Probability of Last Passenger Getting Their Seat

This classic seating puzzle describes a boarding process where one early disruption propagates randomness through a long sequence of otherwise deterministic choices. Candidates must track how a single initial mistake in choosing seats interacts with a fixed rule followed by everyone else, and reason about the implications for the final passenger's fate. The setting feels combinatorial and potentially messy, with many possible seat-allocation paths, which tempts brute-force thinking. It often appears in interviews for quantitative trading and research roles, as well as big-tech data science and algorithmic positions, because it tests whether someone can see through apparent complexity to a simple underlying structure.

The solution leans on recognizing symmetry, recursive structure, and invariance in a sequential random process. Strong answers reframe the evolving situation in terms of a small set of "effective states" rather than 100 separate passengers, and argue about those states qualitatively. Interviewers watch for resistance to simulation-style reasoning, the ability to construct and justify a recursive or inductive view, clarity in handling conditional probabilities, and a clean explanation of why intermediate details do not affect the final probability.

What it tests

This problem class is governed by the principle of symmetry and invariance in sequential random processes with absorbing states. When a process involves a sequence of choices that recursively reduce to the same situation but with fewer options, the ultimate outcome often depends only on the endpoints or key states, not on the intermediate steps. Here, the process is memoryless in the sense that, after each random selection, the remaining problem is structurally identical but with fewer participants. The symmetry between the two 'absorbing' seats (seat 1 and seat 100) ensures that, regardless of the path taken, the probability of either being the first to be chosen remains equal. This holds because, at every recursive step, the decision reduces to the same binary choice, preserving the 50-50 split.