Group Russian Roulette - The Easy Part

1. The probability that a person will be shot at by one other person is $\displaystyle \frac{1}{n-1}.$ The probability that a person will not be shot at equals $\displaystyle \left(1-\frac{1}{n-1}\right)^{n-1}.$ This of course tends to $\displaystyle \frac{1}{e}$ as $n\to\infty.$

2. Let's consider the problem of exactly $m$ survivors, $0\le m\le n.$ Questions #2 and #3 correspond to $m=0$ and $m=1,$ respectively.

   For convenience, denote the probability of individual survival from part 1 as $\displaystyle p= \left(\frac{n-2}{n-1}\right)^{n-1}.$ Then the probability of $m$ survivors is given by the binomial distribution:

   $\displaystyle P(\text{there are }m\text{ survivors})={n\choose m}p^m(1-p)^{n-m}.$

   For $m=0,$ $P(\text{there are no survivors})=(1-p)^{n}.$

3. For $m=1,$ $P(\text{there is }1\text{ survivor})=np(1-p)^{n-1}.$

Answering the second question, the probability of no survivors is the number of derangements $!n$ - as each person chooses somebody else to shoot at and no person is shot at twice, times the probability that every person was shot at:

$\displaystyle P(\text{there are no survivors})=!n\left(\frac{1}{n-1}\right)^n=\frac{!n}{(n-1)^n}.$

The problem in the first part can be iterated: the survivors of the first shooting repeat the action on another signal; and the process may go on until either one person remains standing or there are no survivors at all. The problem is especially interesting in that - and this is the omitted hard part - the probability of no survivors has no limit but oscillates as the number $n$ tends to infinity. Here's a graph from THE ASYMPTOTICS OF GROUP RUSSIAN ROULETTE by TIM VAN DE BRUG, WOUTER KAGER, AND RONALD MEESTER:

This is a problem from P. Winkler's Mathematical Puzzles: A Connoisseur's Collection (A K Peters, 2004, p 33.) It was discussed at the math.stackexchange where there are additional references.