Answer: The Absent Ticket Inspector

Key Idea / Intuition

The answer is exactly $\frac{1}{2}$, independent of $n$. This is shocking — with 100 or 1,000,000 passengers, the last person still has a 50/50 chance. The key insight is that throughout the entire process, the "fate" of the last seat is determined by a single competition: at every displacement, the displaced passenger randomly chooses either seat 1 (your seat) or seat $n$ (their seat), or they pass the problem along. The last seat is eventually "claimed" by one of these two outcomes, and by symmetry they are equally likely.

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Formal Proof / Solution

Setup: Label the seats $1, 2, \ldots, n$. You (passenger 1) sit randomly. Passenger $k$ sits in seat $k$ if free, otherwise picks randomly among free seats. We want $P(\text{passenger } n \text{ sits in seat } n)$.

Key observation: At any moment during boarding, the only "contested" seats are seat $1$ (yours, the one you stole) and seat $n$ (the last seat). All other seats will eventually be resolved.

Elegant symmetry argument:

Consider the following invariant: whenever a displaced passenger must choose a random free seat, seat 1 and seat $n$ are always equally likely to be chosen (since neither is ever "preferentially" occupied during the random choices).

More precisely, we prove by induction: the last passenger sits in seat $n$ with probability $\frac{1}{2}$.

Base case $n = 2$: You sit randomly in seat 1 or seat 2, each with probability $\frac{1}{2}$. If you sit in seat 1 (your own), passenger 2 gets seat 2. If you sit in seat 2, passenger 2 is displaced and gets seat 1. So $P(\text{last in own seat}) = \frac{1}{2}$. ✓

Inductive step: Suppose the result holds for $n-1$ passengers. With $n$ passengers:

• With probability $\frac{1}{n}$, you sit in seat 1 (your own) → everyone boards correctly → passenger $n$ gets seat $n$. ✓
• With probability $\frac{1}{n}$, you sit in seat $n$ → passenger $n$ is immediately displaced → passenger $n$ cannot sit in seat $n$. ✗
• With probability $\frac{k}{n}$ (for seats $2 \leq k \leq n-1$, i.e., probability $\frac{n-2}{n}$), you sit in seat $k$ for some $2 \leq k \leq n-1$.

In the last case, passengers $2, \ldots, k-1$ sit normally. Passenger $k$ is displaced and now faces the same problem with $n - k + 1$ remaining seats (including seat 1 and seat $n$). By the symmetry of the random choice, this sub-problem has exactly the same structure as the original with fewer passengers.

By summing over all $k$:

$P_n = \frac{1}{n} \cdot 1 + \frac{1}{n} \cdot 0 + \sum_{k=2}^{n-1} \frac{1}{n} \cdot P_{n-k+1}$

where $P_j$ is the probability for $j$ passengers. Using $P_2 = \frac{1}{2}$ and assuming $P_j = \frac{1}{2}$ for all $j < n$:

$P_n = \frac{1}{n} + 0 + \sum_{k=2}^{n-1} \frac{1}{n} \cdot \frac{1}{2} = \frac{1}{n} + \frac{n-2}{n} \cdot \frac{1}{2} = \frac{1}{n} + \frac{n-2}{2n} = \frac{2 + n - 2}{2n} = \frac{n}{2n} = \frac{1}{2}.$

The cleanest way to see it: The process terminates when either seat 1 or seat $n$ is chosen by a displaced passenger. At that moment, the choice is uniform between these two seats (both are always free until one is taken). So:

$P(\text{seat } n \text{ chosen first}) = \frac{1}{2}.$

Answer: $\boxed{\dfrac{1}{2}}$, for all $n \geq 2$.