Question
In the Monty Hall problem, you are on a game show with 3 doors. Behind one door is a car; behind the other two are goats. You pick a door. The host (who knows what’s behind each door) opens one of the remaining doors to reveal a goat. Should you switch to the other unopened door or stay? Does it matter?
Solution — Step by Step
- 3 doors: 1 car, 2 goats
- You pick door 1 (say)
- Host opens a door (say door 3) showing a goat
- You are offered the choice: switch to door 2 or stay at door 1
The question: what is your probability of winning if you switch vs if you stay?
The surprising answer: switch — your probability doubles from 1/3 to 2/3.
When you initially pick door 1:
- P(car behind door 1) = 1/3
- P(goat behind door 1) = 2/3
If you stay, you win only if your initial pick was correct: P(win by staying) = 1/3.
The host’s action (revealing a goat) doesn’t change the probability of your initial pick — you chose before any information was revealed.
If your initial pick was wrong (P = 2/3): The host must reveal the remaining goat (only one door to open). Switching takes you to the car. You win.
If your initial pick was correct (P = 1/3): Host reveals one goat. Switching takes you to the remaining goat. You lose.
Therefore: P(win by switching) = P(initial pick was wrong) = 2/3.
Summary:
- Stay: P(win) = 1/3
- Switch: P(win) = 2/3
Always switch!
| Car location | You pick | Host opens | Result if stay | Result if switch |
|---|---|---|---|---|
| Door 1 | Door 1 | Door 2 or 3 | Win | Lose |
| Door 2 | Door 1 | Door 3 | Lose | Win |
| Door 3 | Door 1 | Door 2 | Lose | Win |
All three cases are equally likely (P = 1/3 each).
Stay wins in 1 out of 3 cases. Switch wins in 2 out of 3 cases. Confirmed.
Why This Works
The host’s action is not random — he always reveals a goat, and he never opens your door. This asymmetry is the key.
When you initially picked, 2/3 of the probability was “elsewhere” (the other two doors combined). The host’s revelation doesn’t spread that probability back to you — it concentrates all the “elsewhere” probability onto the one remaining door.
Think of it with 100 doors: you pick one, host opens 98 goat doors, leaves one unopened. Should you switch? Obviously yes — the one remaining door has ~99/100 probability of having the car. The Monty Hall problem is the same principle with 3 doors.
Alternative Method — Conditional Probability (Formal)
Let’s use Bayes’ theorem. Suppose you pick door 1 and host opens door 3.
P(car at door 2 | host opened door 3) = P(host opens door 3 | car at door 2) × P(car at door 2) / P(host opens door 3)
= 1 × (1/3) / (1/2) = 2/3
P(car at door 1 | host opened door 3) = P(host opens door 3 | car at door 1) × P(car at door 1) / P(host opens door 3)
= (1/2) × (1/3) / (1/2) = 1/3
This confirms: switch has P = 2/3, stay has P = 1/3.
Common Mistake
Thinking “there are 2 doors left, so it’s 50-50.” This is the classic error. The two remaining doors are NOT equally likely because the host’s information-laden choice broke the symmetry. Your door was chosen randomly from 3; the remaining door was chosen by the host after eliminating a guaranteed goat. These are fundamentally different situations.