Can someone finally explain the Monty Hall problem?

Question

Ive read like 10 explanations and I STILL dont get why you should switch.

The setup: 3 doors, 1 car, 2 goats. You pick a door. Monty (who knows where the car is) opens a different door revealing a goat. Now you can switch to the remaining door.

Everyone says switching gives $\frac{2}{3}$ chance to win vs staying gives $\frac{1}{3}$ .

But heres why I think its 50/50: After Monty opens a door with a goat, there are 2 doors left. The car is behind 1 of them. So its $\frac{1}{2}$ no matter what you do.

I know I must be wrong because mathematicians and Mythbusters proved it. But my brain refuses to accept it. Can someone give an explanation that finally clicks for someone whos not good at probability?

Does the 100 doors version help? Like if theres 100 doors, you pick 1, Monty opens 98 goats, you switch to the last door. That one makes more sense but I still cant reconcile the 3-door case.

Emily Park · Accepted Answer

The 100-door version is the key to understanding. 100 doors version: 1. You pick 1 door (1% chance of car) 2. Monty opens 98 doors that all have goats 3. One door remains unopened (besides yours) Your initial pick has 1% chance. The remaining door has 99% chance. Would YOU switch? OF COURSE! Now reduce to 3 doors: 1. You pick 1 door (33.3% chance) 2. Monty opens 1 door with a goat 3. One door remains Your initial pick has 33.3% chance. The remaining door has 66.7% chance. Why its NOT 50/50: The probability doesnt redistribute equally between the two unopened doors because Montys action gives you information. He KNOWS where the car is and deliberately avoids opening it. His knowledge affects the conditional probability. Pr ⁡ ( car behind door 1 ∣ Monty opens door 2 ) = Pr ⁡ ( Monty opens door 2 ∣ car behind door 1 ) ⋅ Pr ⁡ ( car behind door 1 ) Pr ⁡ ( Monty opens door 2 ) If the car is behind your door (door 1), Monty can open either goat door with 50% probability each. If the car is behind door 3, Monty MUST open door 2 (cant open the car). This asymmetry creates the 2/3 probability. Whats stopping you from accepting it? Is it the feeling that "two doors left = 50/50"? Because thats the gamblers fallacy in disguise.

Alex Kim · Answer

I was in your boat for YEARS. What finally made it click for me:

Imagine theres a million doors. You pick one. Monty opens 999,998 goats. Theres your door and one other door left.

Either:
A) You got lucky and picked the car initially (1 in a million)
B) The car is behind the other door (999,999 in a million)

Are you feeling lucky?

For 3 doors: Either you picked right (1 in 3) or the car is behind the other door (2 in 3).

The "two doors = 50/50" is wrong because your initial pick has already established a probability that Montys action doesnt erase.

Can someone finally explain the Monty Hall problem?

2 Answers

Can someone finally explain the Monty Hall problem?

2 Answers