Monty Hall Problem

There are 3 doors: 1 hides a car, 2 hide goats. You choose one door. The host, who knows where the car is, opens another door and reveals a goat. Now you must decide: stay with your original choice or switch to the other unopened door.

Why switching works

Suppose you initially pick Door A. At that moment:

🚪
Door A
Your original pick
P(car behind A) = 1/3
🚪
Door B
Not chosen
Part of the other 2/3
🚪
Door C
Not chosen
Part of the other 2/3

Your first choice is correct only 1/3 of the time. So the probability that the car is in one of the other two doors is 2/3.

When the host opens one of those two other doors and shows a goat, that whole 2/3 probability effectively gets concentrated onto the single remaining unopened door.

Stay

1/3

Your original door never got more likely.

Switch

2/3

The remaining unopened door carries the other probability mass.

Three equally likely cases

Assume your first pick is Door A.

Case 1: Car behind A

Probability = 1/3

Staying wins. Switching loses.

Case 2: Car behind B

Probability = 1/3

Host opens C. Switching wins.

Case 3: Car behind C

Probability = 1/3

Host opens B. Switching wins.

Stay wins in 1 case; switch wins in 2 cases.

Simulation

Run repeated random games. Over many trials, the results approach: stay ≈ 33.3% and switch ≈ 66.7%.

Games played
0
Stay win rate
—
Switch win rate
—
Run a game to see what happens.

Final answer

You should switch.

The probability of winning the car is:

P(win by staying) = 1/3     P(win by switching) = 2/3

So switching doubles your chance of winning.