The standard rules
The problem is based on a television game show format. There are three doors. Behind one is a prize (a car in the usual telling) and behind the other two are goats. The contestant picks a door. The host, who knows where the prize is, then opens one of the other two doors, always one that hides a goat, and offers the contestant the chance to switch to the remaining closed door. Should the contestant switch?
Under these rules, switching wins the prize two times in three, and staying wins one time in three. The puzzle became widely discussed after it appeared in a magazine column in 1990, and many readers, including some with mathematical training, wrote in to disagree. The disagreement is understandable. The answer is correct, but only given the exact rules below.
The rules that matter
The standard answer depends on these assumptions:
- The prize is equally likely to be behind each door at the start.
- The host always opens a door the contestant did not pick.
- The host always opens a door with a goat, and knows which doors those are.
- If the host has a choice of two goat doors, he picks one at random.
- The host always makes the offer to switch.
Change any of these and the answer can change. The knowledge in rules 3 and 4 is what makes the puzzle work.
The full outcome table
Suppose you pick door 1. This costs no generality, since the doors are interchangeable. There are three equally likely cases for where the prize is, each with probability 1/3.
| Prize is behind | Host opens | If you stay | If you switch |
|---|---|---|---|
| Door 1 | Door 2 or 3 (host’s choice) | Win | Lose |
| Door 2 | Door 3 (forced) | Lose | Win |
| Door 3 | Door 2 (forced) | Lose | Win |
Reading down the table:
- Staying wins only in the first row. That is 1 case out of 3, or 1/3.
- Switching wins in the second and third rows. That is 2 cases out of 3, or 2/3.
The two cases where the prize is behind a door you did not pick are exactly where the host has no freedom: he must open the one remaining goat door, which leaves the prize as the only other closed door. Switching takes you to it.
Why “50/50” feels right and is wrong
After the host opens a door, two closed doors remain. It feels as if the prize must be behind either with equal probability. This is the same mistake as in the Bertrand box paradox: counting the remaining options instead of asking how likely each was.
Your first pick had a 1/3 chance of being right, and the host’s action does not change that, because the host would always be able to open a goat door whatever you picked. The remaining 2/3 of probability, which was spread over the two doors you did not pick, is now concentrated on the one door the host left closed. The host has not given you new information about your own door. He has given you information about the others.
Another way to see it is to imagine 100 doors. You pick one, and the host opens 98 goat doors, leaving yours and one other. Your door was a 1-in-100 shot. The other closed door carries the remaining 99/100. Few people would stay in that version, and the three-door game is the same logic in a smaller setting.
Checking with the conditional formula
You can confirm the result with Bayes theorem. Say you pick door 1 and the host opens door 3. Let P(prize at 1) = P(prize at 2) = P(prize at 3) = 1/3.
- If the prize is at door 1, the host picks door 2 or 3 at random, so he opens door 3 with probability 1/2.
- If the prize is at door 2, he must open door 3: probability 1.
- If the prize is at door 3, he cannot open door 3: probability 0.
The chance of seeing door 3 opened is (1/3 × 1/2) + (1/3 × 1) + 0 = 1/6 + 1/3 = 1/2. The chance that the prize is at door 1 given this is (1/6) ÷ (1/2) = 1/3, and at door 2 it is (1/3) ÷ (1/2) = 2/3. So switching to door 2 wins with probability 2/3.
When the answer changes
If the host does not know where the prize is and opens a door at random, there is a chance that he reveals the prize. If he happens to reveal a goat, then the two remaining doors are equally likely, and switching and staying are each 1/2. The result depends on why a goat was revealed.
Similarly, if the host offers the switch only when the contestant has picked the prize, then switching always loses. The standard answer assumes the offer is always made. Real show formats were not necessarily run under these rules, so the puzzle is best read as a question about a precisely stated game, not as a claim about any television programme.
Testing it yourself
Because the table is so small, the problem is a favourite for experiments. You can play it with three cards, one marked as the prize, many times, and record how often switching wins. The result should drift towards two thirds as the number of games grows, though a short run of games can easily look different. A computer version makes this quicker, and a Monty Hall simulation is a classic first programming exercise.
The puzzle is a lesson in reasoning, not a gambling strategy. Real games of chance rarely give you a host who knows the outcome and acts under fixed rules, and a game with a house edge cannot be turned into a winning proposition by a clever choice of door.