Expected number of people getting their own hat given that at least one of them gets his hat. Suppose there are $N$ people at a party. Their hats get mixed and when leaving they grab a hat at random.
Let $\displaystyle X_i=I(\text{$i$th person gets his own hat})$ and $X=\displaystyle\sum_{i=1}^N X_i$.
It follows, from the linearity of expectation that the expected number of people who get their own hat is $1$. 
I want to know how to compute $E(X\mid X\neq 0)$, without having to compute $P(X\neq 0)$. I guess I can't get an exact expression, but is it possible to get a non-trivial lower bound?
My method:
$$E(X\mid X\neq 0)= \sum_{i=1}^N P(X_i=1\mid X\neq 0).$$
To get a lower bound on $P(X_i=1\mid X\neq 0)$, I want to use the the fact that $P(X_i=1\mid X_j=1)=1/(n-1)$. 
Is it true that $P(X_i=1\mid X\neq 0)>P(X_i=1\mid X_j=1)$ for $i\neq j$ ?
 A: Let $E=1$ if $X>0$, $E=0$ elsewhere. We want $E(X \mid E=1)$
Then $1=E(X)= E[E[X\mid E]]=E(X \mid E=0)P(E=0)+E(X \mid E=1)P(E=1)$   
But $E(X \mid E=0)=0$, hence $E(X\mid E=1)=1/P(E=1)$
Further, calling $!n$ the subfactorial or derangements count:
$$P(E=1)=1 - \frac{!n}{n!}\approx 1-\frac{1}{e}$$
So $E(X \mid E=1) \approx \frac{e}{e-1} = 1.5819767...$
If you want strict bounds: $ n!/e-1/2 \le \,!n\le n!/e+1/2$, hence
$$ \frac{e}{e-1+\frac{e}{2 \, n!}} \le E(X \mid E=1) \le \frac{e}{e-1-\frac{e}{2 \, n!}} $$
A: Note This method finds $E(X|X\neq0)$ by computing $P(X\neq 0)$, which you didn't want to do. Sorry, got carried away.
We have that
$$
P(X_i=1|X\neq0)=\frac{P(X_i=1\cap X\neq0)}{P(X\neq0)}=\frac{P(X_i=1)}{P(X\neq0)}
$$
The last equality holds since $X_i=1$ implies that $X\neq 0$. Adding these up, we get
$$
E(X|X\neq0)=\sum_{i=1}^n \frac{P(X_i=1)}{P(X\neq0)}=\sum_{i=1}^n \frac{1/n}{P(X\neq0)}=\frac1{P(X\neq0)}
$$
All that remains is to compute $P(X\neq0)$. Let $E_i$ be the event $X_i=1$. Using the Principle of Inclusion/Exclusion, then $X\neq0$ is the union of the events $E_i$, so that
$$
P(X\neq 0)=\sum_{i}P(E_i)-\sum_{i<j}P(E_i\cap E_j)+\sum_{i<j<k}P(E_i\cap E_j\cap E_k)-\dots
$$
There are $\binom{n}{k}$ copies of each $P(E_{i_1}\cap\dots\cap E_{i_k})$ term, and these probabilities are each $\frac{(n-k)!}{n!}$ (if $k$ people get their hats, there are $(n-k)!$ ways to choose how the other hats are arranged), so this boils down to
$$
P(X\neq 0)=\sum_{k=1}^n \binom{n}{k}\frac{(n-k)!}{n!}(-1)^k=\sum_{k=1}^n\frac{(-1)^k}{k!}
$$
This is just the first $n$ terms of the taylor series for $e^{-1}$, so that $P(X\neq 0)\approx 1/e$, and
$$
E(X|X\neq0)=\frac1{1/e}=e\qquad(!!!)
$$
The conditional expectation is just the number $e$! If you restrict your attention to the cases where at least one person gets their hat back, on average 2.718 people will have their hats. Weird!
