# Monty Hall problem: why switching wins two times in three > Three doors, one car, and a host who opens a goat door. Switching wins 2/3 of the time. Three solutions, and why the answer depends on what the host knows. Series: Probability (https://www.monocurl.com/monocurl-essays/probability/) Canonical: https://www.monocurl.com/monocurl-essays/probability/monty-hall/ Keywords: Monty Hall problem, should you switch doors, Monty Hall explained, Monty Hall Bayes theorem, 100 doors, conditional probability puzzle, three prisoners problem [Figure 1, animation: The Monty Hall game, and a simulation of 1000 plays of it. You pick door 1, the host, who knows where the car is and never reveals it, opens door 3 to show a goat, and you may switch to door 2. In the simulation, staying with the first pick wins about a third of the games and switching wins about two thirds.] You are on a game show, facing three closed doors. Behind one is a car, and behind the other two are goats. You pick a door, say door 1. The host, who knows where the car is, opens one of the other two doors, say door 3, and shows you a goat. Then he makes an offer: you may stay with door 1 or switch to door 2. Does it matter what you do? **Pause and try.** Decide before reading on: is the car more likely behind door 1, more likely behind door 2, or equally likely behind either? Write down a probability for each. **Answer.** Most people say the two closed doors are equally likely, $\frac12$ each, so the offer is worthless. Under the rules stated below, door 1 has probability $\frac13$ and door 2 has probability $\frac23$. Switching doubles your chance of winning. The puzzle is named after Monty Hall, the host of the American show *Let's Make a Deal*. It became notorious in 1990, when the columnist Marilyn vos Savant gave the answer $\frac23$ in *Parade* magazine and received thousands of letters, many from readers with doctorates, telling her she was wrong. She was right. The simulation above plays the game 1000 times and agrees with her: staying won 334 games and switching won 666. By [the law of large numbers](@probability/law-of-large-numbers), shares over that many games sit close to the probabilities. An answer this counterintuitive deserves more than one argument. We will solve the problem three ways, and then change one detail of the story and watch the answer change to $\frac12$, which shows what the correct argument depends on. ## The rules of the game The answer depends on how the host behaves, so the first job is to say exactly what we assume. Informal tellings usually leave this out, and that omission is responsible for much of the arguing. **Definition.** The **standard Monty Hall game** follows these rules. - The car is equally likely to be behind each of the three doors, with goats behind the other two. - You choose a door with no information about where the car is. - The host knows where the car is. He always opens one of the two doors you did not choose, and it always reveals a goat. - If both doors you did not choose hide goats, the host opens each of them with probability $\frac12$. You then either stay with your door or switch to the one other door that is still closed, and you win whatever is behind your final door. Rule 3 is the heart of it: the host is not exploring. He cannot open your door and he cannot reveal the car. Rule 4 only matters for one of our three arguments, and we will see what happens when it is dropped. ## First solution: list the cases Suppose you pick door 1. (The doors are only labels, so nothing is lost.) By rule 1 there are three equally likely cases for where the car is, and in each of them we can follow what the host must do and how each strategy fares. [Figure 2, steps: The three equally likely cases when you pick door 1. In each row the host opens a goat door that is not yours. Staying wins only in the row where the car was behind door 1, and switching wins in the other two rows.] - step: You pick door 1. The car is behind door 1, 2 or 3 with probability 1/3 each: three rows, one per case. - step: The host opens a goat door that is not yours. If the car is behind door 2 or door 3 he has only one door he can open. - step: If you stay with door 1, you win only in the first row. - step: If you switch to the other closed door, you win in the second and third rows: two cases out of three. Staying wins in one case out of three and switching wins in two. The table also shows why, in a form worth saying aloud. If your first pick was the car, switching loses. If your first pick was a goat, the host is forced to open the only other goat door, so the door he leaves closed is the car, and switching wins. Your first pick is a goat with probability $\frac23$. **Key idea.