The Monty Hall calculator above computes the probability of winning by switching and by staying, for any number of doors, any number of host reveals, and four different accounts of what the host is doing. The famous answer — switching wins two thirds of the time — is correct, but it is correct only under a specific set of assumptions about the host that the puzzle's usual wording never states.
Arb Digital publishes it as a companion to our other conditional-probability tools, and the reason for building it this way is the same reason our boy or girl paradox calculator presents several readings side by side. In both problems the arithmetic is easy and the disagreement is about what generated the evidence. Getting into the habit of asking that question is worth considerably more than memorising two thirds.
What This Monty Hall Calculator Does
It computes exact probabilities, not simulations. For each host rule it works out the probability that the prize is behind your original door given that you have received the offer to switch, and the probability that it is behind any one of the doors still closed. It generalises to n doors with h reveals, so you can watch the effect grow as the numbers do.
All four rules are plotted together on every calculation. That is deliberate: showing one answer would let you leave with the impression that the puzzle has one, and it does not. The classic two-thirds figure belongs to the standard rule and to no other.
The projection over a run of games converts the probabilities into expected wins, which is often the form that makes the size of the effect land. Over 900 plays of the standard three-door game, switching is worth 300 extra wins.
How to Use It
- Pick the host rule that matches the game you are actually being offered. If nobody has told you the rule, that is the finding, and the answer is genuinely indeterminate.
- Set the number of doors. Three is the television version. A hundred is the version that convinces sceptics in one line.
- Set how many doors the host opens. The tool leaves at least one door besides yours; opening more concentrates the probability further.
- Read the ratio. Switching versus staying is the number that matters, and it is not always greater than one.
- Compare the bars before concluding anything. The four rules give answers from certain loss to certain win.
The Formula and How It Is Calculated
Take n doors, one prize, and your initial pick. Before the host does anything, your door holds the prize with probability 1 divided by n, and the remaining n − 1 doors hold it with probability (n − 1) divided by n between them.
Under the standard rule the host knows where the prize is and deliberately opens h doors that do not hide it. Because the host was constrained to avoid the prize, opening those doors tells you nothing about your own door — the host could always have done it whatever you picked. Your door stays at 1 divided by n, and the whole of (n − 1) divided by n is now concentrated on the n − 1 − h doors still closed. Each of those holds the prize with probability (n − 1) divided by n(n − 1 − h). With n = 3 and h = 1 that is two thirds against your one third, the result Wolfram MathWorld states on its Monty Hall problem page.
Under the ignorant-host rule the arithmetic changes because the reveal is now informative. A host opening doors blindly could have exposed the prize and did not, and that near miss is evidence. Conditioning on it, your door's probability rises from 1 divided by n to 1 divided by (n − h), and each remaining closed door sits at exactly the same figure. Switching and staying become equally good, which for the three-door game means one half each rather than two thirds and one third.
Why The Host's Knowledge Changes The Answer
This is the part that most retellings skip, and it is the entire content of the problem. Two hosts open the same door and reveal the same goat. In one case switching wins two thirds of the time; in the other it wins half. The physical situation you are looking at is identical. What differs is the set of things that could have happened and did not.
The knowing host was never going to open the prize door. That constraint means the reveal carries no information about your door — it was guaranteed regardless. The blind host might have opened the prize door, and the fact that this did not happen is evidence, and the evidence points slightly towards your original pick, because worlds in which you picked correctly were worlds in which the host could not possibly have blundered.
The malicious and generous rules push this to the extreme. A host who only offers a switch when you have already won turns the offer itself into a proof that you should decline; a host who only offers when you have lost turns it into a guarantee that you should accept. Neither is contrived — a game show has every commercial reason to behave strategically, and the original puzzle says nothing that rules it out.
What The Puzzle's Wording Leaves Out
Richard Gill's paper "The Monty Hall Problem is not a Probability Puzzle (it's a challenge in mathematical modelling)", published in Statistica Neerlandica in 2011, makes this argument formally. Gill's position is that the standard textbook treatment, which makes a set of "obvious" or "natural" assumptions and then computes a conditional probability, is a classic case of a solution driving the choice of problem rather than the other way around.
The assumptions being smuggled in are at least these four: the host always opens a door; the host always opens a losing door; the host always offers the switch regardless of your pick; and where the host has a choice of losing doors, the choice is made with some specified probability. Change any one of them and the answer moves. Leave any one of them unstated and the answer is not determined.
