# Russian roulette, how many people left

I have a questian about a game similar to russian roulette. Suppose that we have n people in a room. Every round, everyone shoots a random person. So it can happen that everbody dies, or if everyone shoots the same person only two people die(the unlucky person and the person that hè shot). I want to know what the expected number of surviving persons is after one round. I have no clue how to approach this problem.

• "Russian roulette" normally involves shooting oneself, not shooting other people. – vadim123 Mar 11 '14 at 23:10

Let $X_i=1$ if $i$ survives, and $X_i=0$ if she does not.
Then the number of survivors is $X_1+\cdots+X_n$. By the linearity of expectation the expected number of survivors is $E(X_1)+\cdots +E(X_n)$.
The probability that $i$ survives is the probability nobody shoots at her. This is $\left(\frac{n-2}{n-1}\right)^{n-1}$, since every person other than herself must shoot at someone other than herself.
The expectation is therefore $n\left(\frac{n-2}{n-1}\right)^{n-1}$, approximately $\frac{n}{e}$ unless $n$ is small.
Hint: pick one person and figure the chance he survives. There are $n-1$ people who might shoot him. Then use the linearity of expectation to get the expected number surviving.