The Monty Hall problem is a puzzle with three doors: a car behind one, goats behind the other two. You pick a door, the host opens another to reveal a goat, and asks if you want to switch. Switching wins 2/3 of the time (66.67%) and staying wins only 1/3 (33.33%) — so always switch.
Named after Monty Hall, host of Let’s Make a Deal, it became famous in 1990 when Marilyn vos Savant answered it in Parade — and thousands of readers, including mathematicians, wrote in to say she was wrong. She was right.
Check the numbers yourself with a free probability calculator — its Monty Hall mode compares switching and staying for 3, 100 or 300 doors.
The Setup: Three Doors, One Car, Two Goats
There are 3 doors: a car behind one, goats behind the other two. You pick a door — say Door 1 — and do not open it yet. The host, who knows where the car is, opens a different door — say Door 3 — and shows you a goat. He will never open the door with the car. Then he asks: do you want to switch to Door 2, or stay with Door 1?
Two doors left feels like 50/50 — but it is not. Switching doubles your chance, from 1/3 to 2/3.
The 3-Door Table: Every Possible Outcome
Suppose you always pick Door 1 first. The car is equally likely to be behind any door, so each row happens 1/3 of the time:
| Car is behind | Host opens | Switching gives you | Staying gives you |
|---|---|---|---|
| Door 1 (1/3 of the time) | Door 2 or 3 (a goat) | A goat — you lose | The car — you win |
| Door 2 (1/3 of the time) | Door 3 (a goat) | The car — you win | A goat — you lose |
| Door 3 (1/3 of the time) | Door 2 (a goat) | The car — you win | A goat — you lose |
Switching wins in 2 of the 3 rows: 2/3 = 66.67%. Staying wins in only 1 row: 1/3 = 33.33%. That is the whole answer — just counting. Switching loses only when your first pick was already right (1/3 of the time). The other 2/3 of the time your first pick was wrong — and then the host’s reveal points straight at the car, because he must open the other goat door.
Why Switching Wins: Your First Pick Never Improves
Your first pick keeps its original 1/3 chance forever, because nothing about your door changed — the new information all lands on the other doors. When you picked Door 1, the car had a 1/3 chance of being behind it and a 2/3 chance of being behind Doors 2 and 3 together. The host then opens a goat door from Doors 2 and 3 — he was always going to open a goat door, whatever you picked. So your door is still a 1/3 shot, but the 2/3 that used to be spread across two doors now sits entirely behind the one door he left closed. The reveal did not create probability; it moved it.
Conditionally, P(car behind Door 2 | host opened Door 3) = (1/3)/(1/2) = 2/3. For the full method, see how to calculate probability — including the P(6 | even) = 1/3 example.
The 100-Door Version: The Intuition That Makes It Click
Imagine 100 doors: 1 car, 99 goats. You pick Door 1 — a 1/100 chance of being right. The host opens 98 doors, all goats, leaving your Door 1 and one other door, say Door 57. Switch or stay? Now it is obvious: your 1/100 guess does not become a good guess just because 98 doors were opened. The car is behind the other 99 doors with probability 99/100, and the host eliminated 98 of them, so the whole 99/100 now sits behind Door 57. Switching wins 99/100 = 99%; staying wins 1/100 = 1%. With 300 doors, switching wins 299/300 = 99.67% — same logic, smaller numbers.
Why Your Gut Says Stay (And Why It Is Wrong)
“Two doors left means 50/50.” The big error: two doors are left, but they are not equally likely. Your door got here by a 1/3 guess; the other door got here by surviving the host’s deliberate choice to leave it closed. Probability follows the history, not the count.
“Switching feels like admitting I was wrong.” Staying feels loyal to your first instinct — but your first pick was a blind 1/3 guess, not an insight. The host just handed you information worth 2/3; refusing it is refusing a better bet.
“I have already won half the time by staying.” People remember the lucky stays and forget the losses. Run the game 300 times: staying wins about 100, switching about 200.
What If the Host Does Not Know?
If the host opens a door at random, without knowing where the car is, and happens to reveal a goat, it really is 50/50. The difference is knowledge: in the real game he chooses a goat door on purpose, and that choice carries information.
Frequently Asked Questions
Should you switch doors in the Monty Hall problem?
Yes — switching wins 2/3 (66.67%), staying wins 1/3 (33.33%). Your first pick is right only 1/3 of the time, so the host’s reveal concentrates the other 2/3 onto the one remaining door.
Is the Monty Hall problem 50/50?
No — not when the host knows where the car is and always reveals a goat. Your door keeps its 1/3; the other holds 2/3. Only a random reveal makes it 50/50.
Why does switching win 2/3 of the time?
Because switching wins whenever your first pick was wrong — and your first pick is wrong 2/3 of the time. The 3-door table shows it: switching wins in 2 of the 3 equally likely rows.
What is the 100-door Monty Hall version?
Switching wins 99/100 = 99%, staying wins 1/100 = 1%. The host opens 98 goat doors, and the one surviving door holds the other 99%.
Who made the Monty Hall problem famous?
Marilyn vos Savant, in her Parade column in 1990. She gave the switch answer, thousands of readers wrote in to disagree — including mathematicians — and she was right.
Does the answer change if the host does not know where the car is?
Yes — then it is genuinely 50/50. A random reveal carries no information, so the two remaining doors are equally likely.
Try It Yourself
Run the numbers: use a free probability calculator and compare switching vs staying for 3 doors (66.67% vs 33.33%), 100 doors (99% vs 1%) and 300 doors (99.67% vs 0.33%). For the arithmetic behind the reasoning, the Fraction Calculator simplifies odds like 2/3, and the Factorial Calculator handles the counting behind combinations.