Expected number of wrong seats on plane

Let $X_k$ be the number of incorrect seats occupied by passenger $k$. Obviously $X_k$ is either $0$ or $1$; by this answer we have $$P(X_k=1)=\frac1{n+2-k}\quad\hbox{for $k=2,3,\ldots,n$}\ ;$$ hence $$E(X_k)=0\cdot P(X_k{=}0)+1\cdot P(X_k{=}1)=\frac1{n+2-k}\ .$$ Similarly, $$E(X_1)=\frac{n-1}n=1-\frac1n\ .$$ The total number of passengers in the wrong seat is $T_n=X_1+\cdots+X_n$, and by linearity its expected value is $$E(X_1)+E(X_2)+E(X_3)+\cdots+E(X_n) =\Bigl(1-\frac1n\Bigr)+\frac1n+\frac1{n-1}+\cdots+\frac12\ ,$$ that is, $$E(T_n)=1+\frac12+\cdots+\frac1{n-1}\ ,$$ provided $n>1$.

Each configuration is associated to a cycle like $1\mapsto 5\mapsto 8\mapsto 72\mapsto 1$, meaning that the first person takes the fifth place, the fifth person takes the eighth place, $\ldots$, the $72$th persone takes the first place. If this occurs any other person besides $1,5,8,72$ takes his/her place. The probability that $1\mapsto 5\mapsto 8\mapsto 72\mapsto 1$ occurs is $\frac{1}{101-1}\cdot\frac{1}{101-5}\cdot\frac{1}{101-8}\cdot\frac{1}{101-72}$ and by this way there are four people in the wrong place.

The probability that $0$ people are in the wrong place is $\frac{1}{100}$.
The probability that just $1$ person is in the wrong place is zero.
The probability that $2$ people are in the wrong place (configurations $1\mapsto m\mapsto 1$) is given by $\frac{1}{100}$ times the sum of $\frac{1}{101-k}$ for $k$ that goes from $2$ to $100$, i.e. by the coefficient of $x^2$ in

$$ g(x)=\frac{x}{101-1}\left(1+\frac{x}{101-2}\right)\cdot\left(1+\frac{x}{101-3}\right)\cdot\ldots\cdot\left(1+\frac{x}{101-100}\right).$$ Similarly, the probability that $k\geq 2$ people are in the wrong place is given by the coefficient of $x^k$ in $g(x)$. In particular the average value of the number of people in the wrong place is given by $$ \sum_{k\geq 2} k\cdot [x^k]g(x) = \left.\frac{d}{dx}\left(g(x)-\frac{x}{100}\right)\right|_{x=1}=g'(1)-\frac{1}{100}=\frac{g'(1)}{g(1)}-\frac{1}{100}.$$ On the other hand $\log(g(x))=\log(x)-\log(100)+\log\left(1+\frac{x}{99}\right)+\ldots+\log\left(1+\frac{x}{1}\right)$, hence

$$ \frac{g'(x)}{g(x)}=\frac{1}{x}+\frac{1}{99+x}+\frac{1}{98+x}+\ldots+\frac{1}{1+x} $$ and $\frac{g'(1)}{g(1)}$ equals the $100$th harmonic number $H_{100}$.
It follows that the wanted average value is $H_{99}\approx 5.17738$.

The same argument works for any other number of seats $s\geq 3$ and it does not depend on the parity of the number of seats. By considering $g''(x)$ you may also compute the variance of the random variable $W$ giving the number of people in the wrong place:

$$\operatorname{Var}[W]=\mathbb{E}[W^2]-\mathbb{E}[W]^2 = -H_{s-1}^2+\sum_{n\geq 2}n^2\cdot [x^n]g(x) $$ equals $$ -H_{s-1}^2+H_{s-1}+g''(1) = H_{s-1}-H_{s-1}^2+H_s^2+\left.\frac{d}{dx}\left(\frac{g'(x)}{g(x)}\right)\right|_{x=1}$$ such that $$\operatorname{Var}[W]=H_{s-1}\left(1+\frac{2}{s}\right)-H_{s-1}^{(2)}. $$ For large values of $s$ the distribution of $W$ is very well approximated by a Poisson distribution.