The game show problem that fooled mathematicians
Tonight's idea
The game: a car is behind one of three doors, goats behind the other two. You pick a door, say door 1. The host, who knows where the car is, opens one of the other doors to reveal a goat, and asks whether you want to switch to the remaining closed door. Should you?
Almost everyone says it makes no difference: two doors left, fifty-fifty. Marilyn vos Savant wrote in Parade magazine that you should switch, because switching wins two times out of three. Thousands wrote in to say she was wrong. She was right. Here are three ways to see it.
Count the cases. You picked door 1. The car is behind door 1, 2 or 3, each equally likely. If it is behind 1, switching loses. If it is behind 2, the host must open 3, and switching to 2 wins. If it is behind 3, the host must open 2, and switching wins. Switching wins in two of three cases.
Nothing changed your first guess. Your first pick was right one time in three. The host opening a goat door was always going to happen, so it tells you nothing about your door. It is still right one time in three, so the other closed door must be right two times in three.
Make it a hundred doors. Pick one of 100. The host opens 98 goat doors, leaving yours and one other. Do you really think your first guess was as good as the door the host carefully left closed?
Try it for real at the Monty Hall demo, a few hundred times.
Watch one: why the host knowing matters
Change one thing: the host does not know where the car is and opens one of the other two doors at random. It happens to be a goat. Now should you switch?
You are on door 1. The car is behind each door with probability \(\tfrac13\), and the host opens door 2 or 3 with probability \(\tfrac12\) each, so every row below has probability \(\tfrac16\).
| Car behind | Host opens | Probability | You see |
|---|---|---|---|
| 1 | 2 | \(\tfrac16\) | goat |
| 1 | 3 | \(\tfrac16\) | goat |
| 2 | 2 | \(\tfrac16\) | car |
| 2 | 3 | \(\tfrac16\) | goat |
| 3 | 2 | \(\tfrac16\) | goat |
| 3 | 3 | \(\tfrac16\) | car |
Keep only the rows where you saw a goat. Staying (door 1) wins in two rows, \(\tfrac16 + \tfrac16\). Switching wins in the other two rows, also \(\tfrac16 + \tfrac16\). Fifty-fifty after all.
The difference is information. A host who knows and deliberately avoids the car is telling you something with the door he leaves shut. A host who got lucky tells you nothing.
Your turn
- Fill in the case table for the standard game with you on door 1: for each place the car could be, which door does the host open, and does switching win?
Car behind Host opens Stay Switch 1 2 3 - In the 100-door version, what is the probability of winning if you stay? If you switch?
- Four doors, one car. You pick one, the host (who knows) opens one goat door among the other three, and you may switch to either of the two remaining closed doors. Work out the chance of winning if you stay, and if you switch to one of the two at random.
- In the random-host version, why do we "throw out" the games where the host revealed the car? What would be wrong with counting them?
- Explain to a friend in one or two sentences why switching helps, without using the word probability.
- Play the demo (or act it out with three cups and a coin) 30 times always switching and 30 times always staying. Record your wins. How close were you to 20 and 10?
Stretch: two children
"I have two children. At least one is a boy. What is the chance both are boys?" Most people say one half. List the four equally likely families in birth order (BB, BG, GB, GG), cross out the ones ruled out by the clue, and find the real answer. Then change the clue to "my older child is a boy" and repeat. Why does naming which child changes the answer?
Answers
- Car behind 1: host opens 2 or 3, stay wins, switch loses. Car behind 2: host opens 3, stay loses, switch wins. Car behind 3: host opens 2, stay loses, switch wins. Switch wins 2 of 3.
- Stay: \(\frac{1}{100}\). Switch: \(\frac{99}{100}\).
- Stay wins \(\frac14\). The car is behind one of the other three doors with probability \(\frac34\); the host removes one goat from those, leaving two doors, and you pick one at random: \(\frac34 \times \frac12 = \frac38\). Switching is still better (37.5% versus 25%), but no longer twice as good, because the host gave away less.
- Because the question asks what to do given that we saw a goat. Games where the car was revealed never reach the decision, so they cannot count toward it. Including them would be answering a different question.
- Something like: "Your first pick is usually wrong, and when it is wrong the host is forced to show you exactly where the car isn't, so the other door is usually right."
- Results vary. Switching should win around 20 of 30 and staying around 10 of 30; anything from roughly 15 to 25 for switching is normal with only 30 games.
Stretch. "At least one boy" rules out only GG, leaving BB, BG, GB. Both boys in 1 of 3: \(\frac13\). "The older is a boy" rules out GB and GG, leaving BB and BG: \(\frac12\). Naming which child removes two families instead of one, because it pins the clue to a specific position. Vague clues leave more cases alive, and that changes the odds.