Monty Hall, 1,000 Times
In 1990, a magazine columnist answered a puzzle about a game show. About 10,000 readers wrote in to tell her she was wrong, nearly 1,000 of them with PhDs. The columnist was right. (The letter counts vary by source.)
The columnist was Marilyn vos Savant. The puzzle is now named after Monty Hall, the host of the TV game show Let's Make a Deal. Read the puzzle, and decide what you would do.
You're on a game show, facing three closed doors. A car is behind one door, and a goat is behind each of the other two. You pick door 1. The host knows what is behind every door. He opens door 3 and shows you a goat. Then he asks: do you want to keep door 1, or switch to door 2?
The puzzle, in our words
Two closed doors, one car: most people call the choice a coin flip.
Make your guess. Should you keep door 1 or switch to door 2?
- Keep door 1
- Switch to door 2
- It doesn't matter: each door is a 50-50 chance
Show answer
B. Switch to door 2
Switching wins the car 2 times out of 3. Keeping door 1 wins only 1 time out of 3. If the answer feels wrong, you're in good company: Paul Erdős, one of the most productive mathematicians in history, reportedly refused to believe the answer until he watched a computer simulation of the game.
Don't take our word for it. Which move wins more often when you actually play?
Play the game
Your goal: play five rounds keeping your door and five rounds switching, then see which row of the scoreboard has more cars.
After ten rounds almost anything can happen, and keeping might even be ahead by luck. How many rounds would it take before luck stops mattering?
Let the computer play
A computer can play thousands of rounds in a few seconds. Here it plays two players: one always switches, and one always keeps the first door.
Your goal: press 100, then 10,000, and watch where the blue line and the grey line settle.
Both lines wobble at first, then settle: blue (always switch) near 2/3, grey (always keep) near 1/3. The more games, the tighter the lines hug those values. This settling down is called the law of large numbers.
Pigeons learn the answer the same way. In a 2010 study, pigeons that played the game day after day ended up switching on almost every round. The college students in the same study switched far less often, even with practice.
So switching really does win about 2 games in 3. But with two closed doors and one car, why isn't the choice a coin flip?
Why not 50-50?
Go back to the moment you picked door 1, before the host did anything. What is the chance that door 1 hides the car?
- 1 in 3
- 1 in 2
- 2 in 3
Show answer
A. 1 in 3
One car, three doors, no clues: door 1 hides the car 1 time in 3. So the car is behind one of the other two doors 2 times in 3.
Here is the key fact: the host knows where the car is and never opens the car's door. One of the two doors you didn't pick always hides a goat, so the host can always show you a goat.
So the goat tells you nothing new about door 1, and door 1 keeps its 1-in-3 chance. Doors 2 and 3 shared the other 2-in-3 chance. Door 3 is now out, so door 2 carries the whole 2-in-3 chance.
Tidy arguments can still be wrong. Can we check this one by listing every possible game?
Every possible game
Say you always pick door 1. The car is behind door 1, door 2 or door 3, each with a 1-in-3 chance. That makes three possible worlds.
Your goal: press "Everyone switches", then "Everyone keeps door 1", and count the worlds that win the car each time.
Switching wins in 2 worlds out of 3. Still, 2/3 against 1/3 can feel like a small edge. Is there a version of the game where the answer is obvious?
Try 100 doors
Make the game extreme: 100 doors and one car. You pick one door. The host, who knows where the car is, opens 98 of the other doors, and all 98 hide goats. One door besides yours is still closed.
How often does switching to the one door the host skipped win the car?
- 1 time in 2
- 2 times in 3
- 99 times in 100
Show answer
C. 99 times in 100
Your first pick had a 1-in-100 chance, and showing goats can't change that chance. The other 99 doors held a 99-in-100 chance together. The host opened 98 of those 99 doors, so the whole 99-in-100 chance sits on the one door he left shut.
Your goal: pick any door, watch the host sweep away 98 goats, and spot the one door he skips. Then try 10 doors, and then 3.
With 100 doors, keeping your first pick feels silly. With 3 doors, keeping feels as good as switching. But one rule covers any number of doors, as long as the host opens all the other doors but one:
| Doors | Keeping wins | Switching wins |
|---|---|---|
| 3 | 1/3 | 2/3 |
| 10 | 1/10 | 9/10 |
| 100 | 1/100 | 99/100 |
| \(n\) | \(1/n\) | \((n-1)/n\) |
Every version so far depends on one rule: the host knows where the car is. What happens if the host doesn't know?
