The Monty Hall Problem: The Three-Door Puzzle That Drew 10,000 Letters

The rules that make switching right

The puzzle is short. A car sits behind one of three doors and goats behind the other two. You pick a door. The host, who knows where the car is, opens one of the doors you did not pick and shows a goat. He then offers you the chance to move to the last closed door. Should you?

The answer depends on three rules, and the standard version fixes all of them:

• The car was placed at random, so each door starts with a 1-in-3 chance.
• The host always opens a door you did not choose, and it always hides a goat.
• The host always offers the switch, whatever you picked.

Under those rules, switching doubles your odds, from one-third to two-thirds. Change any rule and the answer can change with it, which is why the problem still fills journals more than 30 years after it reached the general public.

The column that drew 10,000 letters

The question reached Marilyn vos Savant from a reader, Craig F. Whitaker, and ran in her "Ask Marilyn" column in Parade on September 9, 1990. She answered that the contestant should switch. Readers revolted.

By her estimate, the magazine received about 10,000 letters, and close to 1,000 of them carried signatures with PhDs, according to John Tierney's front-page report in The New York Times on July 21, 1991. Most insisted she was wrong. "You are utterly incorrect," wrote E. Ray Bobo, a professor of mathematics at Georgetown University. The figure of nearly 1,000 PhDs is hers, not an audited count, and it should be quoted that way.

Tierney took the problem to Monty Hall, the host of Let's Make a Deal. At his dining table, Hall played ten rounds in which the contestant never switched: four cars, six goats. Then ten rounds with switching: eight cars, two goats. "So her answer's right: you should switch," Hall said.

The puzzle was not new. Steve Selvin, a statistician, posed it in a 1975 letter to The American Statistician and gave it Hall's name. Martin Gardner had published an equivalent riddle, the three prisoners problem, in Scientific American in 1959. Paul Erdős, one of the most prolific mathematicians of the 20th century, reportedly stayed unconvinced until a colleague showed him a computer simulation.

The math, case by case

Say you pick door 1. The car is equally likely to be behind any of the three doors. List the cases and the answer falls out.

The key word is "forced." In two of the three cases, the host has no choice: one of his doors hides the car, so he must open the other. His action tells you where the car is two times in three. Staying ignores that information. Switching uses it.

Bayes' rule gives the same result. Suppose the host opens door 3. If the car is behind door 1, he opens door 3 half the time. If it is behind door 2, he opens door 3 every time. The weights are 1/3 x 1/2 against 1/3 x 1, or 1 to 2. Door 2 now holds two-thirds of the probability.

We played it a million times

Arguments persuade few people. Counting persuades more. We wrote a short Python simulation that places the car at random, picks a door at random and has a host follow the standard rules. Fixed random seed, 1,000,000 games per strategy.

The last row matters most. When the host does not know where the car is and only happens to reveal a goat, the advantage vanishes. The two-thirds edge comes from his knowledge, not from the doors.

Why smart people get it wrong

The usual error is to see two closed doors and assign each a 50 percent chance. That reasoning treats the host's choice as random. It is not. He is barred from opening your door and barred from showing the car, so his move carries information about the other doors and none about yours.

The statistician Persi Diaconis told the Times in 1991: "Our brains are just not wired to do probability problems very well, so I'm not surprised there were mistakes."

The cleanest cure is to scale up. Imagine 100 doors. You pick one. The host, who knows everything, opens 98 others and shows 98 goats, skipping just one door. Few people would stay with their first pick at 1 in 100. The three-door game is the same logic with smaller numbers.

Probability puzzles of this kind sit close to the reasoning items on intelligence tests, which is one reason they turn up in collections of IQ test questions and answers. They reward slow, explicit counting over a fast hunch, the same habit that cracks Einstein's riddle.

The host's rules decide everything

Monty Hall himself raised the sharpest objection. On the real show, he told Tierney, he did not have to offer a switch, and he could tempt contestants with cash to keep their door. A host who offers the swap only when you already hold the car turns switching into a guaranteed loss. "My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home," he said.

