, two goats and a car. You pick one, and , who knows what is behind each, opens another to reveal a goat and offers you the remaining unopened door. Two doors are left, so the instinct is that it cannot matter which you take.
It doubles your chances. Sticking leaves you with the you started with, because the host will reveal a goat whatever you picked first, so the opened door tells you nothing about your own. Switching wins two times in three, counterintuitive as that seems.
Assume a room with three doors. Behind two are goats, behind the third a shiny new car, and you are asked to pick a door and win whatever is behind it. Say you pick door one.
The host then acts under rules, not whim. Before your door is opened, he opens one of the other two to reveal a goat, and offers you the remaining unopened door. He does this regardless of what you chose.
Sticking cannot help. You keep the one in three you began with, because the host was always going to show you a goat whether or not your first pick was right. The reveal carries no information about your own door.
Switching is the better play. There is a chance you win the car by moving to the other door, counterintuitive though it seems, and the problem generalizes to four doors and beyond.
wat. Stick and win one time in three; switch and win two.
wat wat
The trap is assuming the host’s choice is random when the rules make it anything but.
How do we know?
Wolfram MathWorld states the problem, the rules binding the host, and the winning probability for each strategy, along with the generalization to four doors.
What is still unknown
The answer depends entirely on those rules: the host always opens a goat door and always offers the switch.
Learn something?
Where to next
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