P(H₃ | C₃) = 0

Using Bayes' theorem:

P(C₂ | H₃) = P(H₃ | C₂) × P(C₂) / P(H₃)
           = (1 × 1/3) / (1/2 × 1/3 + 1 × 1/3 + 0 × 1/3)
           = (1/3) / (1/2)
           = 2/3

See the rules of conditional probability on the statistics formulas page.

An Insider Reference: The Column That Started a War

The puzzle is named after Monty Hall, host of the TV game show Let's Make a Deal. It was first posed in this form by statistician Steve Selvin in two letters to The American Statistician in 1975.

It became famous in September 1990, when a reader asked Marilyn vos Savant about it in her "Ask Marilyn" column in Parade magazine. She answered correctly: switch. She later reported receiving around 10,000 letters, including nearly 1,000 signed by PhDs, many insisting she was wrong.

Even Paul Erdős, one of the most prolific mathematicians who ever lived, reportedly refused to accept the answer until his colleague Andrew Vázsonyi showed him a computer simulation. After thousands of trials, switching kept winning about two-thirds of the time.

When the Answer Changes

The 2/3 result depends on the host's rules. Change them, and the answer changes:

This is why careful wording matters in probability. The same observation can mean different things depending on how it was produced.

Two Concepts Worth Knowing

Conditional Probability

Conditional probability P(A | B) is the probability of A given that B happened. The Monty Hall problem is really a lesson in how new information should update probabilities.

Simulation

When intuition and algebra disagree, a Monte Carlo simulation, playing the game thousands of times with random numbers, settles the question empirically. That's exactly what finally convinced Erdős.

Quick Answer: Should You Switch in the Monty Hall Problem?

Yes. Your first pick has a 1/3 chance of being the car. The host always reveals a goat from the other two doors, so the 2/3 chance that the car was elsewhere moves entirely to the remaining unopened door. Switching wins 2/3 of the time; staying wins 1/3.

Try Them Yourself

• Random Number Generator: simulate 30 rounds with numbers 1–3
• Statistics Formulas: conditional probability and Bayes' theorem
• Number 3: the number of doors
• Division Tables: fractions like 1/3 and 2/3
• The Birthday Paradox: another probability shock
• Mastering Truth Tables: list every case systematically

Play the game with a friend using three cups and a coin. Play 30 rounds switching and 30 staying, and count wins. The numbers won't lie.