A host who guesses
Picture a new host who has forgotten where the car is. You pick door 1. The forgetful host opens door 3 at random, and by luck door 3 hides a goat. The stage looks exactly like the original puzzle.
With the forgetful host, should you switch to door 2?
- Yes: switching still wins 2 times in 3
- It doesn't matter: each door wins 1 time in 2
- No: keeping door 1 wins more often
Show answer
B. It doesn't matter: each door wins 1 time in 2
Each door now wins half the time. A guessing host sometimes opens the car's door by accident. Those games end early, and the games that survive split evenly between door 1 and door 2.
A guessing host gives six equally likely games: three spots for the car, times two doors he might open.
Only two games match what you saw: the car behind door 1, and the car behind door 2. At 1/6 each, door 1 and door 2 are tied at 1/2.
Same doors, a different host, a different answer. Twice now we have counted games by hand. Is there one rule that does the counting for us?
The math: Bayes' rule
There is: Bayes' rule, which updates a chance after you see new evidence. In the original puzzle, the evidence is "the host opened door 3".
Write \(C_2\) for "the car is behind door 2" and \(H_3\) for "the host opens door 3". We want \(P(C_2 \mid H_3)\), read "the probability of \(C_2\) given \(H_3\)".
Draw every possibility as one rectangle, like the guessing host's grid. This time the host knows where the car is.
The shaded area adds up to \(1/6 + 1/3 = 1/2\), and door 2 owns 1/3 of it: two-thirds of the shaded area. Bayes' rule says the same thing in one line:
Show the full derivation (try it yourself first)
- Before the host acts. The car is equally likely to be behind any door:
\[ P(C_1) = P(C_2) = P(C_3) = \frac13 \]2. How likely is "the host opens door 3" in each world? You hold door 1.
\[ \begin{aligned} P(H_3 \mid C_1) &= \tfrac12 \\ P(H_3 \mid C_2) &= 1 \\ P(H_3 \mid C_3) &= 0 \end{aligned} \]If the car is behind door 1, the host picks door 2 or door 3 at random. If the car is behind door 2, door 3 is the host's only goat door. If the car is behind door 3, the host never opens door 3.
- The total chance that the host opens door 3:
\[ P(H_3) = \frac13 \cdot \frac12 + \frac13 \cdot 1 + \frac13 \cdot 0 = \frac12 \]4. Bayes' rule for door 1 and door 2:
\[ P(C_1 \mid H_3) = \frac{\frac12 \cdot \frac13}{\frac12} = \frac13 \]\[ P(C_2 \mid H_3) = \frac{1 \cdot \frac13}{\frac12} = \frac23 \]5. Read off the answer. Keeping door 1 wins with probability 1/3, and switching to door 2 wins with probability 2/3. The two chances add to 1 because door 3 is out of the game.
- The host who guesses. Let \(G_3\) mean "the host opens door 3, and door 3 hides a goat". A guessing host opens door 3 half the time in every world, but in world 3 door 3 hides the car:
\[ \begin{aligned} P(G_3 \mid C_1) &= \tfrac12 \\ P(G_3 \mid C_2) &= \tfrac12 \\ P(G_3 \mid C_3) &= 0 \end{aligned} \]\[ P(G_3) = \frac13 \cdot \frac12 + \frac13 \cdot \frac12 + 0 = \frac13 \]\[ P(C_2 \mid G_3) = \frac{\frac12 \cdot \frac13}{\frac13} = \frac12 \]
The letters, answered
Now you can answer the 10,000 letters. Two closed doors look equal, but the host's choice is a clue. He knows where the car is, so the door he leaves shut collects the chance of every door he opened.
You saw that clue at work in your own rounds, in 10,000 computer games, in three worlds, behind 100 doors and in Bayes' rule. With a host who guesses, the clue disappears, and then the letter writers would have been right.
Ask Phy
- Why does a host who guesses make switching a 50-50 choice? Walk me through a tree diagram.
- With 4 doors, the host opens just one goat door. Should I still switch, and what are my chances?
- Give me an AP Statistics free-response question that uses a tree diagram and conditional probability.