Monty Hall: Switch vs Stay

The Monty Hall problem

Picture three doors. One hides a prize, the other two hide goats. You pick a door, and then the host Monty, who knows exactly what sits behind each one, opens a different door to reveal a goat. Now the question: do you switch or stay? It feels like a coin flip, but it isn't. Switching wins with probability 2/3 and staying only 1/3. Think about why. Your first pick was right just 1/3 of the time, which means the remaining unopened door soaks up the other 2/3. The puzzle blew up in 1990 after Marilyn vos Savant printed the answer in Parade magazine. Thousands of letters poured in saying she was wrong, plenty of them from people with mathematics PhDs. She wasn't. Stretch it to n doors and switching wins with probability (n−1)/n.

Applications

It's a workhorse in teaching probability, the go-to example for conditional probability and Bayesian updating. It also makes a great case study in cognitive bias and how badly intuition can fail. Beyond the classroom it shows up in decision theory under asymmetric information, in lessons on Monte Carlo simulation, and in any situation where evidence from an informed agent ought to shift your beliefs, such as auctions, security games and multi-armed bandits.

FAQ

Why isn't it 50/50 after one door opens? Monty doesn't pick at random. He always opens a goat door, never the prize. That deliberate choice leaks information, and the information breaks the apparent symmetry between the two doors still closed.

What if Monty opens a door at random? Then, on the occasions a goat happens to turn up, the odds really do collapse to 50/50. That's the "Monty Fall" variant, and it shows that the host's knowledge is doing all the work.

Does it work with 100 doors? It does, and this is where it clicks for most people. You pick one door, Monty throws open 98 goats, and switching wins 99/100 of the time. The bigger n gets, the harder the gain is to deny.