Chance to Take One's Own Seat?

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 $\frac{1}{2}$.