Statistics (expectation)

Question

Each of $n\ge 2$ people puts his or her name on a slip of paper (no two have the same name). The slips of paper are shuffled in a hat, and then each person draws one (uniformly at random at each stage, without replacement). Find the average number of people who draw their own names.

Answer 1

Let $I_i$ be a random variable that designates if the $i$th person got his name, namely $$ I_i \sim Ber(1/n), \quad i=1,...,n . $$ To be sure, you can check that $$ P(I_i=1)=\frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{n-i}{n-(i-1)}\frac{1}{n-i} =\frac{1}{n}. $$ The total number of people that got their names are $X = \sum_{i=1}^n I_i $, hence, $$ EX=nEI_i= \frac{n}{n}=1. $$