Should you switch doors?

The rules: one car and N − 1 goats are placed at random behind N doors, and you pick one. The host knows where the car is. He always opens N − 2 of the other doors, every one a goat and never yours, so one other door stays closed. If your first pick is the car, he leaves a random goat door closed. Then you stick or switch.

Sticking wins only if your first pick was the car: P(stick) = 1/N. Switching wins every other time, because the host had to leave the car closed: P(switch) = (N − 1)/N. With 3 doors that is 1/3 vs 2/3; with 100 doors, 1/100 vs 99/100. Because the host picks at random when he has two goats to choose from, the same holds for whichever door he actually opened (Bayes: odds 1 : 2 for your door vs the other).

The answer depends on these rules. If the host didn’t know and opened doors at random that just happened to show goats, switching would win only half the time. If he could choose whether to offer a switch at all, the math changes again. Real TV game shows didn’t follow these rules.

Every tally is real games, played with a random number generator. Auto-play scores each game both ways, as if one player always sticks and another always switches; in each game exactly one of them wins. Your own games count only for the choice you made. Changing the number of doors starts a fresh tally. The film’s tally is 10,000 games with 100 doors from a fixed random seed, so it replays identically: sticking won 94 (0.9%) and switching 9,906 (99.1%).

Sources: Steve Selvin, “A problem in probability” and “On the Monty Hall problem,” letters to The American Statistician 29 (1975), JSTOR 2683689 and JSTOR 2683443; Marilyn vos Savant, “Ask Marilyn,” Parade, 9 September 1990, and follow-ups (archived column and replies); Monty Hall problem (Wikipedia); J. S. Rosenthal, “Monty Hall, Monty Fall, Monty Crawl”, Math Horizons, September 2008.