** When the host knows where the car is and must show a goat, switching wins exactly when your first pick was wrong, and your first pick is wrong two times in three. The host's reveal does not change whether your first pick was right. It tells you where the car is in the cases where it was wrong. The $\frac12$ answer comes from counting closed doors: two remain, so each gets half. But "there are two possibilities" does not make them equally likely. As [the lesson on sample spaces](@probability/sample-spaces) stressed, dividing by the number of outcomes is only valid when the outcomes are equally likely, and here the two doors have very different histories. Door 1 is closed because you chose it. Door 2 is closed because the host, who knows where the car is, passed over it. ## Second solution: Bayes' theorem A careful reader may object to the table. It shows that the *strategy* of switching wins two games in three overall. But at the moment of decision you know more: you saw the host open door 3 in particular. The first row of the table mixes games where he opens door 2 with games where he opens door 3. Is the chance still $\frac23$ given exactly what you saw? This is a question about conditional probability, and the tool for it is Bayes' theorem from [the lesson on conditional probability](@probability/conditional-probability-bayes). To restate what we need: $P(A \mid B) = P(A \cap B)/P(B)$ is the probability of $A$ given that $B$ occurred. If $B_1, \dots, B_n$ are events of positive probability, exactly one of which occurs, and $A$ is an event with $P(A) > 0$, then $$ P(B_j \mid A) = \frac{P(A \mid B_j)\,P(B_j)}{\sum_{i=1}^{n} P(A \mid B_i)\,P(B_i)}. $$ Here the hypotheses are $C_1$, $C_2$, $C_3$: the car is behind door 1, 2 or 3. You picked door 1. The evidence is the event $E$: the host opens door 3. **Pause and try.** Using the four rules, find $P(E \mid C_1)$, $P(E \mid C_2)$ and $P(E \mid C_3)$: the probability that the host opens door 3 in each of the three cases. Then apply Bayes' theorem to get $P(C_2 \mid E)$. **Answer.** If the car is behind door 1, both other doors hide goats and by rule 4 the host opens door 3 with probability $\frac12$. If the car is behind door 2, door 3 is the only door he may open: probability 1. If the car is behind door 3 he cannot open it: probability 0. So $$ P(C_2 \mid E) = \frac{1 \cdot \frac13}{\frac12 \cdot \frac13 + 1 \cdot \frac13 + 0 \cdot \frac13} = \frac{1/3}{1/2} = \frac23. $$ [Figure 3, steps: Bayes' theorem for the Monty Hall game as areas in a unit square, after you pick door 1. The columns are the three places the car can be. Each column is divided according to which door the host opens. Conditioning on "the host opens door 3" keeps the two purple regions, whose areas are 1/6 and 1/3.] - step: You pick door 1. The car is behind each door with probability 1/3, so the square is cut into three equal columns. - step: Divide each column by what the host does. Car at 1: he opens door 2 or 3, half each. Car at 2: he must open 3. Car at 3: he must open 2. - step: He opened door 3, so only the purple regions remain. The car-at-2 region is twice the car-at-1 region. The picture shows where the intuition goes wrong. Before the host acts, each door has $\frac13$. The host then opens door 3 in half of the games where the car is behind door 1, and in all of the games where it is behind door 2. Seeing door 3 open is therefore twice as likely when the car is behind door 2 as when it is behind door 1, and the evidence favors door 2 by two to one. There is a second thing to read off. $P(C_1 \mid E) = \frac13$, the same as $P(C_1)$ before the host did anything. Under rule 4 the host opens door 3 with probability $\frac12$ whether or not the car is behind door 1, so what he does is independent of whether your first pick was right, and it cannot change your door's probability. The $\frac13$ that belonged to door 3 does not get shared out. All of it goes to door 2. ## Third solution: a hundred doors The first two arguments are correct, and many people who follow every line still feel that the two doors ought to be even. For them there is a third argument, which changes the numbers until the intuition gives way. Play the same game with 100 doors: one car, 99 goats. You pick a door. The host, who knows where the car is, opens 98 of