Calculating Probability in Russian Roulette Game | Homework Questions

[PsychonautQQ]

Homework Statement

1.In the ``game'' of Russian roulette, the player inserts a single cartridge into the drum of a revolver, leaving the other five chambers of the drum empty. The player then spins the drum, aims at his/her head, and pulls the trigger. 1.What is the probability of the player still being alive after playing the game $$N$$ times?
2.What is the probability of the player surviving $$N-1$$ turns in this game, and then being shot the $$Nth$$ time he/she pulls the trigger?
3.What is the mean number of times the player gets to pull the trigger?

The Attempt at a Solution

(I'm in the process of learning LaTex, but for now...)

1. The chances of the person still being with 5 chambers is

$$\sum_{n=zero}^4 \frac{1}{5-n}$$
series from n=0 to n=4 (1/(5-n)
right? This would equate to 1/(5-0) + 1/(5-1) + 1/(5-2) etc etc...

i'm a bit stumped at question #2 however.

 

[statdad]

First write out the chance that when the trigger is pulled the bullet is not under the hammer: this will also give you the chance the bullet is under the hammer.
Now for the first question: if the player is still alive after N trigger pulls, the bullet was not under the hammer any of those times (not the first AND not the second AND ... AND not the Nth): how would you find the probability of that joint event?

For the second: if the player is not shot until trigger pull N, that means the first N-1 tries did not have the bullet under the trigger AND the Nth trigger pull did: find the probability of that joint event.

As a start to see what happens, you might try working out each solution in the very special case N = 3, just to see how the numbers work out.