The last passenger's seat

Idea. The whole process ends as soon as someone sits in seat $1$ (the first passenger's seat) or seat $100$ (the last passenger's seat). Every random choice treats those two seats the same way, so each is equally likely to be taken first.

Step 1: what decides the outcome. Call seat $1$ the first passenger's seat and seat $100$ the last passenger's seat.

• If someone sits in seat $1$ before seat $100$ is taken, every later passenger finds their own seat free, so the last passenger gets seat $100$.
• If someone sits in seat $100$ first, the last passenger cannot get their seat.

When passenger $100$ boards, exactly one seat is left, and it is seat $1$ or seat $100$: any other seat $k$ would have been taken by passenger $k$ at the latest.

Step 2: symmetry. Each random choice is uniform over the free seats. As long as neither seat $1$ nor seat $100$ has been taken, both are free, and a random choice picks seat $1$ exactly as often as seat $100$. Choices of other seats only postpone the decision. So the first of the two seats to be taken is equally likely to be either.

Step 3. The last passenger gets their own seat with probability $\frac{1}{2}$, whatever the number of passengers (at least $2$).

Check with 3 passengers. Passenger $1$ picks seat $1$ (prob. $\frac{1}{3}$, success), seat $3$ (prob. $\frac{1}{3}$, failure), or seat $2$ (prob. $\frac{1}{3}$), in which case passenger $2$ picks seat $1$ or $3$ with probability $\frac{1}{2}$ each. Total: $\frac{1}{3}+\frac{1}{3}\cdot \frac{1}{2}=\frac{1}{2}$.