The puzzle is named after Monty Hall, the host of the American television game show Let's Make a Deal, which began in 1963. In 1990 a reader asked Marilyn vos Savant, who wrote a question-and-answer column in Parade magazine, whether to switch. She said yes: 2/3 for switching, 1/3 for staying. About ten thousand letters arrived, close to a thousand of them from people with PhDs, most of them insisting she was wrong. She was right. The mathematician Paul Erdős, one of the most prolific in history, reportedly refused to accept the answer until a colleague showed him a computer simulation, which is exactly what the first board on this page is. Monty Hall himself, asked about it in 1991, played the game with a reporter and got the same result, and then pointed out that on the real show he was under no obligation to offer a switch at all.
The three doors
Three doors. One hides a bicycle, two hide goats. You pick a door, Robo opens a different door to show you a goat, and then asks: stay or switch? Nearly everyone says it makes no difference. Nearly everyone is wrong, and the reason is not the doors. It is what Robo knows.
Here is the game. Three closed doors. Behind one is a bicycle. Behind each of the other two is a goat. Robo is the host, and Robo knows exactly where the bicycle is. You pick a door. Robo then opens one of the other two doors, always one with a goat behind it, and asks: do you want to stay with your door, or switch to the one closed door that is left?
Most people say it cannot matter. Two doors left, one bicycle, so it is fifty-fifty either way. That is what almost everyone thinks the first time, including a great many people with mathematics degrees. Play by hand for a while and watch the two tallies, one for the rounds where you stayed and one for the rounds where you switched. Then, before you let Robo play a thousand rounds for you, make a prediction. The prediction is the star.
The three doors
Click a door. Robo opens a goat door. Stay or switch.Before Robo plays a thousand rounds for you: what fraction of the time does switching win?
Why it is two thirds
Forget the switch for a moment and look at your first pick. Three doors, one bicycle, and you know nothing yet. Your first pick is right 1 time in 3 and wrong 2 times in 3. Nothing Robo does afterwards can reach back and change that.
Now follow each case to the end. If your first pick was right (1 time in 3), both other doors hide goats. Robo opens one of them, and switching takes you to the other goat. Switching loses. If your first pick was wrong (2 times in 3), the other two doors hide one bicycle and one goat. Robo is not allowed to open the bicycle, so Robo is forced to open the goat. The only door left to switch to is the bicycle. Switching wins.
So switching wins exactly when your first pick was wrong, and your first pick is wrong 2 times in 3. That is the whole proof. Here it is as a table, with you picking door 1. Picking door 2 or door 3 works the same way; just rename the doors.
| Bicycle behind | You pick | Robo opens | Switching gets | Staying gets |
|---|---|---|---|---|
| Door 1 | Door 1 | Door 2 or door 3 (Robo may choose; both are goats) | the other goat | the bicycle |
| Door 2 | Door 1 | Door 3 (forced: door 2 is the bicycle) | the bicycle | a goat |
| Door 3 | Door 1 | Door 2 (forced: door 3 is the bicycle) | the bicycle | a goat |
The three rows are equally likely, and switching wins in two of them. The fifty-fifty feeling comes from treating the two closed doors as the same kind of thing. They are not. Your door was chosen blind. The other door survived an inspection: Robo looked at two doors, knowing where the bicycle was, and left that one shut. Two times in three, Robo had no choice about which door to leave.
If it still feels slippery, make the game bigger. A hundred doors, one bicycle. You pick one. Robo, who knows where the bicycle is, opens 98 doors, every one of them a goat, and leaves exactly one other door closed. Now: your door, or the one Robo carefully stepped around?
A hundred doors
Pick one. Robo opens every other goat door but one. Stay or switch.It only works because the host knows
Here is the part most explanations skip, and it is the part that makes the puzzle honest. Everything above depends on Robo knowing where the bicycle is. Change one rule. Robo does not know, and simply opens one of the two other doors at random. Sometimes Robo opens the bicycle by accident. Call that round spoiled: you never got your choice, so it counts for neither staying nor switching. But suppose the random door shows a goat, exactly like before. Same three doors, same goat, same question. Now switching wins only half the time.
Count the cases over 300 games. Your first pick is right in about 100 of them and wrong in about 200. When your pick is right, both other doors hide goats, so a guessing Robo shows a goat every time: 100 goat games. When your pick is wrong, a guessing Robo opens the bicycle half the time and the goat half the time: about 100 spoiled games and 100 goat games. Throw out the spoiled games. Among the 200 games that survive, 100 had a right first pick and 100 had a wrong one. Switching wins in the wrong-first-pick games: 100 out of 200, one half.
Compare the knowing Robo. It never spoils a game, so all 200 wrong-first-pick games survive, and switching wins 200 out of 300. The guessing Robo's accidents land entirely on the games where switching would have won. That is why the rate drops. Run it, then answer the question at the bottom of the board.
The guessing host
Robo opens one of the other doors at random. If it is the bicycle, the round is spoiled.Why did the switching rate drop from 2/3 to 1/2?
Run at least 1000 rounds with a guessing Robo first, then choose. One answer is right.
Conditional probability, gently
What you just did has a name. Whenever you ask "of the cases where B happened, what fraction are also cases where A happened," you are computing a conditional probability (say: kon-DISH-un-ul), written P(A given B). Mathematicians write the "given" as a vertical bar: P(A | B). It is a plain fraction. Count the B-cases, count how many of them are also A-cases, and divide.
