In the game show “Let’s Make a Deal”, a car is hidden behind one of three doors and goats behind the other two. You choose a door. The host, Monty Hall, who knows where the car is, opens one of the other doors to reveal a goat and offers you the chance to switch. Should you? When Marilyn vos Savant answered “yes, switching doubles your chances” in 1990, thousands of readers – including many mathematicians – wrote in to say she was wrong. She was right.
Your first pick is correct with probability 1/3, and nothing the host does changes that, because he can always open a goat door. So the remaining 2/3 must sit on the other closed door. Switching wins whenever your first pick was wrong, which happens two times in three.
Play the game here to feel the decision, or switch to simulation mode and run thousands of games. You can also add doors: with 100 doors and the host opening 98 goats, switching wins 99% of the time, which makes the logic obvious. The table and chart compare the simulated win rates with the exact values.
How to use the Monty Hall Problem Calculator
- Mode
- Which door do you choose?
- Your decision after the host opens a door
- Number of doors
- Goat doors the host opens
- Number of games to simulate
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
In play mode the car is placed at random, the host opens a goat door you did not pick and your decision reveals the outcome; the game stays the same until you press “Play again” or pick another door. In simulation mode each game places the car and your pick at random, the host opens the requested number of goat doors and the switching player picks one of the remaining closed doors at random.
Worked example
Suppose you pick door 1. The car is behind door 1 with probability 1/3 – then the host opens door 2 or 3 and switching loses. The car is behind door 2 with probability 1/3 – the host must open door 3 and switching wins. The car is behind door 3 with probability 1/3 – the host must open door 2 and switching wins. Switching wins in two of three equally likely cases: 2/3 ≈ 66.67%. With 10,000 simulated games expect roughly 6,667 wins for switching and 3,333 for staying.
Win rates with more doors
| Doors | Goats opened | Stay wins | Switch wins |
|---|---|---|---|
| 3 | 1 | 33.33% | 66.67% |
| 4 | 1 | 25% | 37.5% |
| 4 | 2 | 25% | 75% |
| 10 | 8 | 10% | 90% |
| 100 | 98 | 1% | 99% |
Formula
Three doors: P(win | stay) = 1/3 · P(win | switch) = 2/3 · d doors, host opens h goats: P(stay) = 1/d · P(switch) = (d − 1)/d × 1/(d − 1 − h)
Worked example
Inputs — Mode: Play the game; Which door do you choose?: Door 1; Your decision after the host opens a door: Not decided yet.
Results — Result = Waiting for your decision; What happened = You picked door 1. The host, who knows where the car is, opens door 2 and shows a goat. Will you stay with door 1 or switch to door 3? Choose in "Your decision".; The doors = [ Door 1 (your pick): ? ] [ Door 2: goat ] [ Door 3: ? ]; Exact chance of winning by staying = 33.33 %; Exact chance of winning by switching = 66.67 %.
Key terms
- Conditional probability
- The chance of an event once you know something else has happened.
- Host strategy
- The rule the host follows; the classic answer needs a host who knows where the car is and always opens a goat door.
- Simulation
- Repeating a random experiment many times on a computer to estimate probabilities.
Frequently asked questions
Should you switch in the Monty Hall problem?
Yes. Switching wins with probability 2/3, staying with 1/3.
Why isn't it 50/50 after a door is opened?
The host's choice is not random – he must avoid the car. Your first pick keeps its 1/3 chance, so the other closed door carries the remaining 2/3.
What if the host opens a door at random and happens to show a goat?
Then switching and staying are both 1/2. The 2/3 answer depends on the host knowing where the car is and always revealing a goat.
How does Bayes' theorem solve it?
You pick door 1 and the host opens door 3. P(host opens 3 | car behind 1) = 1/2, P(host opens 3 | car behind 2) = 1, so the posterior for door 2 is (1 × 1/3) ÷ (1/2) = 2/3.
What happens with more doors?
With 10 doors and 8 goats revealed, switching wins 9/10. With d doors and all but one other door opened, switching wins (d − 1)/d.
How many games do I need to see the effect?
After a few hundred simulated games the switch rate is usually between 62% and 72%; after 10,000 games it is within about 1 percentage point of 66.67%.