the other 99 doors, all goats, leaving your door and one other closed. [Figure 4, animation: The Monty Hall game with 100 doors. You pick one door (blue), which hides the car with probability 1 in 100. The host, who knows where the car is, then opens 98 goat doors and leaves exactly one other door closed (gold). That door carries the remaining 99 in 100.] Nobody watching the host walk down the line, opening door after door and conspicuously skipping one, believes the skipped door is a coin flip against their own. Your first pick was a 1 in 100 shot and still is. In the other 99 games out of 100 the car is among the doors you did not pick, and the host, who is not allowed to reveal it, has no choice but to leave it closed. **Pause and try.** With $n$ doors, of which the host opens $n - 2$, what is the probability that switching wins? Check your formula at $n = 3$ and $n = 100$. **Answer.** Switching wins exactly when the first pick is wrong, which has probability $\frac{n-1}{n}$: that is $\frac23$ for three doors and $\frac{99}{100}$ for a hundred. The three-door game is the same game with the contrast turned down. The general statement uses only rules 1 to 3. **Theorem.** Let $n \ge 3$. A car is equally likely to be behind each of $n$ doors, with goats behind the rest. You choose a door with no information. The host, who knows where the car is, opens $n - 2$ doors, none of them yours and none hiding the car, leaving your door and one other closed. Then staying wins with probability $\frac1n$, and switching to the other closed door wins with probability $\frac{n-1}{n}$. **Proof.** Staying wins exactly when the car is behind the door you first chose. That door was chosen with no information, so this has probability $\frac1n$, and nothing the host does afterwards changes which door you chose or where the car is. Suppose instead that your first choice hides a goat, which has probability $\frac{n-1}{n}$. The car is then behind one of the other $n - 1$ doors. The host opens $n - 2$ of those and may not open the car's door, so the one he leaves closed is the car's door, and switching wins. If your first choice hides the car, switching loses. So switching wins exactly when the first choice was wrong. Notice that this proof says nothing about *which* goat door the host opens when he has a choice. Rule 4 was not used. The theorem is about how often each strategy wins over many games. The Bayes calculation answered a finer question, the probability given the particular door you saw opened, and that one does depend on rule 4, as the worked examples will show. [Figure 5, image: The chance of winning by staying (orange) and by switching (green) when the host opens all but one of the other doors, for games with 3, 4, 10 and 100 doors. Staying wins 1 time in n and switching wins the rest.] ## What the host's knowledge does Every argument so far leaned on rule 3. To see how much weight it carries, change it. Suppose the host does *not* know where the car is. He opens one of the two doors you did not pick by flipping a coin, and it happens to reveal a goat. From where you stand the scene is identical: your door, one open door with a goat, one other closed door. Should you still switch? **Pause and try.** With an ignorant host there are six equally likely cases: three places for the car, times two doors the host might open. You picked door 1. List the six cases, cross out those in which the host reveals the car, and find the probability that switching wins among the rest. **Answer.** Car at 1, opens 2: goat. Car at 1, opens 3: goat. Car at 2, opens 2: car revealed. Car at 2, opens 3: goat. Car at 3, opens 2: goat. Car at 3, opens 3: car revealed. Four cases survive and they are equally likely. In two of them the car is behind your door and in two it is behind the other closed door. Switching wins with probability $\frac24 = \frac12$. [Figure 6, steps: The game with a host who does not know where the car is and opens door 2 or door 3 at random, after you pick door 1. Of the six equally likely cases, two reveal the car and are discarded. In the four that remain, staying and switching each win twice.] - step: You pick door 1. The car has three possible places and the host's coin picks door 2 or door 3 to open: six equally likely cases. - step: In two of the six cases the host's random door reveals the car. That did not happen, so those cases are ruled out. - step: Four equally likely cases remain. Staying wins in two of them and switching wins in two: an even chance. Bayes' theorem