← Schonfeld Interview Insights
I'd seen a version of this before so I wasn't totally lost, but I still fumbled explaining it cleanly under pressure.
Start by simplifying the problem to small numbers of passengers (e.g., 2, 3, 4) to identify a pattern, then generalize. Use symmetry or induction to argue that the probability is 1/2, and explain the key insight that only the first and last seats matter.
Pro tip: Emphasize that the problem is equivalent to a random walk between seat 1 and seat 100, and that the drunk passenger's choice effectively 'passes the buck' until someone sits in seat 1 or seat 100. This shows you understand the underlying structure, not just the answer.
Clarify the rules: Passenger 1 picks a random seat. Each subsequent passenger sits in their assigned seat if available, otherwise picks randomly among remaining seats. Ask for the probability that Passenger 100 gets their own seat.
Compute probabilities for n=2,3,4 passengers to see a pattern. For n=2, probability is 1/2; for n=3, also 1/2; for n=4, still 1/2. This suggests the answer is always 1/2 regardless of n.
Recognize that the process continues until someone sits in either seat 1 or seat 100. At that point, if seat 1 is taken, all remaining passengers (including 100) get their own seats; if seat 100 is taken, passenger 100 does not. By symmetry, these two outcomes are equally likely.
Use induction or a symmetry argument: At each step, the set of available seats includes seat 1 and seat 100 (unless one is already taken). The first time a 'special' seat (1 or 100) is chosen, it is equally likely to be either. Thus probability = 1/2.
Conclude that the probability is 1/2 (50%). Optionally, mention that this holds for any number of passengers n ≥ 2.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.