The lost boarding pass

Question

100 passengers board a full flight in order. Passenger 1 has lost their boarding pass and sits in a uniformly random seat. Every subsequent passenger sits in their own seat if it's free, and otherwise picks a free seat uniformly at random. What is the probability that passenger 100 ends up in their own seat?

Accepted Answer

The answer is \dfrac{1}{2}, and a symmetry argument shows it cleanly. Only seats 1 and 100 matter. Consider every moment someone chooses a seat at random (passenger 1, or any displaced passenger). At that moment: - If they pick seat 1, every later passenger (including 100) finds their own seat free: passenger 100 wins. - If they pick seat 100, passenger 100 loses. - If they pick any other free seat, the problem repeats with fewer passengers and both special seats still free. At each random choice, seat 1 and seat 100 are both free (until the process ends) and hence equally likely to be chosen…

The lost boarding pass

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