When boarding the plane, a queue of p passengers forms, each with a ticket for one of $\mathrm{n}$ seats. The first in line is a crazy old lady. She rushes into the cabin and sits in a random seat (possibly even her own). Next, passengers take their seats in order, and if their seat is already occupied, they sit in a random available seat. What is the probability that the last passenger will sit in their own seat?
Let the last passenger sit in the wrong seat. Then at the moment when some passenger was taking the last passenger's seat, he could have taken the old lady's seat as well.

## Solution

Let's consider a seating arrangement where the last passenger sits in the wrong seat (we will call such an arrangement unsuccessful). Before the last passenger arrives, his seat is occupied by passenger A (A could be the crazy old lady). At the moment when passenger A arrives, he has a choice of which seat to take. In the considered arrangement, he takes the last passenger's seat. However, with the same probability, he could have taken the old lady's seat, and all subsequent passengers, including the last one, would have taken their own seats. (Of course, it is necessary to explain why, at the moment when passenger A arrives, the old lady's seat is still free. But this is indeed the case— it is not difficult to trace that as long as the old lady's seat is free, among all the passengers who have not yet entered, there is exactly one whose seat is already occupied. As soon as some passenger takes the old lady's seat, all the others will sit only in their own seats.) Thus, each unsuccessful seating arrangement corresponds to a successful one that can happen with the same probability. This means that in exactly half of the cases, the seating arrangement will be unsuccessful.

## Answer

## Problem