Q:

Mean number of times a player can play roulette

In the game of Russian roulette (not recommended by the author), one inserts a single cartridge into the drum of a revolver, leaving the other five chambers of the drum empty. One then spins the drum, aims at one's head, and pulls the trigger.
(a) What is the probability of being still alive after playing the game N times?
(b) What is the probability of surviving (N-1) turns in this game and then being shot the Nth time one pulls the trigger?
(c) What is the mean number of times a player gets the opportunity of pulling the trigger in this macabre game?

A:

(1) $({5\over6})^N$ you need to not get shoot for $N$ consecutive turns.
(2) $({5\over6})^{N-1}{1\over6}$
(3) The expected number is $6$ as the probability is $1\over6$.
As for a rigorous proof using the definition, the expected number is
$1({1\over6})+2({5\over6})({1\over6})+3({5\over6})^2({1\over6})+...$
$=(({1\over6})+({5\over6})({1\over6})+({5\over6})^2({1\over6})+...)+(({5\over6})({1\over6})+({5\over6})^2({1\over6})+...)+(({5\over6})^2({1\over6})+...)+...$
$={1\over6}({1\over1-{5\over6}})+{1\over6}({5\over6})({1\over1-{5\over6}})+{1\over6}({5\over6})^2({1\over1-{5\over6}})+...$
$={1\over6}({1\over1-{5\over6}})({1\over1-{5\over6}})$
$=6$

