Monty Hall Problem Calculator

Three doors. Behind one, a car; behind the other two, goats. You pick a door. The host — who knows where the car is — opens a different door, always revealing a goat, and offers you the switch. Marilyn vos Savant’s 1990 answer (“switch — you win two times in three”) drew ten thousand angry letters, including from PhDs. This calculator settles it with exact probability for any door count and any number of doors the host opens, and lets the classic case’s arithmetic speak for itself: your first pick was right one time in three, and that fraction never moves — the host’s reveal pours all the remaining probability onto the one door he conspicuously avoided.

The Classic Case, Proved Three Ways

By cases. Your pick (say door 1) hides the car with probability 1/3 — then switching loses. It hides a goat with probability 2/3 — and then the host, who must show a goat and cannot open your door, is forced to leave the car’s door as the only switch target, so switching wins. Switching wins exactly when your first pick was wrong: 2/3 of the time. By symmetry-breaking. Before the reveal, the two unchosen doors split 2/3 evenly. The host’s informed choice is not a coin flip — he can always show a goat, and he avoids the car deliberately, so his reveal transfers the pair’s whole 2/3 to the surviving door. Probability is not divided by the reveal; it is concentrated by it. By the hundred doors. With 100 doors you pick one (1% car). The host opens 98 goats, sparing your door and one other. Did your 1% pick really nail it — or is the spared door, the one a knowing host tiptoed around 98 times, holding the 99%? The calculator’s general formula is this intuition made algebra: stay = 1/N, switch = (1 − 1/N)/(doors left closed besides yours).

Worked Numbers Beyond Three Doors

The general case punishes lingering intuition in larger arenas. Ten doors, host opens 7 goats: stay holds its original 10%; the other nine doors’ 90% now sits on the two doors left closed, so switching to a specific one wins 45% — this page’s default-adjacent example, and already a 4.5-to-1 edge. Ten doors, host opens only 1: switch = 90%/8 = 11.25% — still better than staying, by a whisker, because only a little information was poured. The lesson generalises cleanly: the switch advantage is exactly the host’s knowledge, quantified — every goat he is forced to step around concentrates the odds further. Real game shows dilute this with hosts who offer the switch selectively (only when you picked the car, in the adversarial version — against which staying is the only defence), which is why the problem’s fine print — the host always opens a goat and always offers the switch — is not pedantry; it is the load-bearing assumption.

Why Smart People Get It Wrong

The 1990 Parade magazine column remains psychology’s favourite exhibit because the error is structural, not stupid. The brain treats the two closed doors at the decision point as a fresh, symmetric choice — fifty-fifty — discarding the history that made them asymmetric: one door was chosen by a 1-in-3 guess, the other was selected by an informed adversary-avoidance process. Information theory’s framing is the cleanest: the host’s action carries information (he could not have opened the car), and probability must be updated on it — Bayes’ theorem with stage makeup on. Vos Savant’s mailbag — nearly a thousand PhD signatories among the corrections, many later writing back to concede after running simulations or classroom trials — is why this calculator leads with simulation-friendly numbers: the problem is now the standard first lesson that intuition about probability is a hypothesis, and enumeration is its referee. Pigeons, in controlled experiments, learn to switch. Undergraduates, initially, mostly do not — then they run the numbers, as you just did.

Variants That Change the Answer

Three twists repay a run through the inputs. The forgetful host: if the host opens a door at random and merely happens to reveal a goat, the information vanishes — conditional on the goat reveal, staying and switching are both 50% in the three-door game, because a random host might have exposed the car and ended the game, and surviving that risk is itself evidence about your door. The generous host with many doors: as N grows with K = N − 2 (he opens everything possible), switch approaches certainty — (N−1)/N on one door — the 100-door argument in the limit. Monty from Hell / Angel Monty: if the host offers the switch only when you have picked the car (hell) or only when you have not (angel), the offer itself is the information and the optimal policy inverts accordingly — a reminder that in any real negotiation, why you are being offered a trade is data. Set the calculator’s N and K to each scenario’s shape and the exact edge is printed before the argument starts.