gives the same answer and shows which number changed. With $E$ now meaning "the host opens door 3 and it shows a goat": $$ P(E \mid C_1) = \tfrac12, \qquad P(E \mid C_2) = \tfrac12, \qquad P(E \mid C_3) = 0, $$ so $P(C_2 \mid E) = \frac{\frac12 \cdot \frac13}{\frac12 \cdot \frac13 + \frac12 \cdot \frac13} = \frac12$. Compare with the knowing host, for whom $P(E \mid C_2)$ was 1. A host who knows the car is behind door 2 opens door 3 every time. A host who is guessing opens door 3 only half the time, and the other half he ruins the game by revealing the car. For the ignorant host, a goat behind door 3 is exactly as expected whether the car is behind door 1 or door 2, so it favors neither. What you see is the same in both stories. What it tells you depends on the process that produced it. The simulation below plays the same 3000 deals with both hosts. [Figure 7, animation: A simulation of 3000 games played with two different hosts. Green: the host knows where the car is and always shows a goat. Purple: the host opens a random door, and only the games where he happens to show a goat are counted. The share of games that switching wins settles near 2/3 for the knowing host and near 1/2 for the ignorant one.] ## Worked examples **Worked example.** **The three prisoners.** Three prisoners, A, B and C, are told that one of them, chosen at random, will be pardoned and the other two executed. Prisoner A asks the warden, who knows the outcome, to name one of the *other* two who will be executed. The warden says "B". Prisoner A reasons that it is now between himself and C, so his chance of a pardon has risen from $\frac13$ to $\frac12$. Is he right? This is the Monty Hall game in different clothes. The pardon is the car, A's own identity is the door he "picked", and the warden is a knowing host who must name someone other than A who is not pardoned. Assume, as in rule 4, that when A is the one pardoned the warden names B or C with probability $\frac12$ each. Let $E$ be "the warden names B". Then $P(E \mid \text{A pardoned}) = \frac12$, $P(E \mid \text{C pardoned}) = 1$ and $P(E \mid \text{B pardoned}) = 0$, so $$ P(\text{A pardoned} \mid E) = \frac{\frac12 \cdot \frac13}{\frac12 \cdot \frac13 + 1 \cdot \frac13} = \frac13, \qquad P(\text{C pardoned} \mid E) = \frac23. $$ A's chance is unchanged, and C's has doubled. *Check:* the two posteriors sum to 1, as they must, since B is out. **Worked example.** **A lazy host.** Suppose the host follows rules 1 to 3 but not rule 4: when he has a choice, he always opens the lower-numbered door. You pick door 1. (a) He opens door 2. What is the probability that the car is behind door 3? (b) He opens door 3. What is the probability that the car is behind door 2? (c) How often does the switching strategy win overall? (a) Let $E_2$ be "he opens door 2". If the car is behind door 1 he opens door 2 for certain, being lazy. If it is behind door 3 he must open door 2. If it is behind door 2 he cannot. So $P(C_3 \mid E_2) = \frac{1 \cdot \frac13}{1 \cdot \frac13 + 1 \cdot \frac13} = \frac12$. (b) He opens door 3 only when he is not allowed to open door 2, which means the car is behind door 2. So $P(C_2 \mid E_3) = 1$. (c) $P(E_2) = \frac23$ and $P(E_3) = \frac13$, so switching wins with probability $\frac23 \cdot \frac12 + \frac13 \cdot 1 = \frac23$, as the theorem promised. Without rule 4 the conditional probabilities are no longer $\frac23$ in each case, but their average still is, and switching never does worse than staying. ## Common mistakes **Common mistake.** **"Two doors are left, so it is fifty-fifty."** Counting the remaining possibilities gives probabilities only when they are equally likely. Here they are not: your door was chosen blind, and the other was selected by someone who knew where the car was and had to avoid it. The hundred-door game makes the difference in status hard to miss. **Common mistake.** **Ignoring how the evidence was produced.** "A goat is behind door 3" is the same fact whether a knowing host showed it to you or a guessing host stumbled on it, and it supports different conclusions: $\frac23$ in the first case, $\frac12$ in the second. A conditional probability depends on the probability of seeing the evidence under each hypothesis, and that depends on the rules by which the evidence was selected. When a puzzle of this kind does not state those rules, it does not have a single answer. ## Stepping back The arithmetic here is three fractions. The difficulty is that the host's action feels like plain information, "door 3 has a goat", when it is the output of a procedure. A knowing host is constrained by where the car is, and his choice therefore leaks information about it. Under the standard rules the leak says nothing about your door and everything about the other one. The reusable habit is the one Bayes' theorem enforces. When you learn something, ask how probable it was that you would learn exactly that under each hypothesis. Hypotheses that made the observation more probable gain, in proportion. Counting the possibilities that remain is not a substitute. **Check yourself.