There are \(N\) seats on a plane. Suppose the first passengers is drunk and he takes a seat randomly. For each of the other passengers, if his/her seat is not taken by other people, then he/she sits on his/her own seat, otherwise he/she takes a seat randomly. What is the probability that the last passenger takes his/her own seat?

Let's use \(P_n\) to stand for the probability that the last passenger takes his/her own seat, given that there're \(n\) seats in total, i.e. \(N=n\). If the first passenger takes his own seat, then the last passenger will take his/her own seat. If the first passenger takes the last passenger's seat, then the last passenger cannot take his/her own seat (we know that he/she must take the first passenger's seat). If the first passenger takes the \(i^{th}, 1<i<n\) passenger's seat, then for the \(j^{th}\) passenger, \(1< j\lt i\), he/she will takes his/her own seat. Now the \(i^{th}\) passenger has to choose a seat randomly. Since we don't care about whether the \(i^{th}\) passenger takes his/her own seat or not, we can pretend that the first passenger's seat is the \(i^{th}\) passengers. From this perspective, the problem has been changed to a same problem with \(N=n-i+1\). So conditioning on the seat that the first passenger takes, we have the following recursive formula:

where \(P_1=1\).

We can use method of generating function to find the formula of general terms. However, if we notice from the recursive formula that \(P_2=\frac{1}{2}\), \(P_3=\frac{1}{2}\), \(P_4=\frac{1}{2}\) and so on. We can easily see that \(P_i=\frac{1}{2}\) for \(i\ge2\) is the unique solution for the general terms.\ So as long as there're more than 1 people, the probability that the last passenger takes his/her own seat is \(\frac{1}{2}\).