doors and goats and cars, oh my

I was just reading about the Monty Hall problem. It’s a seemingly simple problem from a gameshow involving 3 doors, 2 goats, and a car. Two doors have goats behind them and one has a car that you can win. You pick a door and the host opens another door, revealing a goat, and asks if you want to switch to the other closed door. It turns out to be pretty complex for me to understand. I’d have figured it didn’t matter whether you switch to the other closed door. Since actual mathematicians were also mistaken about the problem in many cases, you now have an excuse to fail in a logic or probability class. 🙂 Anyway, I think this is the best explanation of the solution:

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 2007). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats — imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player’s door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch.

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