What Is the Monty Hall Problem?
Switching doors doubles your chance of winning. Thousands of people with mathematics degrees have written in to say that is wrong. Run it ten thousand times and watch.
Why switching wins twice as often
Junior level — plain language, no maths
Three doors. A car behind one, a goat behind each of the others. You pick a door. The host - who knows where the car is - opens one of the two you did not pick, always revealing a goat. Now you may keep your door or switch to the remaining one. Should you?
You should switch, and it wins twice as often. Almost nobody believes this the first time, and the reason is worth understanding: it feels like two doors are left, so it feels like fifty-fifty. But the two doors are not equivalent, because one of them was protected.
Think about your first pick. You had a one in three chance of being right, and that has not changed - the host opening a goat door tells you nothing about your own door, because he was always going to be able to open a goat door whatever you chose. So your door is still one in three. The other two doors together were two in three, and they still are - except now all of that is concentrated behind the single door left standing.
If that still does not land, use the slider to make it ten doors. Pick one, and the host opens eight goats, leaving yours and one other. Now the feeling matches the mathematics: your door was a one in ten guess, and the host has just carefully avoided one specific door eight times in a row. Then run ten thousand games and watch the two lines separate.
Things worth knowing
- When Marilyn vos Savant gave the correct answer in 1990, she received around 10,000 letters telling her she was wrong, including about 1,000 from people with PhDs.
- With ten doors, switching wins 90% of the time. The mechanism is identical to the three-door version - it is only easier to feel.
- Paul Erdős, one of the most prolific mathematicians of the century, refused to accept the answer until he was shown a computer simulation.
Conditional probability, and why the host's knowledge is the whole problem
Student level — the core equations
Label the doors 1, 2, 3, say you pick door 1, and let \(C_i\) be the event that the car is behind door \(i\), each with prior \(1/3\). Let \(H_3\) be the event that the host opens door 3. The likelihoods are what carry the argument: \(P(H_3|C_1) = 1/2\), because with the car behind your door the host may open either remaining door; \(P(H_3|C_2) = 1\), because he is forced; \(P(H_3|C_3) = 0\), because he never reveals the car.
Bayes then gives \(P(C_1|H_3) = \frac{(1/3)(1/2)}{(1/3)(1/2)+(1/3)(1)} = 1/3\) and \(P(C_2|H_3) = 2/3\). The asymmetry comes entirely from the likelihoods, not the priors - the host's action is twice as probable under the hypothesis that the car is behind the other door, so the evidence favours switching by a factor of two.
Everything hinges on the host's constraints, and the puzzle is famously ambiguous when they are left unstated. If the host does not know where the car is and opens a door at random, then conditional on his having happened to reveal a goat, the odds really are even and switching gains nothing. The identical observed event - a goat behind an opened door - carries different information depending on the process that produced it. This is the most valuable lesson in the problem and the one usually skipped.
The generalisation is clean. With \(n\) doors, one car, you pick one and the host opens \(n-2\) goat doors, switching wins with probability \(\frac{n-1}{n}\). At \(n = 3\) that is the familiar \(2/3\); at \(n = 100\) it is \(99\%\), which is the version to reach for when someone is not persuaded. In every case your original door keeps its prior of \(1/n\), and the entire remaining \(\frac{n-1}{n}\) is funnelled onto one door.
Key Formulas
| Prior | \(P(C_i) = \tfrac{1}{3}\) | |
|---|---|---|
| Likelihoods | \(P(H_3|C_1)=\tfrac12,\; P(H_3|C_2)=1,\; P(H_3|C_3)=0\) | |
| Posterior, stay | \(P(C_1|H_3) = \tfrac{1}{3}\) | |
| Posterior, switch | \(P(C_2|H_3) = \tfrac{2}{3}\) | |
| With n doors | \(P(\text{switch wins}) = \dfrac{n-1}{n}\) | |
Things worth knowing
- If the host chooses at random and happens to reveal a goat, switching gains nothing. Same visible outcome, different generating process, different information.
