The Monty Hall Problem: Why Switching Doors Is Always Better

Three doors. Behind one is a car. Behind the other two, goats. You pick a door, say door 1. Before opening it, the host, who knows exactly where the car is, opens a different door, say door 3, and shows you a goat. Then he asks: do you want to switch to door 2, or stay with door 1?

Most people's gut says it doesn't matter. Two doors left, fifty-fifty, why bother switching?

That gut feeling is wrong. Switching wins two times out of three. Staying wins one time out of three. Switching doubles your odds, and once you see why, you won't be able to unsee it.

Why "fifty-fifty" feels so obviously right

Here's the trap. Once the host opens door 3, it really does look like a fresh problem: two closed doors, one car, pick one. Coin flip, right?

But the two doors aren't in the same position, and that's the whole story. Door 1 is the door you picked before you had any information. Door 2 is the door that survived the host's filtering. Those are two very different histories, even though right now they look like identical closed doors.

To see the difference, we need to go back to the moment you first pointed at door 1, before the host did anything at all.

Your first guess is almost always wrong

When you point at door 1, there's a 1 in 3 chance the car is behind it, and a 2 in 3 chance it's behind one of the other two. Nothing surprising there. Three doors, one car, even odds each.

Now group the doors into two piles: "the one you picked" and "the other two." The first pile has a $\frac{1}{3}$ chance of holding the car. The second pile has a $\frac{2}{3}$ chance, combined.

Here's the part that matters: the host's reveal doesn't touch your pile at all. He never opens your door. He only opens a losing door from the other pile, the one he knows won't have the car. So that pile still holds its full $\frac{2}{3}$ chance, it's just that now only one door in it is still closed. All $\frac{2}{3}$ of that pile's probability has nowhere to go except onto that one remaining door.

Play it out below. Pick a door, watch the host reveal a goat, then decide whether to stay or switch. Do it a bunch of times and watch which pile of dots grows faster.

Notice the switch tally climbing roughly twice as fast as the stay tally. That's not luck balancing out over a handful of rounds. That's the $\frac{1}{3}$ versus $\frac{2}{3}$ split showing up directly in your own data.

Watching the probability move

Let's slow the whole thing down into three frozen moments: right after you pick, right after the host reveals, and what that leaves behind. No random doors this time, just the bookkeeping.

Step through it. At stage one, all three doors carry an equal $\frac{1}{3}$ weight. At stage two, the host removes a losing door from the "other two" group, but crucially, he doesn't touch the weight, only the door. At stage three, that whole $\frac{2}{3}$ block has nowhere to sit except on the one door left in that group.

The host's knowledge is what makes this work. If he opened doors at random and just happened to reveal a goat, the odds really would end up fifty-fifty, because there'd be a chance he could have ruined the game by revealing the car. But he never does that. He always knows, and he always avoids the car. That guarantee is what lets the $\frac{2}{3}$ survive intact and pile onto a single door.

Does this actually hold up over many games, or is it a trick with small numbers?

Fair question. Three doors and a handful of rounds could just be noise. So let's not stop at a handful. Let's run this game thousands of times, automatically, and watch what each strategy's win rate settles into.

Hit play. Early on, with only a few games played, both lines jump around unpredictably, pure small-sample noise. But keep watching. As the number of games climbs into the hundreds and thousands, the stay line settles in right around $\frac{1}{3} \approx 0.333$, and the switch line settles in right around $\frac{2}{3} \approx 0.667$.

This is the law of large numbers at work: individual games are unpredictable, but the long-run average locks onto the true probability. And the true probability here just is $\frac{1}{3}$ for staying and $\frac{2}{3}$ for switching, exactly what we derived from the door-grouping argument, not a coincidence.

What if there were 100 doors instead of 3?

If the doubled odds still feel like a mathematical sleight of hand, this version usually settles it for good. Imagine the same game, but with 100 doors. You pick one. The host, who knows where the car is, opens 98 other doors, all goats, leaving your door and exactly one other door closed.

Would you switch?

Almost everyone says yes immediately, and their reasoning is exactly right: your original pick had a 1-in-100 chance of being correct. The host just spent enormous effort filtering 98 losing doors out of the other 99, concentrating essentially all of that group's probability onto the single door he left closed. Switching there feels obviously smart, because the imbalance is so extreme it's impossible to miss.

The 3-door version is the exact same logic. It just hides better, because the imbalance, $\frac{1}{3}$ versus $\frac{2}{3}$, is small enough to feel like noise.

Drag the door count down from 100 toward 3 and watch the switch-win probability shrink from "obviously yes" toward "twice as good, but easy to doubt." The math never changes. Only how convincing it feels does.

The general rule, in one line

For a game with $n$ doors, one car, where the host always reveals $n - 2$ losing doors from the group you didn't pick:

Plug in $n = 3$: $\frac{2}{3}$. Plug in $n = 100$: $\frac{99}{100}$. As $n$ grows, switching approaches a certainty, and staying approaches a near-guaranteed loss.

This is the same reasoning that shows up any time new information arrives without disturbing your original guess. It's a close cousin of Bayes' theorem: you start with a prior belief spread across possibilities, evidence arrives that rules some possibilities out, and the surviving possibilities absorb the probability that used to belong to the ones that got eliminated. If you want the fully general version of that update rule, that post walks through it from scratch.

The short version

Your first pick locks in a $\frac{1}{3}$ chance of being right and a $\frac{2}{3}$ chance of being wrong, and nothing the host does afterward can change that split, because he never opens your door.

When he opens a losing door from the other two, he's not creating new fifty-fifty odds. He's revealing which single door absorbs the entire $\frac{2}{3}$ chance that your first pick was wrong.

Staying wins $\frac{1}{3}$ of the time. Switching wins $\frac{2}{3}$ of the time. Not close, not a matter of taste, just arithmetic that plays out the same way every single time you run the game, whether you run it three times or three thousand.