Three closed doors hide one car and two goats. You pick door 1, and the host, who knows where the car is, opens door 3 to show a goat and offers you door 2 instead. Two doors are left, so it is tempting to call the choice fifty-fifty, but is it? This page lists every way the game can go under the host’s rules, shows that switching wins the car with probability , and shows why the answer falls to when a host who does not know where the car is happens to reveal a goat.
Key terms
Monty Hall problem
Definition The Monty Hall problem is a probability puzzle about a game with three closed doors. One door hides a car and the other two hide goats, and the car is equally likely to be behind each door, so each door has probability . The player picks one door. The host knows where the car is and follows three rules. He always opens one of the two doors the player did not pick, the door he opens always has a goat behind it, and he always offers the player the chance to switch to the other closed door. When the player’s door hides the car, both other doors hide goats, and the host chooses between them with probability each. The player then stays or switches and wins whatever is behind the final door. Under these rules staying wins the car with probability and switching wins it with probability . The same holds after the player sees which door was opened. For example, if the player picks door 1 and the host opens door 3, the car is behind door 2 with probability . The answer comes from the host’s rules, not from the number of closed doors, and a host who follows different rules can make switching win with a different probability.
In plain words Your first pick is wrong two times in three. Whenever it is wrong, the host’s rules force him to open the only other goat door, so the door he leaves closed hides the car. Switching therefore wins exactly when the first pick was wrong.
Example Fix the first pick at door 1 and go through the three equally likely places for the car.
| Car behind | Host opens | Staying wins | Switching wins |
|---|---|---|---|
| door 1 | door 2 or door 3 | yes | no |
| door 2 | door 3 | no | yes |
| door 3 | door 2 | no | yes |
Each row has probability . Staying wins in one row and switching wins in two, so staying wins with probability and switching with probability . In the first row the host has a choice, and his choice decides which door he opens but not who wins, since staying wins whichever goat door he opens. In the other two rows the host has no choice at all, because he may not open the player’s door or the car’s door.
Sample space
Definition The sample space of a random experiment is the set of all its possible outcomes, described in enough detail that every run of the experiment produces exactly one of them. An event is a set of outcomes. When the sample space is finite, each outcome has a probability, the probabilities are zero or positive and add up to , and the probability of an event is the sum of the probabilities of its outcomes. An event made of outcomes has probability only when all outcomes are equally likely.
In plain words The sample space is the full list of ways the experiment can turn out. Writing it down forces you to say what is random and how likely each possibility is, and that is where the host’s rules enter.
Example Let the player pick door 1, and record an outcome as the pair (car door, opened door). Under the host’s rules there are four outcomes,
with probabilities , , and . The car is behind door 1 with probability , and the host then opens door 2 or door 3 with probability each, which gives for each of the first two outcomes. When the car is behind door 2 or door 3, the host has only one door he may open, so each of those outcomes keeps the full . Pairs such as , in which the host would show the car, are not in this sample space at all. The four outcomes are not equally likely, so treating them as four equal chances would give wrong answers.
Conditional probability
Definition For events and with , the conditional probability of given is
where is the event that and both happen. It is the probability of once we know that happened. To compute it, keep only the outcomes in and divide their probabilities by , so that they add up to . The event has to describe exactly what was learned, in the sample space of the experiment that was actually run. The lesson Conditional Probability in the probability course develops the idea further.
In plain words Learning that happened rules out every outcome outside . The outcomes that remain keep their relative weights, so an outcome that was twice as likely as another before the news is still twice as likely after it.
Example The player picks door 1 and the host opens door 3. Let be the event that the car is behind door 2 and the event that the host opens door 3. In the sample space above, the outcomes in are with probability and with probability , so and
The car is behind door 1 with conditional probability . With the car behind door 2 the host must open door 3, while with the car behind door 1 he opens it only half the time, so an open door 3 is twice as likely to come from the case with the car behind door 2. This step uses the rule that the host chooses fairly when he has a choice. A different tie-breaking rule changes what this particular reveal tells you, although the overall chances of the two strategies stay and .
