Monty Hall Problem Simulator

The problem, exactly as the host plays it

One of three doors hides a car, the other two goats. You pick one. The host, who knows where the car is, opens one of the other two to show a goat and offers you the third door instead. You should switch: switching wins the car 2 times in 3, and staying wins it 1 time in 3.

That answer holds for a host who follows four rules in every game, and this simulator plays by all four:

• He knows where the car is, and never opens its door, or yours.
• He always opens goats: with N doors he opens N − 2 of them, every one a goat.
• He always offers the switch, to the one other door he leaves closed, whatever you picked.
• He chooses at random when he has a choice: if you picked the car, the door he leaves closed is any of the others with equal chance.

Most arguments about the puzzle come from changing one of these rules. The statistician Steve Selvin posed it in a letter to The American Statistician in February 1975, and named it the Monty Hall problem, after the host of the television game show it was modelled on, in a second letter that August. It became famous when Marilyn vos Savant answered it in her Ask Marilyn column in Parade on 9 September 1990 and received thousands of letters, many insisting she was wrong.

Using the simulator

The page opens on the auto-run, which plays the game by those rules with a first pick made at random and scores both strategies in every game: amber where switching wins, blue where staying wins, and green for the car itself. The first six games go slowly enough to watch every door, then the run speeds up to 50 games a second and stops at 1,000. Run carries it on to 10,000 and then 100,000, and the speed control sets anything from 1 game a second to as fast as possible.

Tap any door, or for doors 1 to 9 press its number, to play a round yourself. The host opens his doors and you choose to stay with yours or switch to his, by tapping a door, by its number, with the arrow keys and Enter, or with the buttons under the stage. Your rounds are kept apart from the auto-run, as dots of their own, coloured by which strategy would have won the car.

The Doors slider runs from 3 to 100, with a button for each end, and the host can be made to open his doors at random instead. The ledger under the doors holds the last 1,000 games, one dot each, with the first hundred drawn large so the opening games can be counted as they arrive. The chart below plots each strategy’s win rate as the games add up, with its exact value dashed and the band where 95 percent of runs land. The seed fixes every game, so a copied link replays the same run for whoever opens it.

Worked example: the nine equally likely starts

Every game starts with a car door and a first pick, each one of three, so there are nine starts, each with a chance of 1 in 9. Set them out as a grid, the car’s door down the side and your first pick along the top, the way the Punnett Square Calculator sets out every pairing of two parents’ alleles:

Staying wins only on the diagonal, where your first pick was the car: 3 starts of 9. Off it, the host has to open the only other goat, so the door he leaves closed is the car, and switching wins: 6 of 9. On the diagonal he has two goats to choose from, which splits each of those three starts in two without changing who wins it, so the count stands at 6 against 3, which is 2/3 against 1/3. The simulator’s opening run, seed 1990, shows the same split in practice: switching won 674 of its 1,000 games, 67.4 percent, and staying won 326.

Why 100 doors make it obvious

Set the doors to 100. You pick one, with a chance of 1 in 100 that it hides the car. The host opens 98 of the others, every one a goat, and leaves one closed. He chose that door knowing where the car was: he kept it shut while showing you every goat but one, so it holds the car whenever your first pick was wrong, which is 99 games in 100. On the stage the whole argument is one frame: a wall of 98 open doors and two still closed, yours in blue and his in amber. With the same seed, switching won 991 of the first 1,000 hundred-door games.

Bayes’ theorem for the host’s door

Bayes’ theorem reaches the same answer from what the host does. Say you pick door 1 and he opens door 3. Before he opens anything, each door has a prior of 1/3. The likelihood of his opening door 3 is 1/2 if the car is behind door 1, since he could have opened either goat; 1 if it is behind door 2, since door 3 is then his only goat; and 0 if it is behind door 3. Each prior times its likelihood gives 1/6, 1/3 and 0, and scaling those to add up to 1 gives:

• P(car behind door 1 | he opens door 3) = (1/6) / (1/6 + 1/3) = 1/3
• P(car behind door 2 | he opens door 3) = (1/3) / (1/6 + 1/3) = 2/3

Your door keeps the 1/3 it started with and his collects the rest. With N doors the same working gives 1/N for yours and (N − 1)/N for his, the figures on the stage’s pills and in the readout The door he left, by Bayes. The same theorem turns a positive test result into a chance of disease, with a test’s sensitivity and false positive rate as the likelihoods, which the Sensitivity and Specificity Visualizer draws.

When the host does not know

Choose the host who opens his doors at random and he becomes the one Jeffrey Rosenthal called Monty Fall in Math Horizons in 2008, who slips and happens to push a door open. He leaves one of the other doors closed at random, so sometimes he opens the car, and that game is void. A game survives only when the car is behind your door or behind the one he left, 2 chances in N, so with three doors he reveals the car in 1 game in 3, and with a hundred in 98 games in 100.

In the games he does not spoil, the two closed doors are alike: each had a chance of 1/N to begin with, and his random choice was as likely to leave either one. So each holds the car half the time, and switching wins only half the valid games. With seed 1990 the three-door auto-run voided 336 of its 1,000 games, and switching won 338 of the 664 left, 50.9 percent. Only the host’s knowledge changed, and without it his choice carries no information about the car.

How fast the simulation settles

A run’s switching rate is a share estimated from n games, so it scatters about its exact value p with a standard error of √(p(1 − p)/n), and about 95 percent of runs land within 1.96 standard errors of it. For p = 2/3 that band, the faint dashed lines on the chart, is:

The band narrows as 1/√n, so each extra decimal place of agreement costs a hundred times as many games: it takes about 8,500 to bring it inside one point either side. The band is the normal approximation, and the share of runs inside it, counted from the binomial distribution, is a little under 95 percent at first. The Gap from the exact value readout gives a run’s distance in standard errors: seed 1990 sits 0.49 above after 1,000 games and 0.74 below after 100,000, when switching has won 66,556 games.

To test a run formally, put its counts into the Chi-Square Calculator against the expected 2 to 1. The opening run’s 674 and 326 against 666.7 and 333.3 give χ² = 0.242 with 1 degree of freedom, and p = 0.62, which is consistent with chance alone. It is the same arithmetic that makes twenty nuclei in the Radioactive Decay Simulator decay in jagged steps while five thousand follow the curve.

Common mistakes

• Two doors left means 50/50. Only if they are alike, and they are not: yours was chosen before anything was known, and his by someone who knew where the car was and had to avoid it. Two closed doors can carry 1/3 and 2/3, or 1/100 and 99/100.
• The host’s choice carries no information. It does, because he must avoid the car. When he could have opened either of two goats, nothing is learnt; when he could open only one, the door he skipped is the car. The ignorant host shows the difference: the same doors, opened at random, leave 1/2 and 1/2.
• Switching only helps sometimes. Switching loses exactly when your first pick was right, 1 time in 3, and nothing tells you when that is. As a rule for every game it wins twice as often as staying, which no single round can show and a few hundred rounds show beyond reasonable doubt.
• One run proves the point, or disproves it. After 100 games, about 95 percent of runs land anywhere from 57.4 to 75.9 percent, so a run at 60 percent is unremarkable: seed 1990 stood at 64.0 percent after 100 games and 67.4 percent after 1,000.