Monty Hall Calculator

The hundred-door explanation hides its own assumption

The standard way to make Monty Hall click is to scale it up: “imagine a hundred doors, you pick one, the host opens ninety-eight, would you switch?” It works, and it works because of a detail it does not say out loud.

The persuasive force comes entirely from the host opening ninety-eight. That concentrates all 99% of the remaining probability onto a single door. Switching wins 99% of the time and the intuition becomes obvious.

Now let the host open one door instead of ninety-eight. Staying still wins 1%. Switching now wins 1.0102%, because that 99% is shared among the ninety-eight doors still closed. Switching is still correct — it is always correct — and the advantage is a factor of 1.0102 rather than 99.

So the intuition pump proves rather more than the puzzle contains. Someone who internalises “more doors means switching is overwhelming” has learned something false about the general case. The tool makes the number of opened doors an input precisely so both versions are visible.

The advantage has a clean closed form: (d − 1) divided by (d − k − 1), where d is the doors and k the ones opened. At d = 3, k = 1 that is 2/1 = 2. At d = 100, k = 98 it is 99/1 = 99. At d = 100, k = 1 it is 99/98 = 1.0102. One formula covers every case, and reading it makes the structure obvious: what matters is how many doors the host removed, not how many there were.

Why your own door does not improve

The sticking point is always the same: after a door opens there are two left, so surely it is fifty-fifty. The answer is that the two remaining doors have different histories.

Your door was chosen when every door was equally likely. It had a 1-in-3 chance then, and nothing has happened to it since. The host did not consider it, could not have opened it, and told you nothing about it.

The other door survived a selection process. The host looked at the two doors you did not pick, knew which one held the prize, and deliberately opened the other. That constraint is where the information lives.

Think in terms of the two-thirds you did not pick. That group had a 2/3 chance of containing the prize. The host then removed a known-empty member of that group without touching its total probability — so the whole 2/3 now sits on the one door left in it.

The 50-50 intuition is a real principle applied out of place. Two outcomes are equally likely when nothing distinguishes them. Here something does: one door was picked blindly and the other was left standing by someone who knew where the prize was.

The seeded simulation in the tool exists for this. Algebra convinces almost nobody of this result; a hundred thousand games landing on 33% and 67% convinces most people immediately.

Change the host’s rule and the answer changes

Every number on this page depends on assumptions about the host that the puzzle usually states carelessly. They are not decoration — each one is load-bearing.

The host must know where the prize is. If the host opens a door at random and it happens to be empty, switching and staying are both worth 1/2 and the entire advantage disappears. The 2/3 comes from the constraint, not from the door count — this is sometimes called the Monty Fall variant, and it is the single most consequential difference.

The host must always open a door. If the host only offers the switch when your first pick was correct, switching loses every time. A host free to choose whether to offer at all can make switching arbitrarily bad, and nothing in the observed game reveals which rule is running.

The host must never reveal the prize. That is what makes the opened door uninformative about your own.

When the host has a preference among empty doors, the answer becomes conditional. If the host always opens the leftmost available empty door, then which door was opened carries extra information, and the switching probability varies between 1/2 and 1 depending on what you saw. The unconditional 2/3 still holds on average, which is why the distinction so rarely surfaces.

Which is the real lesson. The puzzle is not about doors; it is about conditioning on how information arrived rather than on what it appears to say. The same trap produces the base rate fallacy in the probability calculator and selection effects in the correlation page.

The same answer through Bayes

The counting argument convinces; the Bayesian one generalises. Both give 2/3, and doing it formally shows exactly which quantity does the work.

Start with the prior. You pick door 1. Each door has probability 1/3 of holding the prize.

Now the likelihoods — the probability the host opens door 3, given each case. If the prize is behind door 1, the host may open either 2 or 3, so the probability is 1/2. If the prize is behind door 2, the host must open door 3, so it is 1. If the prize is behind door 3, the host cannot open it, so it is 0.

Those likelihoods are the whole argument. Door 2 is twice as likely as door 1 to have produced what you saw, so after multiplying by equal priors the posterior is 1/3 for your door and 2/3 for door 2.

The asymmetry is created by the host’s constraint, not by the prior. The priors were equal; the likelihoods were not. That is why a random-opening host gives 1/2 — the likelihoods become 1/2 and 1/2, and equal likelihoods leave equal priors untouched.

This is the general lesson worth carrying away. Evidence updates a belief in proportion to how much more likely it was under one hypothesis than another. When people go wrong on Monty Hall they are treating the opened door as evidence about the prize, when it is evidence about the host’s constraint.

Why so many people got it wrong in public

The puzzle is famous less for its answer than for the reaction to it, and the episode is worth knowing because the objections were not stupid ones.

Marilyn vos Savant published the correct answer in 1990 and received around ten thousand letters disagreeing, roughly a thousand of them from people with doctorates. Several were published, confidently and dismissively. The answer was right.

Paul Erdős, one of the most prolific mathematicians of the century, refused to accept it until he was shown a simulation. That detail is the useful one: the algebra did not persuade him, and watching the frequencies did.

The strongest objections were about the problem statement, not the arithmetic. The original wording did not specify that the host always opens a door, always avoids the prize, and always offers the switch. Without those, the answer genuinely is not 2/3 — so some of the disagreement was correct about a different problem.

That is the durable lesson. Most probability disputes are disputes about the model, not the maths. Two people confident of different answers are usually solving different problems, and the productive question is which assumptions each is making rather than which arithmetic is wrong.

Sources and methodology

References for the puzzle and its variants.

• Selvin, “A Problem in Probability”, The American Statistician 29(1), 1975 — the first published statement
• Morgan et al., “Let’s Make a Deal: The Player’s Dilemma”, The American Statistician 45(4), 1991 — on the conditional version and host preference
• Rosenthal, “Monty Hall, Monty Fall, Monty Crawl”, Math Horizons 16(1), 2008 — the variants where the answer stops being 2/3
• Probability Calculator — including Bayes and the base rate fallacy this shares its structure with
• Beta Distribution Calculator — for updating a belief from observed counts
• How we calculate — methodology

Method. The probabilities are exact closed forms rather than simulation output: staying wins 1/d and each remaining unpicked door wins (1 − 1/d)/(d − k − 1), so the reported advantage is (d − 1)/(d − k − 1) at every configuration. The simulation shown alongside is seeded, which means the figure quoted on this page is the figure you will see — it exists because this is the one probability result that algebra reliably fails to settle and frequencies reliably do. The suite asserts the classic 1/3 against 2/3, that the 100-door pump gives 99% only when the host opens 98, that opening a single door of a hundred gives an advantage of 99/98 rather than 100/99 — a form I had wrong before the suite caught it — and that stay plus every remaining door sums to exactly 1 across four configurations. That engine is verified on every change against 99 assertions. The count and the per-case breakdown are published on the formula verification page.

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