Bayes’ Theorem in One Line of Arithmetic

For readers who trust Bayes more than intuition, the three-door case collapses into a single update. You picked door 1; the host opens door 3. Compare the two surviving hypotheses. If the car is behind door 2, the host was forced to open door 3 — probability 1. If the car is behind your door 1, he chose door 3 freely between two goats — probability ½. Weight each hypothesis by its prior (⅓ each) and normalise: the posterior for door 2 is (1 × ⅓) / ((1 × ⅓) + (½ × ⅓)) = ⅔. The host’s forced-versus-free asymmetry is the entire mystery, expressed as a likelihood ratio of 2:1 — the reveal was twice as likely to happen in the world where switching wins. Every version of the problem, at every door count, is this ratio being counted carefully; the calculator’s general formula is Bayes with the doors relabelled.

The Classroom Experiment, and What It Actually Shows

The standard classroom demonstration — thirty students, playing cards for doors, fifty rounds each of always-stay and always-switch — produces the two win rates near 33% and 67% with a scatter that itself teaches the second lesson: at fifty trials, a 67% strategy can easily show 58% or 76% on a given afternoon (the standard error of a proportion at n = 50 is about 7 points), and small samples of the right strategy can lose to small samples of the wrong one. Teachers who stop at “switching won” have taught a fact; the full lesson is the pair of numbers plus their wobble — probability governing both the long run and the noise around any short one. It is the same wobble the RSE Calculator prices for surveys, met here in a game show’s clothing.

Why Polls Keep Finding Stayers

Decades of classroom replications find the same distribution: offered the switch cold, most people stay. The reasons stack instructively. There is the false symmetry of the final two doors. There is status-quo bias — the pick feels owned, and losing after abandoning it stings more than losing with it, an asymmetry of regret rather than of probability. And there is a misread of the host as a tempter: game-show framing invites players to model the offer as a trick, importing a suspicion that the canonical problem’s rules explicitly forbid. The calculator’s role in this psychology is humble and final — it cannot make the switch feel safe, but it removes every numerical excuse for the feeling: 66.67% is printed in the same size type as 33.33%, and arithmetic is the one participant in the game with no motive to deceive you.

Related Tools

More from the Statistics Calculators hub: the Coin Toss Streak Calculator prices runs in random sequences, the Implied Probability Calculator converts betting odds into the probabilities inside them, and the Descriptive Statistics Calculator covers the data-summary side of the hub.

Frequently Asked Questions

So I should always switch?

Under the classic rules — host knows, always reveals a goat, always offers the switch — yes: switching wins 2/3 in the three-door game, and the general advantage grows with the host’s reveals. Change any clause of the rules and the answer must be recomputed, which the door-count inputs above let you do.

Why is it not 50/50 between the two closed doors?

Because the doors are not symmetric in history: yours was picked blind at 1/3, the other survived an informed elimination process that removed only goats. Probability attaches to how each door arrived at the final pair, not to the pair’s appearance at the end.

What if the host opens a door randomly?

Then, given that he happened to reveal a goat, the two remaining doors are equally likely — 50/50 — in the three-door case. The switch advantage is entirely the host’s knowledge and deliberate avoidance; randomise him and it evaporates.

How does the formula work for N doors?

Stay keeps 1/N. All other doors together hold (N−1)/N; after the host opens K goat doors among them, that mass rests equally on the N−1−K unopened switch targets, giving switch = ((N−1)/N) ÷ (N−1−K). At N=3, K=1: (2/3)/1 = 66.67% — the classic answer.

Did the real Monty Hall play it this way?

Not exactly — on Let’s Make a Deal the offers were theatre, varied case by case, and Hall himself noted the contestant’s psychology was the real game. The mathematical problem (posed by Steve Selvin in 1975, popularised by vos Savant in 1990) fixes the rules precisely so the probability has one answer; the calculator solves that canonical version and its generalisations.

Can I verify 2/3 by simulation?

Yes, and you should: deal the car randomly, pick randomly, always switch, and tally a few hundred rounds — the win rate settles near 66–67%, wobbling by the square-root law that governs all such tallies. The convergence of brute enumeration on the one-line proof is the subject’s happiest property.