- The problem is equivalent to Bertrand's box paradox from 1889 and to the three prisoners problem from 1959. It is repeatedly rediscovered because the intuition fails reliably.
- The general answer is (n−1)/n. The three-door case is the hardest to feel precisely because 2/3 is the smallest the advantage ever gets.
Protocol dependence, the Monty Fall variants, and why the intuition fails
Scholar level — full mathematical depth
01The problem is underdetermined until the protocol is fixed
The classic statement does not specify the host's rules, and the answer genuinely depends on them. Under the standard protocol - the host always opens a door, always a goat, never yours, and chooses uniformly when he has a choice - switching wins with probability \(2/3\). Under Monty Fall, where the host slips and opens a door at random which happens to be a goat, the posterior is \(1/2\) and switching is worthless. Under Monty Crawl, where a lazy host always opens the lowest-numbered permissible door, the answer depends on which door he opened: sometimes \(1/2\), sometimes \(1\). The observed data are identical in all three; only the sampling mechanism differs.
02Non-uniform host choice, and the game-theoretic version
Suppose the host, when free to choose, opens door 3 with probability \(q\). Then \(P(C_1|H_3) = \frac{q}{q+1}\), ranging from \(0\) to \(1/2\) as \(q\) runs from 0 to 1. Switching is therefore never worse than staying and is strictly better unless \(q = 1\), so it weakly dominates for every host policy. Treating it as a zero-sum game against an adversarial host, the contestant's minimax strategy is to switch, guaranteeing \(2/3\), and the host's minimax is \(q = 1/2\) - the uniform rule is the host's optimal play, which is a satisfying justification for the conventional assumption.
03Why the intuition fails, experimentally
The dominant account is the equiprobability bias: when the outcome space has been reduced to two, people default to assigning equal probabilities regardless of the process that produced the reduction. A second factor is a number-of-cases heuristic, counting states rather than weighting them. The finding that makes it more than a curiosity is that the error persists under repeated play with feedback - people do not reliably learn to switch even after dozens of trials, which is unusual for a task with immediate outcome information, and suggests the failure is representational rather than a matter of insufficient evidence.
04Pigeons, and what that implies
Herbranson and Schroeder trained pigeons on the problem with food rewards. Within about thirty sessions the pigeons converged on switching at rates above 90%; humans in the comparable condition stayed near 60% and improved slowly. The interpretation is that pigeons are behaving as pure empirical maximisers, matching their choices to observed reward frequencies, while humans deploy a causal model of the situation and their model is wrong. Having a theory is normally an advantage, and here it is precisely what does the damage.
05The general principle worth extracting
The transferable lesson is that likelihoods depend on the data-generating mechanism, not on the data. This is the same structure as selection bias in trials, as the difference between missing-at-random and missing-not-at-random data, and as the boy-or-girl paradox where the answer depends on how the information was obtained. Monty Hall survives as a teaching example because it is the shortest problem in which the observation is fixed, the answer moves, and the only thing that changed is how the observation came to be made.
Key Formulas
| Bayes | \(P(C_i|H_3) = \dfrac{P(H_3|C_i)P(C_i)}{\sum_j P(H_3|C_j)P(C_j)}\) | |
|---|---|---|
| Biased host | \(P(C_1|H_3) = \dfrac{q}{q+1}\) | q = host preference for door 3 |
| Monty Fall | \(P(C_1|H_3,\text{random}) = \tfrac{1}{2}\) | same data, no information |
| Minimax | \(\text{switch guarantees } \tfrac{2}{3};\quad q^{*}=\tfrac12\) | |
Things worth knowing
- Pigeons learn to switch within about thirty sessions and reach over 90%. Humans given the same feedback stay near 60%, apparently because they have a causal model and it is wrong.
- If the host prefers door 3 with probability q, staying wins with probability q/(q+1). Switching weakly dominates for every possible host policy.
- Treated as a zero-sum game, the contestant's minimax strategy is to switch and the host's is to choose uniformly - which is exactly the assumption the classic version makes.