Monty Hall Problem Explained - Why Switching Wins 2/3 of the Time

The setup

Three doors: one car, two goats. You pick a door. The host - who knows where the car is - opens one of the OTHER doors to reveal a goat, then offers you a switch. Do you take it?

Short answer: Switch. Staying wins 1/3 of the time; switching wins 2/3.

Follow your first pick

Your original door hides the car with probability 1/3, and nothing the host does changes that - he was always going to open a goat door, whatever you picked. So the remaining 2/3 of probability belonged to the other two doors together, and after he eliminates one of them, all of it sits on the single closed door left.

Switching is a bet that your FIRST pick was wrong, and your first pick is wrong 2/3 of the time.

The detail everything hinges on

The host knows. He never opens the car and never opens your door, so his reveal carries information about the doors you did not pick. If instead a door were opened at random and merely happened to show a goat, the two remaining doors really would be 50/50 - different process, different answer.

For intuition, play it with 100 doors: you pick one, the host - always avoiding the car - opens 98 goat doors. Sticking is betting your 1-in-100 first guess was right; switching wins 99 times in 100.

Why interviews ask it

The interview version usually probes whether you can articulate WHY the host's constraint matters - candidates who can state the random-host variant and why it differs demonstrate actual conditional reasoning rather than a memorised answer.