Video 1, chapter 7

Monty Hall

How the Monty Hall problem is settled by writing out the sample space under the host's rules, why the switching strategy wins with probability two thirds, and how conditional probability depends on what the host knows.

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Video 1, chapter 7

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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 2/32/3, and shows why the answer falls to 1/21/2 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 1/31/3. 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 1/21/2 each. The player then stays or switches and wins whatever is behind the final door. Under these rules staying wins the car with probability 1/31/3 and switching wins it with probability 2/32/3. 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 2/32/3. 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 behindHost opensStaying winsSwitching wins
door 1door 2 or door 3yesno
door 2door 3noyes
door 3door 2noyes

Each row has probability 1/31/3. Staying wins in one row and switching wins in two, so staying wins with probability 1/31/3 and switching with probability 2/32/3. 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 11, and the probability of an event is the sum of the probabilities of its outcomes. An event made of kk outcomes has probability k/Nk/N only when all NN 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,

(1,2),(1,3),(2,3),(3,2),\begin{gathered} (1, 2), \quad (1, 3), \\[0.4em] (2, 3), \quad (3, 2), \end{gathered}

with probabilities 1/61/6, 1/61/6, 1/31/3 and 1/31/3. The car is behind door 1 with probability 1/31/3, and the host then opens door 2 or door 3 with probability 1/21/2 each, which gives 13⋅12=16\tfrac13 \cdot \tfrac12 = \tfrac16 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 1/31/3. Pairs such as (2,2)(2, 2), 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 AA and BB with P(B)>0P(B) > 0, the conditional probability of AA given BB is

P(A∣B)=P(A∩B)P(B),P(A \mid B) = \frac{P(A \cap B)}{P(B)},

where A∩BA \cap B is the event that AA and BB both happen. It is the probability of AA once we know that BB happened. To compute it, keep only the outcomes in BB and divide their probabilities by P(B)P(B), so that they add up to 11. The event BB 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 BB happened rules out every outcome outside BB. 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 C2C_2 be the event that the car is behind door 2 and H3H_3 the event that the host opens door 3. In the sample space above, the outcomes in H3H_3 are (1,3)(1, 3) with probability 1/61/6 and (2,3)(2, 3) with probability 1/31/3, so P(H3)=1/2P(H_3) = 1/2 and

P(C2∣H3)=1/31/2=23.\begin{aligned} P(C_2 \mid H_3) &= \frac{1/3}{1/2} \\[0.4em] &= \frac{2}{3}. \end{aligned}

The car is behind door 1 with conditional probability (1/6)/(1/2)=1/3(1/6)/(1/2) = 1/3. 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 1/31/3 and 2/32/3.

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 2/32/3 and staying with probability 1/31/3, whichever door the player picks first and whatever rule the host uses to choose between two goat doors. More generally, take n≥3n \ge 3 doors, one car placed at random, and a host who knows where the car is and always opens n−2n - 2 of the unpicked doors, all with goats, before offering the one unpicked door left closed. Then switching wins with probability (n−1)/n(n - 1)/n. 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 100100 doors the player picks one, and the host opens 9898 of the other 9999, all with goats. The first pick is right with probability 1/1001/100, so switching wins with probability 99/10099/100. 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 1/21/2, and sometimes reveals the car. With the first pick at door 1, the six pairs (car door, opened door) each have probability 1/61/6, and in two of them, (2,2)(2, 2) and (3,3)(3, 3), the car is revealed. Keep only the four pairs in which a goat appears. The car is behind door 1 in (1,2)(1, 2) and (1,3)(1, 3), and behind the other closed door in (2,3)(2, 3) and (3,2)(3, 2), so once a goat appears, staying and switching each win with probability 1/21/2. 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

  1. 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 1/41/4. Switching can win only when the first pick hides a goat, which has probability 3/43/4. 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 1/21/2. So

P(switching wins)=34⋅12=38.\begin{aligned} &P(\text{switching wins}) \\[0.4em] &\quad = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}. \end{aligned}

Switching is still better, since 3/83/8 is more than 1/41/4, but the advantage is smaller than with three doors. The remaining probability 1−1/4−3/8=3/81 - 1/4 - 3/8 = 3/8 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 3/43/4.

  1. Change what the host knows. With 100100 doors the player picks one, and a host who does not know where the car is opens 9898 of the other 9999 doors at random. By luck, all 9898 show goats. Should the player switch?
Show the answer

Opening 9898 of the 9999 unpicked doors at random is the same as choosing at random the one unpicked door to leave closed, each with probability 1/991/99. All 9898 opened doors show goats exactly when the car is behind the first pick or behind the door left closed. The first case has probability 1/1001/100. The second needs the car to be elsewhere, with probability 99/10099/100, and the host to happen to leave its door closed, with probability 1/991/99, so it has probability

99100⋅199=1100.\frac{99}{100} \cdot \frac{1}{99} = \frac{1}{100}.

The two cases are equally likely, so once all 9898 goats have appeared, staying and switching each win with probability 1/21/2. Switching does no harm but gives no advantage. This lucky run of goats happens in only 2/100=1/502/100 = 1/50 of such games, while the informed host produces it every time. The uninformed host’s goats rule out 9898 doors but favor neither of the two that remain.

  1. 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 1/31/3 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 1/31/3 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 1/21/2. 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 11. Door 3 is opened with probability 2/32/3 and door 2 with probability 1/31/3, so overall

23⋅12+13⋅1=23.\frac{2}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot 1 = \frac{2}{3}.

The tie-breaking rule changes what a particular reveal tells you, but the strategy of always switching still wins with probability 2/32/3, and after either reveal switching is at least as good as staying.