The Monty Hall Problem, Explained Three Ways
Why switching doors doubles your odds, and why almost everyone gets this wrong the first time.
I first encountered the Monty Hall problem in a probability class and got it wrong, then I argued with the professor. Then I wrote a simulation to prove him wrong. The simulation proved him right. This is a fairly common sequence of events, the problem has a way of making intelligent people very confident in an incorrect answer.
The setup: you're on a game show. There are three doors. Behind one is a car; behind the other two are goats. You pick a door. The host, who knows what's behind each door, opens one of the other two doors to reveal a goat. He then offers you the chance to switch to the remaining unopened door. Should you?
The intuitive answer is that it doesn't matter, right. Two doors remain, one has the car, so it's 50/50. This is wrong. You should switch. Switching wins two-thirds of the time.
Way one: think about what the host knows. When you picked your door, you had a one-in-three chance of being right. That means there's a two-in-three chance the car is behind one of the other two doors. The host then eliminates one of those doors, but he always eliminates a goat, never the car. So the two-in-three probability doesn't disappear. It collapses onto the one remaining door the host didn't open. Your original door is still one-in-three. The other door is now two-in-three.
Way two: enumerate all cases. You pick door 1. Case A: car is behind door 1 (probability 1/3). Host opens door 2 or 3. You switch, you lose. Case B: car is behind door 2 (probability 1/3). Host must open door 3. You switch to door 2, you win. Case C: car is behind door 3 (probability 1/3). Host must open door 2. You switch to door 3, you win. Switching wins in two of three equally likely cases.
Way three: scale it up. Imagine 100 doors. You pick one. The host opens 98 doors, all goats, leaving just your door and one other. Would you switch? Obviously yes, the host's behavior has concentrated the probability onto that one remaining door. The three-door version is the same logic, just compressed enough that the intuition fails to track it.
What I find interesting about this problem isn't the answer, it's what it reveals about intuition. We're bad at reasoning about conditional probability. We don't naturally account for the information embedded in the host's action. The problem is a clean demonstration that our gut-level probability estimates are often systematically wrong, and that the only reliable fix is to slow down and count.