Here B is "Robo showed a goat" and A is "switching wins", which is the same thing as "your first pick was wrong". The two tables hold the counts for 300 imagined games with each kind of host.
| Shows a goat | Shows the bicycle | |
|---|---|---|
| First pick right (100) | 100 | 0 |
| First pick wrong (200) | 200 | 0 |
| Goat games | 300 | 0 |
| Shows a goat | Shows the bicycle | |
|---|---|---|
| First pick right (100) | 100 | 0 |
| First pick wrong (200) | 100 | 100 (spoiled) |
| Goat games | 200 | 100 |
With the knowing host, P(first pick wrong | goat shown) = 200 / 300 = 2/3. With the guessing host, P(first pick wrong | goat shown) = 100 / 200 = 1/2. Same goat on the screen, different fraction, because the goat was produced by a different process. Information is not only what you see. It is what you see together with how it came to be shown to you.
This way of updating what is likely after seeing evidence is called Bayes' rule (say: BAYZ), after Thomas Bayes, an English clergyman of the 1700s who worked out the idea. It is how a spam filter decides what a suspicious email probably is, and how a doctor works out what a positive test result really means. The count table is the whole idea. The formula people learn later is the count table written in letters.
The two strategies. Switching wins exactly when the first pick was wrong. Staying wins exactly when it was right.
P(stay wins) = P(first pick right) = 1/3
Expected value. From Chapter 4: multiply each outcome by its probability and add. Count a bicycle as 1 and a goat as 0.
EV(stay) = 1/3 × 1 + 2/3 × 0 = 1/3 of a bicycle per game
Over 300 games, switching collects about 200 bicycles and staying about 100.
More doors. With n doors, one bicycle, and a knowing host who opens every other goat door, your first pick is wrong (n − 1) times in n, and every one of those times the host is forced to leave the bicycle closed.
For n = 3 that is 2/3. For n = 100 it is 99/100. The hundred-door board takes any number of doors from 3 to 100, so you can check both.
Conditional probability as counts.
Knowing host: 200 ÷ 300 = 2/3. Guessing host: 100 ÷ 200 = 1/2. Every conditional probability in this chapter is one of these divisions, and if you can count the cases you never need anything fancier.
The Monty Fall problem. In 2008 the statistician Jeffrey Rosenthal gave the guessing-host version a name: the host slips, falls against a door, and it happens to swing open on a goat. Same doors, same goat, and the answer is 1/2. The point of the name is that the answer depends on how the door came to be opened, not only on which door it was.
What you know changes what is likely, and the safe way to work out how is to count cases. List every way the situation could have come about, keep only the ways that agree with what you saw, and see what fraction of those are the ones you care about. The same goat, shown by a host who knows and a host who guesses, means two different things.
1. Four doors, one bicycle, three goats. You pick door 1. Robo, who knows, opens one goat door among the other three and offers you a switch to either of the two other closed doors. What is the chance of winning if you stay? If you switch to a particular one of the two?
Answer
Staying wins 1/4: your first pick was right 1 time in 4, and nothing that happens afterwards changes it. Switching: your first pick is wrong 3/4 of the time, and in those games the bicycle is behind one of the two doors Robo left closed, each equally often. So a particular other door wins 3/4 × 1/2 = 3/8. Better than 1/4, but not the 3/4 you might have hoped for, because Robo removed only one goat, not all of them.
2. A sneaky host. Robo knows where the bicycle is and offers you the switch only in rounds where your first pick was right. In the other rounds Robo just opens your door. Should you ever take the switch?
Answer
Never. If the switch is offered, your first pick was right, so switching loses every time. This is why the rules have to say what the host always does. The famous puzzle assumes the host always opens a goat and always offers the switch. Change the host's rule and you change the answer.
3. In the hundred-door game, what is P(stay wins)? Bonus: what if Robo opens only 50 goat doors instead of 98, leaving 49 other doors closed?
Answer
P(stay wins) = 1/100, no matter how many doors Robo opens. Bonus: your first pick is wrong 99/100 of the time, and then the bicycle is equally likely behind each of the 49 other closed doors, so a particular one of them wins 99/100 × 1/49 = 99/4900, about 2%. Still twice as good as staying.
Set up three cups and a coin, and ask a friend to be the host. Before you start, they decide in secret whether they will peek under the cups before each round or never peek at all, and they keep to that choice for the whole game. You must not watch the setup. Play 30 rounds, switching every time, and write down what happens in each round. Did your friend peek? How can you tell from the tally?
Answer
Look for spoiled rounds first. A host who peeks never turns over the coin by accident. A host who does not peek does it about 1 round in 3, so in 30 rounds you expect around 10 spoiled ones. Then look at the switching wins among the rounds that survived: around 2/3 for a peeker, around 1/2 for a guesser. With only 30 rounds the win rate alone can fool you, since chance wobbles by several rounds either way, but the spoiled count is hard to miss. A single spoiled round proves your friend did not peek. Thirty rounds with no spoils at all and a guessing host happens less than 1 time in 100,000, so no spoils means they peeked.
Stars in this chapter
Earn them by doing the clever thing, not by clicking around.
- Predict 2/3 before the first auto-run, then prove it over 1000 rounds
- Run 1000 rounds against a guessing Robo and explain why the rate changed