In $n$ envelopes, numbered from $1$ to $n$, cards with the corresponding number are putted (the number on the card matches the number of the envelope). Cards is removed, shuffle (all permutations are equally likely) and put back into envelopes. Find the expected number of cards that will be in his old envelopes.
What I do: $\sum\limits_{i=1}^n \frac{i{{n}\choose{i}}}{n!} = \sum\limits_{i=1}^n \frac{i}{i!}= \sum\limits_{i=1}^n \frac{1}{i!}-1$
Let $X_i$ be the random variable such that $X_i = 1$ if the $i$th card lies in its original envelope, and $X_i = 0$ otherwise. Since the cards are uniformly distributed among the envelopes, $\Bbb E[X_i] = \Bbb P(X_i = 1) = \frac{1}{n}$.
Now, $X_1 + \dots + X_n$ is the total number of cards in their original envelope. By linearity of $\Bbb E$, it satisfies $$ \Bbb E[X_1 + \dots + X_n] = \Bbb E[X_1] + \dots + \Bbb E[X_n] = \frac{n}{n} = 1. $$
