Expected number of people sitting in the right seats. There was a popular interview question from a while back: there are $n$ people getting seated an airplane, and the first person comes in and sits at a random seat. Everyone else who comes in either sits in his seat, or if his seat has been taken, sits in a random unoccupied seat. What is the probability that the last person sits in his correct seat?
The answer to this question is $1/2$ because everyone looking to sit on a random seat has an equal probability of sitting in the first person's seat as the last person's. 
My question is: what is the expected number of people sitting in their correct seat?
My take: this would be $\sum_{i=1}^n p_i$ where $p_i$ is the probability that  person $i$ sits in the right seat.. 
$X_1 = 1/n$
$X_2 = 1 - 1/n$
$X_3 = 1 - (1/n + 1/n(n-1))$
$X_4 = 1 - (1/n + 2/n(n-1) + 1/n(n-1)(n-2))$
Is this correct? And does it generalize to $X_i$ having an $\max(0, i-1)$ term of $1/n(n-1)$, a $\max(0, i-2)$ term of $1/n(n-1)(n-2)$ etc?
Thanks.

Correct answer to incorrect question: please see second answer
The answer is $\frac{1}{2}$ as was said.
The general pattern is that for $n$ people, there is a $\frac{1}{n}$ probability of success, $\frac{1}{n}$ probability of failure, and an $\frac{n-2}{n}$ probability that the problem repeats itself on the $n-1$ scale.
Case $n=2$: The probability that the first picks the correct seat is $\frac{1}{2}$, and then the last person sits in proper seat with probability 1. The probability that the first picks the wrong seat is $\frac{1}{2}$, and then the last person sits in proper seat with probability 0. So the probability in total is:
$$
\frac{1}{2}\cdot1+\frac{1}{2}\cdot0=\frac{1}{2}
$$
Case $n=3$: The probability that the first picks the correct seat is $\frac{1}{3}$, and then the last person sits in proper seat with probability 1. The probability that the first picks the last person's seat is $\frac{1}{3}$, and then the last person sits in proper seat with probability 0. The probability that the first picks the middle person's seat is $\frac{1}{3}$, and now there is a $\frac{1}{2}$ that the second person picks the last seat or the first seat (Case 2) since two non-last people switching seats means everyone else takes their own seat. So:
$$
\frac{1}{3}\cdot1+\frac{1}{3}\cdot0+\frac{1}{3}\cdot\frac{1}{2} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
$$
Proof by induction.
Assume it holds true for $n$ that the probability of the last person sitting in the proper seat is $\frac{1}{2}$. Now if there are $n+1$ people, we have a $\frac{1}{n+1}$ chance that the last-seat probability is 1 (correct seat), a $\frac{1}{n+1}$ chance that the last-seat probability is 0 (last seat), and an $\frac{n-1}{n+1}$ chance that the probability is $\frac{1}{2}$, since we know the $n$ case.
$$
\frac{1}{n+1} + \frac{0}{n+1} + \frac{n-1}{2(n+1)}\\
=\frac{2}{2(n+1)}+\frac{n-1}{2(n+1)}\\
=\frac{n+1}{2(n+1)}\\
=\frac{1}{2}
$$
QED