There is a further subtlety even within the standard rule. If you picked the prize, the host has two losing doors to choose between, and if that choice is biased — say the host prefers the left-hand door — then the probability of winning by switching depends on which door was opened. It is still never worse than two thirds and never better than one, so switching remains the right call, but the specific number requires the bias to be specified.
The Hundred-Door Version
The most effective way to convince someone who has decided the answer must be one half is to raise the number of doors. Set the tool to 100 doors with 98 reveals. You pick one door out of a hundred; the host, who knows, opens 98 losing doors; one other door remains.
Almost nobody claims their original one-in-a-hundred guess has become a coin flip. The prize is behind the other door 99 times out of 100, and the reason is visible: your door was fixed at one percent before the host acted, and the host's constraint prevented that number from moving. All the probability that was spread across the other 99 doors has been swept onto the single survivor.
Run the same setup under the ignorant rule and the contrast is sharp. A blind host opening 98 doors and missing the prize every time is an extraordinary coincidence, and conditioning on it leaves your door and the survivor at one half each. Same doors, same reveals, entirely different answer, because a different process produced them. Our conditional probability calculator handles the general form of this updating step.
Where This Sits Among The Classic Puzzles
Three well-known problems share this shape, and it is worth seeing them together. Bertrand's box paradox turns on the fact that a box was chosen and then a ball drawn from it, rather than a ball being chosen directly; our Bertrand's box paradox calculator enumerates it. The boy or girl paradox turns on how you learned that one child is a girl. And this one turns on what the host was and was not allowed to do.
The birthday problem is instructive by contrast: its answer is uncontested and only the intuition is wrong, which our birthday paradox calculator demonstrates. That is a different kind of surprise. Here the intuition is not merely wrong; it is answering a question the wording never pinned down. To convert either into a decision you also need the payoffs, which our expected value calculator handles, and the general updating machinery lives in our Bayes theorem calculator.
Arb Digital asks what was filtered out before the numbers reached you, because the selection rule usually matters more than the numbers.
Browse All Free Tools Talk To Our TeamCommon Mistakes to Avoid
- Quoting two thirds without stating the host rule — that figure belongs to the standard rule alone, and three other reasonable rules give three other answers.
- Assuming two closed doors means fifty-fifty — the number of remaining options never determines the probabilities. What produced the reduction does.
- Treating a blind reveal as equivalent to a deliberate one — a host who could have exposed the prize and did not has given you information a constrained host cannot.
- Ignoring that the offer itself can be evidence — if the switch is offered selectively, accepting it can be strictly worse than declining.
- Confusing the long-run rate with a single game — switching wins two thirds of the time in the standard game, which means it loses a third of the time and always will.
Related Free Tools From Arb Digital
Work the general updating step with the conditional probability calculator, set likelihoods explicitly with the Bayes theorem calculator, build event probabilities from scratch with the probability calculator, compare a puzzle with several defensible answers using the boy or girl paradox calculator, or check a puzzle whose answer is uncontested with the birthday paradox calculator. The free online tools hub lists every probability tool we publish.
Frequently Asked Questions
Under the standard rule, yes: switching wins two thirds of the time against one third for staying. Under other host rules the answer changes, and under a malicious host switching loses every time.
Because the number of remaining doors does not set the probabilities. Your door was fixed at one third before the host acted, and a host constrained to open a losing door cannot change it.
Then the reveal is informative, because the host could have exposed the prize and did not. Conditioning on that, staying and switching are equally good at one half each in the three-door game.
A host who offers the switch only when you have already chosen the prize. The offer then proves you have won, so accepting it loses every time. Nothing in the usual wording rules this out.
Yes, and the effect grows. With 100 doors and 98 opened, staying wins one percent of the time and switching wins 99 percent under the standard rule.
Under the standard rule with an unbiased host, no. If the host has a known preference between two losing doors, the exact probability of winning by switching depends on which was opened, though switching is never worse than two thirds.
Richard Gill argues it is better seen as a modelling problem, because the standard solution depends on assumptions about the host that the wording never states and that the reader is expected to supply.
This page explains a probability puzzle for educational purposes. Every figure it reports depends on the host rule you select, and no rule is implied by the puzzle's usual wording.
