Monty Hall Problem
One car, a lot of goats. Pick a door, then Monty, the host, lets you stay or switch. Try 100 doors once you get the hang of it.
Try your luck
The car is placed randomly. Monty knows where it is. After your pick, he opens every other door except one, always revealing goats. He always offers you a switch.
Pick a door.
Your round, step by step
Why aren’t the two remaining doors equally likely?
Monty’s choice depends on your pick and the car’s location. When your first pick is wrong, he must leave the car as your switching option. When your first pick is right, he leaves a goat. Your first pick is wrong more often than it is right.
With three doors, suppose you pick door 1 and Monty opens door 3. If the car is behind door 1, Monty opens door 3 half the time. If the car is behind door 2, he opens door 3 every time. This reveal is twice as likely when the car is behind door 2, so that door has twice the probability.
This assumes Monty chooses uniformly among his legal options. If he instead opens an unchosen door at random without knowing where the car is, he might reveal the car. In the three-door game, conditional on that random reveal showing a goat, the remaining doors really do have equal chances.