Statisticians have argued over wording ever since. A 1991 paper by J. P. Morgan and colleagues in The American Statistician showed that if the host has a preference between two goat doors, the odds after a particular door is opened can range from one-half to certainty. Vos Savant's answer holds for the standard rules, which is how nearly every textbook now states the problem. Treat the puzzle as a lesson in modeling: before computing a probability, write down exactly how the information was produced. That skill also underpins game theory.

Ten variants to practice, with worked answers

Each item below is our own variant. Assume the standard rules unless the item changes them. We checked every numerical answer by simulation or exact count.

1. The original. Three doors, a host who knows and always shows a goat. Stay or switch?
   Answer: Switch, 2/3 against 1/3. Your door wins only if your first guess was right, a 1-in-3 event.
2. Four doors. One car, three goats. The host opens one goat door and lets you move to either of the two other closed doors. What is the chance for one of them?
   Answer: 3/8 each, against 1/4 for staying. The 3/4 chance that the car is elsewhere splits evenly across two doors.
3. A hundred doors. The host opens 98 goat doors. Switch?
   Answer: Yes. Switching wins 99 times in 100; we measured 99.02 percent.
4. The ignorant host. The host opens one of your two unchosen doors at random, and it happens to show a goat. Should you switch?
   Answer: It does not matter: 1/2 each. A goat appears every time when you hold the car, but only half the time when you do not, which cancels the 1-to-2 prior.
5. The host with a favorite door. You pick door 1. Whenever he can, the host opens door 2. This time he opens door 3. Switch?
   Answer: Switch and win with certainty. He skipped his favorite, so door 2 must hide the car. Had he opened door 2, the odds would be 1/2.
6. The spiteful host. He offers a switch only when your first pick is the car. He offers. What now?
   Answer: Stay. The offer itself proves you hold the car.
7. Two cars, two goats. Four doors. The host opens one goat door among the three you did not pick, and you move to one of the remaining two at random.
   Answer: Switching wins 3/4; staying wins 1/2. If you began on a goat, both remaining doors hide cars.
8. The three prisoners. One of three prisoners, A, B or C, will be pardoned at random. The warden tells A truthfully that B will be executed. What is A's chance now, and C's?
   Answer: A stays at 1/3; C rises to 2/3. The warden could never name A, so A learns nothing about his own fate.
9. Bertrand's box (1889). Three boxes hold two gold coins, two silver coins, and one of each. You draw a gold coin from a random box. What is the chance the other coin in that box is gold?
   Answer: 2/3, not 1/2. There are three gold coins you could have drawn, and two of them sit in the gold-gold box. Our simulation returned 66.6 percent.
10. The case-opening show. Twenty-six sealed cases, one top prize. You keep one case and open 24 others yourself, at random, and none holds the top prize. Should you swap for the last case?
    Answer: 1/2 either way. Nobody with knowledge steered the openings, so this is the ignorant-host game, not Monty Hall.

Practice builds familiarity with a format. It does not raise intelligence. Our free IQ test is a practice screener, not a clinical measurement.

Marilyn vos Savant and the 228 figure

Part of the fury in 1990 came from the columnist's billing. Guinness listed vos Savant under "Highest IQ" from 1985 to 1989, citing a childhood score of 228. By her account, she took the 1937 Stanford-Binet in September 1956, at age 10, and her mental age was measured at 22 years and 10 months. Mental age divided by actual age, times 100, gives 228.

That is a ratio IQ, an old formula that inflates at the extremes. Modern tests use deviation IQ, with a mean of 100 and a standard deviation of 15, and stop near 160. Our page on an IQ of 228 shows why the figure cannot exist on today's scales, and Guinness retired the category in 1990. Our profile of Marilyn vos Savant's IQ covers the record in full.

The Monty Hall episode is a better advertisement for her reasoning than any number. She held a correct position against thousands of objections, many signed by credentialed experts. For how scores are built and what they mean, start with our IQ guides and what counts as a high IQ.

Editorial standard: IQ figures for historical figures are retrospective estimates and are labeled as such; scores for living people appear only when the person disclosed them or a reputable outlet reported them. Corrections: contact the editors.