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The Monty Hall problem: why switching wins two times in three

There are three doors. Behind one is a car, behind the other two goats. You pick a door; the host, who knows what is behind each, opens one of the other two doors to show a goat. Then he asks: "Do you want to switch?" Should you? Many people answer: "Two doors remain, so it's fifty-fifty; switching makes no difference." The correct answer is switching wins 2/3 of the time and staying wins 1/3. The problem is famous for dividing even mathematicians. This page shows in three ways why it is not fifty-fifty and makes clear which assumptions the conclusion rests on.

Where the problem comes from

The problem is generally credited to the biostatistician Steve Selvin, who posed and solved it in a 1975 letter to the journal The American Statistician; the name "Monty Hall problem" came from his follow-up letter. It became widely known when Marilyn vos Savant answered a reader's question in her Parade magazine column on 9 September 1990 by saying "switch". The answer is reported to have drawn a great many letters of objection. Richard Gill has argued that this is not just a probability puzzle: assumptions about the host's behaviour decide the result. We confirmed this history only through summaries in secondary sources and did not read the papers in full, so the rest of this page concentrates on the mathematics, not the history.

First, run it yourself

Choose the number of doors and press the button to play 20,000 games with each strategy. With three doors, staying wins about 33% and switching about 67%.

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With 5, 10 or 100 doors, staying wins 1/doors in theory (20%, 10%, 1%) and switching wins (doors − 1)/doors (80%, 90%, 99%). The simulator uses the generalised rule in which the host opens every goat door except the one door he leaves closed.

Method 1: list every case

Suppose the car is behind door 1 and consider the three equally likely first picks of door 1, 2 or 3. Each case below has probability 1/3.

Car behind door 1: outcomes by my first pick
My first pickHost opensIf I stayIf I switch
Door 1 (car)door 2 or 3 (goat)winlose
Door 2 (goat)door 3 (goat)losewin
Door 3 (goat)door 2 (goat)losewin

Staying wins only in the first case; switching wins in the second and third. That is one third against two thirds. The key is that the host did not open a door at random; of the two doors I did not pick, he chose one with a goat. If my first pick was a goat (probability 2/3), the host has no choice but to open the other goat door, and the remaining door must hold the car.

Method 2: where does the information collect?

The door I chose has a 1/3 chance of being right, and the other two doors together have 2/3. The host opens one of those two and shows it is a goat. The "2/3 of the two doors" is now concentrated on the one that remains. My own door received no information, so it stays at 1/3; what carried information was the host's action on the other door. With 100 doors the intuition sharpens. If I pick one and the host opens 98 goat doors, leaving one, the chance that the remaining door holds the car is 99/100. It is hard then to believe "it's one of two, so fifty-fifty".

Method 3: conditional probability

Written as conditional probability for three doors: I picked door 1 and the host opened door 3. The host's rule is "open a goat door, choosing at random if there are two".

The total probability of "the host opened door 3" is 1/6 + 1/3 = 1/2. Of that, the share with the car behind door 1 is (1/6) ÷ (1/2) = 1/3, and behind door 2 it is (1/3) ÷ (1/2) = 2/3. Switching means choosing door 2, which wins with 2/3. This is the same framework, Bayes' theorem, used for reading a positive test result.

The assumptions behind the answer

The problem caused controversy because assumptions not written in the question change the result. The answer 2/3 presumes:

  1. The host knows where the car is.
  2. The host always opens a door I did not pick that hides a goat (he never opens the car).
  3. When he can choose between two goat doors, he picks either with equal probability.
  4. The host always offers the chance to switch.

When these break one by one, the answer changes.

So the accurate statement is not "switching always wins" but "if the assumptions above hold, switching wins with probability 2/3". "Holds" refers to the 2/3 average rate, not a guarantee of winning any single game.

Why intuition fails here

Links with other articles

The lesson of the Monty Hall problem shares a root with the birthday problem we saw before: intuition errs by counting outcomes while leaving out how the events were generated. The conditional-probability calculation here is used unchanged in the next article for interpreting test results. In lotteries the opposite structure appears in the gambler's fallacy: there, earlier results give no information about later ones, whereas in Monty Hall the host's action does. Whether information is given or not is the difference between the two problems.

Practice questions

  1. There are 4 doors and one car. I pick a door and the host opens one goat door among the other three. Three closed doors remain, including mine. What is the chance of winning if I stay? If I switch to one of the other two?
  2. With the same 4 doors, if the host opens two goat doors, leaving one, what is the chance of winning by switching?

Answers: (1) Staying wins 1/4. The 3/4 chance that I was wrong is split equally between the other two doors, so switching to one of them wins 3/4 × 1/2 = 3/8. Adding my door and the two others, 3/8 + 3/8 + 1/4 = 1. (2) 3/4. This is the generalised rule of the simulator (one door left closed); with 4 doors it gives (4−1)/4 = 3/4.

What the practice shows is that the Monty Hall result is not the number "switching always gives 2/3" but the principle of where the host's information concentrates. The more doors are opened, the more information collects on the closed doors that remain. The same principle explains why switching wins 99% with 100 doors.

Limits

Sources and verification