The Monty Hall Problem

The so-called Monty Hall problem is one of the best-known examples of a probability result that conflicts with our spontaneous intuition.

The initial situation

Behind three closed doors there is one car and two goats. A contestant first chooses one door. The host then opens another door, behind which there is a goat. The contestant may now keep the original choice or switch to the only other closed door.

Four rules are crucial for the classical solution:

• The host knows which door hides the car.
• He never opens the door initially chosen.
• He always opens a door with a goat.
• He always offers the contestant the opportunity to switch afterwards.

The common 50:50 intuition

After the goat door has been opened, only two doors remain closed. It therefore seems natural to think that the car must now be equally likely to be behind either one. This reasoning overlooks the fact that the host did not open a door at random. He knows where the car is and, under the rules of the game, is forced to show a goat.

The host's information therefore does not retrospectively change the probability of the first choice. At the moment of choosing, that probability was one third, and it remains one third. The two doors that were initially not chosen had a combined probability of two thirds. Once the host deliberately removes a goat door from that group, the full two-thirds probability remains on the only other closed door.

Solution 1: All possible initial choices

Suppose the car is behind Door 1. Since each of the three doors is equally likely to be chosen first, there are three equally likely cases:

A contestant who stays with the first choice wins only if the car was chosen immediately: one case out of three. A contestant who always switches wins whenever the first choice was wrong: two cases out of three.

Solution 2: Probabilities of groups of doors

At the first choice, the chance of selecting the car is one third. Conversely, the chance that the car is behind one of the two other doors is two thirds.

The host then provides additional information, but only within the group of the two unchosen doors: he deliberately reveals a goat there. The two-thirds probability cannot remain on the opened door, because we now know with certainty that there is no car there. The two thirds therefore remain with the only other closed door.

Put another way: the switching strategy wins exactly when the first choice was wrong. Since the first choice is wrong with probability two thirds, switching also wins with probability two thirds.

The 100-door thought experiment

With one hundred doors the difference is usually easier to see:

1. A car is behind one of 100 doors.
2. You choose one door. The chance that it is correct is only 1%.
3. The informed host opens 98 other doors and reveals a goat behind each one.
4. Only the door first chosen and one additional door remain closed.

The first door has not suddenly become more likely merely because the other doors were opened; it still has only a 1% chance of winning. The remaining 99% lies on the only other closed door. Under these rules, switching is obviously attractive.

Checking with a Python simulation

A simulation does not replace the mathematical proof, but it makes the result experimentally visible. The following program simulates the complete process, including the host's rule-compliant choice:

import random

rng = random.Random(2026)
doors = (0, 1, 2)
games = 100_000

stay_wins = 0
switch_wins = 0

for _ in range(games):
    car = rng.choice(doors)
    first_choice = rng.choice(doors)

    # The host may open neither the chosen door
    # nor the door with the car.
    possible_doors = [
        door for door in doors
        if door != first_choice and door != car
    ]
    opened = rng.choice(possible_doors)

    # After the host opens one door, exactly one
    # other closed door remains besides the first choice.
    switch_door = next(
        door for door in doors
        if door not in (first_choice, opened)
    )

    if first_choice == car:
        stay_wins += 1

    if switch_door == car:
        switch_wins += 1

print(f"Stay:   {stay_wins / games:.3%}")
print(f"Switch: {switch_wins / games:.3%}")

With the fixed seed 2026, the result is reproducible:

Stay:   33.195%
Switch: 66.805%

With other seeds the values vary slightly. As the number of games grows, however, they approach the theoretical values of one third and two thirds more and more closely.

When would the calculation be different?

The solution depends on the host's behaviour. If he opens a door at random, does not know where the car is, does not always offer a switch, or follows some other selection rule, a different probability model results. The statement “switching wins with probability two thirds” therefore does not apply to every possible variant, but only to the classical rules listed above.

Why the problem is instructive

The Monty Hall problem shows that probabilities cannot be judged solely by counting how many possibilities visibly remain. What matters is also how the observed information was generated. The deliberate action of an informed host is different from a random selection.

Gero von Randow's book

Gero von Randow uses the Monty Hall problem as a starting point for an accessible discussion of probability, chance and typical errors of reasoning. The book shows why statistical thinking often conflicts with our immediate intuition — and why a formal analysis is therefore so useful.

Reference: Gero von Randow, Das Ziegenproblem – Denken in Wahrscheinlichkeiten, Rowohlt.