Explain the Monty Hall problem and why switching wins two-thirds of the time.
In the Monty Hall problem, a host who knows where the prize is always opens an empty door — this action transfers probability mass to the remaining unopened door, making switching win with probability 2/3 and staying win only 1/3. The key is that the host's action is not random.
How to think about it
The Monty Hall problem is a canonical test of conditional probability intuition. The answer — switch — is counterintuitive, so walk through it rigorously rather than hand-waving.
Setup
Three doors: one hides a car, two hide goats. You pick door 1. The host, who always knows where the car is, opens one of the other doors to reveal a goat, then offers a switch.
Why 1/3 vs 2/3?
Enumerate your initial pick by whether you originally chose the car:
Case 1 — You initially picked the car (prob 1/3). Host opens either goat door. Switching loses.
Case 2 — You initially picked a goat (prob 2/3). Host is forced to open the other goat door — only one option. Switching wins.
So switching wins exactly when your initial pick was wrong, which happens with probability 2/3.
Bayesian confirmation
After the host opens door 3:
P(car at door 2 | host opens door 3) = P(host opens 3 | car at 2)×P(car at 2) / P(host opens 3)
= 1 × (1/3) / (1/2) = 2/3Meanwhile P(car at door 1 | host opens door 3) = (1/2 × 1/3) / (1/2) = 1/3.
Simulation sanity check
In 9 000 simulated games:
- Staying wins approximately 3 000 times (~33 %).
- Switching wins approximately 6 000 times (~67 %).
These numbers converge reliably even after a few hundred iterations — try it in five lines of Python.