In this famous probability puzzle, named after the long-time host of the American game show "Let's Make a Deal," a contestant picks one of three doors (one hides a prize, two hide nothing), the host then reveals an empty door among the remaining two, and the contestant is offered the chance to switch their choice. The correct answer — that switching doubles your chances of winning — is so persistently counterintuitive that its 1990 publication provoked an enormous flood of outraged letters, including from mathematicians with PhDs insisting the widely-published correct solution was wrong.
You're a contestant on a game show. There are three closed doors: behind one is a car, behind the other two, nothing. You pick a door — say, Door 1. Before opening it, the host (who knows exactly where the car is) opens a different door — say, Door 3 — revealing it to be empty. The host then asks: "Would you like to switch your choice to Door 2, or stick with Door 1?"
The key detail almost everyone initially overlooks: the host's action isn't random. The host knows where the car is and deliberately always opens an empty door among the two you didn't pick. This means the host's choice carries genuine information — it's not simply "one door removed at random," which would indeed leave a 50-50 split, but rather "the specific empty door the host was forced to select, given what you originally picked and where the car actually is."
Imagine instead 100 doors, with a car behind just one of them. You pick a door. The host — who knows exactly where the car is — then opens 98 of the remaining 99 doors, all revealed to be empty, leaving just your original door and one other door still closed. Should you switch? Almost everyone's intuition agrees immediately here: your original 1-in-100 guess was almost certainly wrong, so the one remaining unopened door the host deliberately left closed is very likely hiding the car. The exact same underlying logic, just less dramatically apparent, is at work in the original 3-door version.
Suppose you always adopt the "always switch" strategy. There are three equally likely scenarios for where the car actually is, relative to your original pick:
Switching wins in exactly 2 out of these 3 equally likely scenarios — a 2/3 probability of winning, precisely twice the 1/3 probability of winning by staying.
The famous 2/3-vs-1/3 result depends critically on the host's behaviour being exactly as described: the host always knows where the car is, is required to always open a door, and always specifically opens an empty one (never accidentally revealing the car, and never opening your originally chosen door). If any of these specific behavioural assumptions change — for instance, if the host sometimes opens doors at random, occasionally risking revealing the car itself — the resulting probabilities can shift substantially, and switching may no longer offer any advantage at all in some modified versions.
Marilyn vos Savant published the problem, together with the correct "always switch" solution, in her weekly Parade magazine column in 1990. The response was a torrent of reader mail — an estimated 10,000 letters, including a substantial number from mathematics professors and even a few from mathematicians holding PhDs — many confidently insisting the correct published solution was simply wrong, illustrating just how persistently and robustly this particular result defies untrained human intuition.
The problem can be worked out with complete rigour using Bayes' Theorem (see that entry). Let A = "car is behind your originally chosen door," and let B = "host opens door 3" (given you picked door 1). Before the host acts, P(A)=1/3. The host's action provides new evidence: if the car IS behind your door, the host could have opened either remaining door (each with probability 1/2); if the car is NOT behind your door, the host is forced to open the one specific remaining empty door with certainty. Applying Bayes' theorem to update your belief given this specific new evidence yields P(car is behind door 1 | host opened door 3) = 1/3, while P(car is behind the other unopened door | host opened door 3) = 2/3 — confirming the switching strategy's advantage through completely explicit, formal probabilistic calculation.
A logically equivalent puzzle (sometimes called Bertrand's box paradox, or a version closely related to it) was discussed by Joseph Bertrand as early as 1889. A version essentially identical to the modern "Monty Hall" framing was also posed and correctly solved by Steve Selvin in a 1975 letter to the journal The American Statistician, considerably predating vos Savant's much more widely publicised and controversial 1990 column — illustrating that the core mathematical insight itself was neither new nor originally controversial among working statisticians; it was primarily the massive, unprecedented scale of public engagement (and public disagreement) with vos Savant's popular column that made the problem famous.
Because the correct solution is so persistently counterintuitive, the Monty Hall problem has become a favourite subject of empirical psychological research into probabilistic reasoning and cognitive biases — including controlled studies with human participants repeatedly playing the actual game, and, in a memorable and widely cited demonstration, studies training pigeons on repeated trials of an analogous choice task. The pigeons, through simple repeated reinforcement learning (with no verbal reasoning or intuition involved at all), reliably converged on the mathematically optimal always-switch strategy over repeated trials — while human participants, despite explicit verbal explanation of the correct reasoning, often persisted substantially longer in the mathematically suboptimal "stay" strategy, a striking illustration of just how robustly counterintuitive this particular result is to explicit human reasoning specifically.
The problem generalises naturally to n doors, with the host opening k of the remaining empty doors (leaving n−k−1 unopened besides your own): switching in this general setting gives a winning probability of (n−1)/(n(n−k−1)), which is always at least as good as, and generally strictly better than, simply staying with your original choice — confirming that the core "switching helps" insight is a robust, general phenomenon, not an artifact special to the specific case of exactly three doors.