Assigned Seating

Any time a random seat must be chosen, either the seat of the first person or the seat of the last person can be chosen with equal probability. Eventually, one of these seats must be chosen. Whichever one of these two seats is chosen first will leave the other seat open for the last person. Cumulatively, each one of the two seats has an equal probability of being chosen. Therefore the probability that the seat of the first person is taken and the seat of the last person is left for the last person is 1/2.

Claim. The probability is $\frac{1}{2}$ (and hence the answer is $\boxed{3}$ ). We proceed by strong induction on the number of people and seats.

Proof. When there are $2$ people and $2$ seats, the second person will sit in their assigned seat if and only if the first person sits in their assigned seat. Thus, the probability is $\frac{1}{2}$ .

Assume the claim holds when there are $n$ people and $n$ seats for all $2\le n<k$ .

Suppose there are $k$ people and $k$ seats. If the first person sits in their assigned seat, then everyone else (including the last person) will sit in the correct seat. If the first person sits in the last person's seat, then the last person certainly won't sit in the correct seat.

If the first person sits in the $m$ -th person's seat for $1<m<k$ , then everyone else before the $m$ -th person will sit in the correct seat. Thus when it is the $m$ -th person's turn to board, we have reduced to the case where there are $k-m+1$ people and seats. By inductive hypothesis, the last person will sit in their assigned seat with probability $\frac{1}{2}$ .

Therefore, when there are $k$ people and seats, the probability that the last person will sit in their assigned seat is $\frac{1}{k}\cdot 1 + \frac{1}{k}\cdot 0 + \frac{k-2}{k}\cdot\frac{1}{2}=\frac{1}{2}.$ QED