At a party n people throw their name tags on a table. The name tags
are mixed up and then each people randomly and simultaneously selects one.
Find the expected number of people who select their own name tag.
Since the people choose simultaneously, I assumed that each has $\frac{(n-1)!}{n!}$ or $\frac{1}{n}$ chance of selecting their own, thus expected number of people who get their own name tag is $ n\cdot \frac{1}{n} = 1 $. Am I correct in this?
Best Answer
Yes, you're right. To make the reasoning a little more precise, for each person $1,\dots,n$ we could define a Bernoulli random variable $X_i$, i.e., a random variable which takes one of only two possible values 0 and 1. We set $X_i$ to equal $1$ if the $i$th person draws their own name tag and $0$ if they do not. As you argue, the probability of any particular person drawing their own name tag is $1/n$. That is, $P(X_i=1)=1/n$. It follows that the expected value of $X_i$ is \begin{align*} E(X_i)&=\sum_{x=0}^1 xP(X_i=x) \\ &= 0\cdot P(X_i=0)+1\cdot P(X_i=1) \\ &= 0+1/n \\ &= 1/n \end{align*} Now the total number of people who select their own name tag is $X_1+\cdots+X_n$, and we can use the linearity property of expected value to conclude that $$E(X_1+\cdots+X_n) = E(X_1)+\cdots+E(X_n) = \frac1n+\cdots+\frac1n = n\cdot\frac1n = 1$$