The Monty Hall Problem

Behind one of three doors is a car. Behind the other two are goats. You pick a door. The host — who knows where the car is — opens one of the remaining doors to reveal a goat, then offers you the chance to switch. Should you?

The classical answer is yes: switching wins with probability 2/3, while staying wins only 1/3.

Where it comes from

The puzzle takes its name from Monty Hall, longtime host of the American game show Let’s Make a Deal (premiered 1963), where contestants chose among curtained options for prizes ranging from a new car to a stuffed donkey. Steve Selvin first posed the puzzle in this exact form in a 1975 letter to The American Statistician, but it became a cultural flash-point in September 1990 when Marilyn vos Savant gave the correct “switch” answer in her Parade magazine column. Roughly ten thousand readers wrote in to tell her she was wrong, including about a thousand from people with PhDs in mathematics. The number theorist Paul Erdős reportedly held out against the result until a colleague ran a computer simulation in front of him.

The intuition is genuinely hard. The game below lets you play your way to it; if the math at the end still doesn’t land, do what convinced Erdős — scroll down to the simulator and watch the empirical fraction converge.

Play

Simulate

Generalized rule. With $N$ doors total, after your initial pick the host opens $N-2$ goat-doors, leaving exactly one other door unopened besides yours. The classical version is the case $N = 3$, where the host opens a single goat-door.

Under this rule, switching wins with probability $(N-1)/N$ and staying wins with probability $1/N$. Try larger $N$ to see how decisively switching dominates — and how the convergence in the chart below sharpens.

Why switching wins

Your initial pick has probability $1/N$ of being the car, so the other $N-1$ doors collectively share probability $(N-1)/N$. The host then eliminates $N-2$ of those goat-doors — but crucially, never the car-door. All of that $(N-1)/N$ probability mass collapses onto the single unopened door that isn’t yours. Switching takes that mass; staying keeps your original $1/N$.

The host’s knowledge is what makes the puzzle work. If the host opened a remaining door at random (sometimes revealing the car, ending the game), the conditional probabilities would be different.

Two ideas at work

Beyond the answer, two foundational probability ideas show up clearly here.

1. Conditional probability and the law of total probability. Two random things happen: which door you initially pick, and whether your strategy wins given that pick. The conditional structure between them is the whole story. Under “switch”:

• Initial pick is the car, with probability $1/N$. Switching lands on a goat, so the conditional win probability is $0$.
• Initial pick is a goat, with probability $(N-1)/N$. The host has eliminated all other goats among the remaining doors, so the unique unopened non-pick must be the car. Switching lands on it, with conditional win probability $1$.

Combining via the law of total probability,

“Stay” reverses the conditional probabilities and lands at $1/N$. A good strategy isn’t about being lucky on the first event — it’s about converting first-event outcomes into second-event wins as efficiently as possible.

2. The law of large numbers. Each round of the simulator is an i.i.d. Bernoulli trial: $X_r = 1$ if the strategy wins on round $r$, otherwise $X_r = 0$, with $\mathbb{E}[X_r] = P(\text{win} \mid \text{strategy})$. The running curve in the chart is the sample mean

The law of large numbers guarantees $\bar{X}_R \to P(\text{win})$ as $R \to \infty$, and the fluctuation around the limit shrinks at rate $1/\sqrt{R}$, the standard error of a binomial fraction. Early rounds therefore swing wildly while the line eventually settles onto the dashed theoretical reference.