Monty Hall problem calculator

The formula

There are n doors with a car behind one. You pick a door. The host, who knows where the car is, opens k of the other doors, all empty, and offers you the chance to switch to one of the n − 1 − k doors still closed.

P(win by staying) = 1 ÷ n P(win by switching) = (n − 1) ÷ n × 1 ÷ (n − 1 − k)

Your first pick is right with probability 1/n, and nothing the host does changes that. The car is behind one of the other doors with probability (n − 1)/n, and after the host opens k of them that probability is shared equally among the n − 1 − k that remain.

A worked example

The default is the classic game: 3 doors, and the host opens 1.

P(stay) = 1/3 = 33.33% P(switch) = (2/3) × (1/1) = 2/3 = 66.67%

Switching is twice as likely to win. A table of the three equally likely places the car can be makes this concrete. Say you pick door 1. If the car is behind door 1, the host opens 2 or 3 and switching loses. If it is behind door 2, the host must open door 3 and switching wins. If it is behind door 3, the host must open door 2 and switching wins. Switching wins in two of the three cases.

With 4 doors and the host opening 1, switching still helps, but less: (3/4) × (1/2) = 3/8 = 37.5% against 25% for staying. If the host opens 2 of the 4, switching wins 3/4.

Checking it by simulation

Choose “Also simulate games” and press Calculate. The calculator plays the game as many times as you ask: it hides the car at random, makes a random first pick, has the host open empty doors at random from those allowed, and records whether staying or switching would have won. The bar chart puts the simulated rates beside the exact ones. Run it a few times; with 10,000 games the rates settle within about a percentage point of 1/3 and 2/3.

How much switching helps as doors are added

The gain from switching depends on how many doors the host opens. If he opens all but one of the other doors (k = n − 2), switching wins (n − 1)/n, so the advantage grows with the number of doors. If he opens only one, switching wins (n − 1)/(n(n − 2)), which is only slightly more than 1/n when n is large. The ratio shown in the result is how many times more likely switching is to win than staying.

Common mistakes

• Treating the last two doors as equally likely. That would be true only if the host picked a door to open at random, without knowing where the car was.
• Thinking the host's reveal “resets” the game. It gives information about the other doors, not about yours.
• Forgetting the rules matter. If the host only offers a switch when you picked the car, switching always loses. The calculation assumes he always opens empty doors and always offers the switch.