Switching strategy
Definition In the Monty Hall problem the switching strategy is the plan, made before the game starts, to accept the host’s offer every time and move to the other closed door. The staying strategy keeps the first pick every time. Under the host’s rules exactly one of the two strategies wins each game, and switching wins exactly when the first pick hides a goat. Hence switching wins with probability and staying with probability , whichever door the player picks first and whatever rule the host uses to choose between two goat doors. More generally, take doors, one car placed at random, and a host who knows where the car is and always opens of the unpicked doors, all with goats, before offering the one unpicked door left closed. Then switching wins with probability . A higher probability does not promise a win in any one game. If the host does not know where the car is, or offers the switch only in some games, these probabilities can change.
In plain words Switching bets that your first pick was wrong, and staying bets that it was right. The host’s reveal does not change whether your first pick was right. It only makes sure that, when it was wrong, the one door left to switch to hides the car.
Example With doors the player picks one, and the host opens of the other , all with goats. The first pick is right with probability , so switching wins with probability . Now return to three doors and change one rule. A host who does not know where the car is opens one of the two unpicked doors at random, each with probability , and sometimes reveals the car. With the first pick at door 1, the six pairs (car door, opened door) each have probability , and in two of them, and , the car is revealed. Keep only the four pairs in which a goat appears. The car is behind door 1 in and , and behind the other closed door in and , so once a goat appears, staying and switching each win with probability . The picture of two closed doors and one goat is the same as in the original game, but the experiment that produced it is different.
Check yourself
- Play with four doors, one car and three goats. The host knows where the car is and opens just one of the three unpicked doors, always a goat, choosing at random when he has a choice. A player who switches then picks one of the two other closed doors at random. Find the probability of winning for staying and for switching.
Show the answer
Staying wins exactly when the first pick hides the car, which has probability . Switching can win only when the first pick hides a goat, which has probability . In that case the car is behind one of the three unpicked doors, the other two unpicked doors hide goats, and the host opens one of those two. The car is then behind one of the two closed doors the player can switch to, and a random choice between them finds it with probability . So
Switching is still better, since is more than , but the advantage is smaller than with three doors. The remaining probability belongs to the closed door the switcher did not take. If the host opened two goat doors instead of one, switching would win with probability .
- Change what the host knows. With doors the player picks one, and a host who does not know where the car is opens of the other doors at random. By luck, all show goats. Should the player switch?
Show the answer
Opening of the unpicked doors at random is the same as choosing at random the one unpicked door to leave closed, each with probability . All opened doors show goats exactly when the car is behind the first pick or behind the door left closed. The first case has probability . The second needs the car to be elsewhere, with probability , and the host to happen to leave its door closed, with probability , so it has probability
The two cases are equally likely, so once all goats have appeared, staying and switching each win with probability . Switching does no harm but gives no advantage. This lucky run of goats happens in only of such games, while the informed host produces it every time. The uninformed host’s goats rule out doors but favor neither of the two that remain.
- Return to three doors and the informed host, but change the tie-breaking rule. When the player’s first pick hides the car, the host now opens door 3 whenever he can. The player picks door 1. Find the probability that switching wins given that the host opens door 3, and given that he opens door 2. What is the overall probability that switching wins?
Show the answer
The car is behind door 1, 2 or 3 with probability each. With the car behind door 1 the host now opens door 3, with the car behind door 2 he must open door 3, and with the car behind door 3 he must open door 2. So door 3 is opened in two cases of probability each, one with the car behind door 1 and one with it behind door 2. Given that door 3 is opened, these two cases are equally likely, and switching wins with probability . Given that door 2 is opened, the car must be behind door 3, since with the car behind door 1 the host would have opened door 3, so switching wins with probability . Door 3 is opened with probability and door 2 with probability , so overall
The tie-breaking rule changes what a particular reveal tells you, but the strategy of always switching still wins with probability , and after either reveal switching is at least as good as staying.