Monty Hall Problem: Puzzle and Step-by-Step Solution
Probability
Easy
Reported at Citadel, Jane Street, SIG, Optiver, Akuna
The puzzle
There are three doors: one hides a car, two hide goats. You pick a door. The host, who knows where the car is, opens one of the other two doors to reveal a goat, then offers you the chance to switch to the remaining closed door. Should you switch?
Short answer: Switch. Staying wins 1/3 of the time; switching wins 2/3.
Step 1: Follow your first pick
Your first pick hides the car with probability 1/3. Nothing the host does can change that, because he was always going to open a goat door whatever you picked. So the other two doors together held the remaining 2/3 of the probability before he opened anything.
When he opens one of those two and shows a goat, the 2/3 does not vanish. All of it now sits on the one closed door you did not pick. Switching is a bet that your first pick was wrong, and your first pick is wrong two times in three.
Step 2: The same answer with Bayes' rule
Say you pick door 1 and the host opens door 3. Let C1, C2, C3 be the events that the car is behind each door, each with prior 1/3. If the car is behind door 1, the host chooses between doors 2 and 3 at random, so P(opens 3 | C1) = 1/2. If it is behind door 2, he must open door 3, so P(opens 3 | C2) = 1. If it is behind door 3, he never opens it, so P(opens 3 | C3) = 0.
The total probability of the host opening door 3 is 1/3 × 1/2 + 1/3 × 1 + 0 = 1/2. Then P(C2 | opens 3) = (1/3 × 1) / (1/2) = 2/3, and P(C1 | opens 3) = (1/3 × 1/2) / (1/2) = 1/3. Switching to door 2 wins with probability 2/3.
Step 3: The detail everything hinges on
The host knows where the car is and never reveals it. That constraint is what makes his choice informative. If instead a door were opened at random and merely happened to show a goat, the two remaining doors really would be 50/50: P(goat shown | C1) and P(goat shown | C2) would both be 1/2, and Bayes gives equal posteriors.
For intuition, play it with 100 doors. You pick one, and the host, always avoiding the car, opens 98 goat doors. Sticking is betting your 1-in-100 guess was right; switching wins 99 times in 100.
Why interviewers ask it
The interview version probes whether you can say why the host's knowledge matters. State the random-host variant and why its answer differs; that shows conditional reasoning rather than a memorised 2/3.
Knowing the proof is half of it. Work Monty Hall Problem interactively, then take it under a clock with the rest of the Outcry simulators.