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.

Now that I have cleaned my glasses, I'll try again. Thank you, Lord Farin.
Some observations.
The answer must be greater than 1. The probability that the first person, regardless of the number of people, sits in the proper seat is $\frac{1}{n}$, and there are $n$ people, so that expectation is 1. Even if the first person sits in the wrong seat, there is non-zero probability of other people sitting in the correct seat, so the answer must be greater than or equal to 1, and I'm pretty sure equality only exists in the case $n=2$.
There is some form of recurrence going on here. Enumerating the possibilities, I get (if I haven't erred):
$$
E_2 = \frac{1}{2}\cdot 2 + \frac{1}{2}\cdot 0 = 1\\
E_3 = \frac{1}{3}\cdot 3 + \frac{2}{3}\left[\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0\right] = 1+\frac{1}{3}=\frac{4}{3}\\
E_4 = \frac{1}{4}\cdot 4 + \frac{3}{4}\left[\frac{1}{3}\cdot2 + \frac{2}{3}\left[\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0\right]\right] = 1+\frac{3}{4}\cdot\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{7}{4}\\
E_5 = \frac{1}{5}\cdot 5 + \frac{4}{5}\left[\frac{1}{4}\cdot3 + \frac{3}{4}\left[\frac{1}{3}\cdot2 + \frac{2}{3}\left[\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0\right]\right]\right] = 1+\frac{4}{5}\left(\frac{3}{4} + \frac{3}{4}\cdot\left(\frac{2}{3}+\frac{1}{3}\right)\right)=\frac{11}{5}\\
$$
Let $E_k$ be the expected number of people in the correct seats when the starting population is k. The relationship seems to be:
$$
E_k = 1 + \left(E_{k-1} - 1 + \frac{k-2}{k-1}\right)\frac{k-1}{k}
$$
The first 1 is the expected value of person 1 taking the correct seat. The next term is broken into two parts. If the first person did not take the correct seat, then the second person can "fix" the error by swapping and taking the previous person's seat, leaving the remaining $n-2$ people their proper seats. If the second person also takes the wrong seat, the problem restarts on the $n-1$ scale. In both the latter two cases, there is "one less" correct seat than the initial, since the previous term used up a seat with an incorrect choice. If the above supposition is correct, the expected value would just be the sum of the recurrence relation applied to $n$ or $\sum_{k=1}^n E_k$. 
Generating the first few terms using this relationship gives:
$$
1, 1, \frac{4}{3}, \frac{7}{4}, \frac{11}{5}, \frac{16}{6}, \frac{22}{7}, \frac{29}{8}, \frac{37}{9}, \frac{46}{10}, \ldots
$$
The numerator is a quadratic and the denominator is just $n$ so the expected value should be:
$$
E_n = \frac{n^2-n+2}{2n}
$$
Edit: The long-term proportion of people sitting in the proper seats intuitively would be $$
\lim_{n \to \infty} \frac{E_n}{n}\\
=\lim_{n \to \infty} \frac{n^2-n+2}{2n^2}\\
=\frac{1}{2}
$$
Which dovetails nicely with the long-term probability of the last person sitting in his or her proper seat.