** In the standard game you pick door 1 and the host opens door 2, showing a goat. What is the probability that the car is behind door 3? Answer: $\frac23$ Your first pick is wrong with probability $\frac23$, and whenever it is wrong the host must leave the car's door closed. **Recap.** - Standard rules: the car is placed at random; you pick blind; the host knows where the car is and always opens a goat door that is not yours, choosing at random if he has a choice. - Under these rules switching wins exactly when the first pick was wrong: probability $\frac23$ with three doors, $\frac{n-1}{n}$ with $n$ doors when the host opens all but one of the others. - By Bayes' theorem, using the random choice in the last rule, $P(\text{car at 2} \mid \text{host opens 3}) = \frac{1 \cdot \frac13}{\frac12 \cdot \frac13 + 1 \cdot \frac13} = \frac23$. Your own door stays at $\frac13$. - If the host opens a door at random and happens to show a goat, the two closed doors are equally likely. - The same observation can support different conclusions, depending on the process that produced it. ## Problems The first three repeat the lesson's arguments with new numbers. The next three change the rules. The last two connect the problem to other settings. **Problem 1.** There are 5 doors with one car. You pick a door, and the host, who knows where the car is, opens 3 goat doors among the other four. What is the probability of winning if you stay, and if you switch to the other closed door? **Solution.** Your first pick is right with probability $\frac15$, and the host's actions do not change that, so staying wins with probability $\frac15$. If the first pick is wrong, which has probability $\frac45$, the car is among the other four doors, and the host must leave it closed while opening the other three. So switching wins with probability $\frac45$. **Problem 2.** Imagine 300 standard games in which you pick door 1 every time, and suppose the car is behind each door in exactly 100 of them and the host's coin splits evenly. In how many games does the host open door 3? In how many of those is the car behind door 2? **Solution.** Car behind door 1 (100 games): the host opens door 2 in 50 and door 3 in 50. Car behind door 2 (100 games): he opens door 3 in all 100. Car behind door 3 (100 games): he opens door 2 in all 100. So he opens door 3 in $50 + 100 = 150$ games, and in 100 of those the car is behind door 2. That is $\frac{100}{150} = \frac23$, the Bayes calculation in whole numbers. **Problem 3.** A friend says: "Once door 3 is open, the situation is symmetric between doors 1 and 2. Each is a closed door that might hide the car. So by symmetry they have equal probability." Identify precisely what breaks the symmetry. **Solution.** Symmetry would require that the two doors were treated alike by the process that led here, and they were not. Door 1 was protected: by rule 3 the host could not open it whatever was behind it, so its staying closed carries no information. Door 2 was exposed: the host was free to open it and would have done so in half the games where the car is behind door 1, but was forbidden to in every game where the car is behind door 2. Its staying closed is evidence. In numbers, "door 2 stays closed" has probability $\frac12$ if the car is behind door 1 and probability 1 if it is behind door 2. **Problem 4.** There are 4 doors with one car. You pick a door, and the host, who knows where the car is, opens *one* goat door among the other three, choosing at random among the goat doors available to him. You then switch to one of the two other closed doors, chosen at random. What is your probability of winning? Is switching still better than staying? **Solution.** Staying wins with probability $\frac14$. If the first pick is wrong, which has probability $\frac34$, the car is behind one of the two other closed doors, and a random choice between them finds it with probability $\frac12$. If the first pick is right, switching loses. So switching wins with probability $\frac34 \cdot \frac12 = \frac38$, which is more than $\frac14$. *Check:* by symmetry the two other closed doors have equal probability, and $\frac14 + \frac38 + \frac38 = 1$. **Problem 5.