The Drunk Passenger Puzzle

100 passengers, 100 seats. Passenger 1 chooses randomly. Everyone else takes their own seat if possible; otherwise they choose randomly from the remaining seats.

Passengers
100
Probability passenger 100 gets seat 100
1/2
Answer
50%

Follow one random flight

Only the chain of displaced passengers matters.

The key idea

Ignore passengers who sit in their own seats normally. Track only the passenger whose seat was stolen. That “displacement chain” continues until it reaches one of two special seats:

Chain reaches
Seat 1
Passenger 100 is safe ✓
↔
Chain reaches
Seat 100
Passenger 100 loses ✗
Suppose passenger k finds their own seat occupied. Among the remaining seats: choosing seat 1 ends the chain in success, choosing seat 100 ends it in failure, and choosing any intermediate seat j simply transfers the problem to passenger j.
P(last passenger gets own seat) = P(seat 1 is hit before seat 100) = 1/2.

Why exactly 1/2?

Seats 1 and 100 are symmetric in the random displacement process. Neither has any advantage. The chain must eventually hit one of them, and exactly one is hit first.

Therefore: P(seat 1 first) = P(seat 100 first) = 1/2.

Monte Carlo check

Simulate the complete boarding process many times.
50%

The theoretical result is exactly 50%.

Mental shortcut

Don’t track all 100 passengers. Track only the “stolen-seat chain.” Eventually it hits either seat 1 or seat 100. Since these two outcomes are symmetric, the last passenger gets their own seat with probability 50%.