Four doors, two chances to switch
One of four doors hides a car and the other three hide goats. You pick a door. The host, who knows where the car is, then opens one of the other doors with a goat behind it (choosing at random if he has a choice), and lets you switch to any other closed door.
After your decision, he again opens a goat door that is neither your current door nor already open, and again lets you switch, now to the only other closed door. Then your door is opened.
What is your highest possible probability of winning the car?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
There are four plans: stay or switch at the first offer, then stay or switch at the second.
The host never opens the door you hold. What does that tell you about the door you held from the start?
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. The host never opens the door you are holding, so if you never leave your first door, he learns nothing about it: it keeps its chance, and at the end the only other closed door has the remaining . Switching early spoils this by putting you on a door the host was allowed to open.
Plan 1: stay, then switch. Your first door hides the car with probability , and the host never touches it. At the end, the two closed doors are yours and one other, and the car is behind one of them. So switching at the end wins exactly when your first door was wrong: .
Plan 2: stay, then stay. You win when the first pick was right: .
Plan 3: switch, then stay. After the first reveal, the car is behind your first door with probability , and the two other closed doors share , so each has . You move to one of them, and the second reveal does not touch it. So staying wins .
Plan 4: switch, then switch. The two closed doors at the end are your new door and one other, and they share all the probability. Switching wins .
Answer. The best plan is to stay at the first offer and switch at the last, winning with probability . A random choice between plans cannot beat the best plan, so this is the maximum.
The trap. Switching every time, by analogy with the three-door game, gives only .