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: The player has a $\dfrac56$ chance of survival per game.
a) To survive $N$ games is $\left(\dfrac56\right)^N$
b) $\left(\dfrac56\right)^{N-1}\dfrac16$
c) The answer to (b) is a geometric distribution, with $p$ being the chance of success, (in this case failure!), so the mean is $\dfrac{1}{\frac16}=6$.
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
$1({1\over6})+2({5\over6})({1\over6})+3({5\over6})^2({1\over6})+...$
$=(({1\over6})+({5\over6})({1\over6})+({5\over6})^2({1\over6})+...)+(({5\over6})({1\over6})+({5\over6})^2({1\over6})+...)+(({5\over6})^2({1\over6})+...)+...$
$={1\over6}({1\over1-{5\over6}})+{1\over6}({5\over6})({1\over1-{5\over6}})+{1\over6}({5\over6})^2({1\over1-{5\over6}})+...$
$={1\over6}({1\over1-{5\over6}})({1\over1-{5\over6}})$
$=6$
A: Another way to understand what needs to be done at letter (c):
On (b) we found out the chance of surviving $N$ turns is:
$p(N) = (\frac{5}{6})^{N-1}(\frac{1}{6})$
By definition, the expected value of a variable is:
$<N> = \sum_{N=1}^\infty p(N) * N$
Here we go from one (died on first shot) to infinity, because the game can go on forever.
So,
$<N> = \sum_{N=1}^\infty (\frac{5}{6})^{N-1}(\frac{1}{6}) N = \frac{1}{6} \cdot \sum_{N=1}^\infty (\frac{5}{6})^{N-1} N$
That's the series you needd to calculate; then just use any method (like this or any other suggested) and find that:
$<N> = \frac{1}{6} \cdot 36 = 6$
(Wolfram confirms it)