** An ignorant host plays the three-door game: he opens one of the two doors you did not pick at random. If he reveals the car, the game is void and you win nothing. If he reveals a goat, you switch. (a) What is the probability that the game is void? (b) What is the probability that you win the car? (c) Would always staying do better? **Solution.** Use the six equally likely cases from the lesson. (a) The car is revealed in 2 of 6 cases: $\frac13$. (b) You win by switching in the cases where the car is behind the other closed door and a goat was shown: car at 2 with door 3 opened, and car at 3 with door 2 opened. That is 2 of 6: $\frac13$. (c) Staying wins when the car is behind door 1 and the game is not void, which is also 2 of 6. The two strategies are equally good, consistent with the conditional probability $\frac12$: each wins half of the $\frac23$ of games that are not void. **Problem 6.** A host follows rules 1 to 3, and when he has a choice between doors 2 and 3 he opens door 3 with probability $q$, where $0 \le q \le 1$. You pick door 1. (a) Find $P(C_2 \mid \text{he opens door 3})$ and $P(C_3 \mid \text{he opens door 2})$ in terms of $q$. (b) Show that both are at least $\frac12$, so switching is never worse than staying whatever you see. (c) Verify that the overall probability that switching wins is $\frac23$ for every $q$. **Solution.** (a) For "opens door 3": the probabilities under $C_1, C_2, C_3$ are $q$, 1, 0, so $P(C_2 \mid \text{opens 3}) = \frac{\frac13}{\frac13 q + \frac13} = \frac{1}{1 + q}$. For "opens door 2": the probabilities are $1 - q$, 0, 1, so $P(C_3 \mid \text{opens 2}) = \frac{1}{(1 - q) + 1} = \frac{1}{2 - q}$. (b) Since $0 \le q \le 1$, both denominators $1 + q$ and $2 - q$ lie between 1 and 2, so both probabilities lie between $\frac12$ and 1. (c) $P(\text{opens 3}) = \frac{1 + q}{3}$ and $P(\text{opens 2}) = \frac{2 - q}{3}$. So the probability that switching wins is $$ \frac{1+q}{3} \cdot \frac{1}{1+q} + \frac{2-q}{3} \cdot \frac{1}{2-q} = \frac13 + \frac13 = \frac23. $$ The case $q = \frac12$ is rule 4, and $q = 0$ is the lazy host of the worked example. **Problem 7.** In the 100-door game, suppose the host does not know where the car is. He opens 98 of the other 99 doors at random, and all 98 happen to show goats. What is the probability that the car is behind your door? **Solution.** Let $E$ be "the 98 opened doors all show goats". If the car is behind your door, every other door hides a goat and $E$ is certain: $P(E \mid \text{yours}) = 1$. If the car is behind a particular other door $j$, then $E$ happens only if door $j$ is the one the host leaves closed. He leaves each of the 99 doors closed with equal probability, so $P(E \mid \text{car at } j) = \frac{1}{99}$. By Bayes' theorem, $$ P(\text{yours} \mid E) = \frac{1 \cdot \frac{1}{100}}{1 \cdot \frac{1}{100} + 99 \cdot \frac{1}{99} \cdot \frac{1}{100}} = \frac{1/100}{2/100} = \frac12. $$ With an ignorant host the two closed doors are equally likely even with 100 doors. Surviving 98 random openings is strong evidence, and it is equally strong for your door and for the other one, since each had to be lucky in the same way. **Problem 8.** (Bertrand's boxes.) Three boxes each hold two coins: one has two gold coins, one has two silver coins, and one has one of each. You choose a box at random and draw one of its coins at random. It is gold. What is the probability that the other coin in the same box is also gold? Explain the connection to the Monty Hall problem. **Solution.** The tempting answer is $\frac12$: the box is either gold-gold or gold-silver. But those two boxes are not equally likely given what you saw. There are three gold coins, each equally likely to be the one you drew: two of them are in the gold-gold box and one is in the mixed box. So the other coin is gold with probability $\frac23$. By Bayes' theorem: $P(\text{gold drawn} \mid GG) = 1$, $P(\text{gold drawn} \mid GS) = \frac12$, $P(\text{gold drawn} \mid SS) = 0$, so $P(GG \mid \text{gold}) = \frac{1 \cdot \frac13}{1 \cdot \frac13 + \frac12 \cdot \frac13} = \frac23$. The numbers $1, \frac12, 0$ are the same three numbers as in the Monty Hall calculation, and the error is the same: two possibilities remain, and one of them was twice as likely to produce the observation.