Random Variable Probability Russian Roulette I am very confused on this one. Any help in how to solve it and what probability rule to use would be appreciated.
Consider the situation in which a person played Russian roulette (one bullet, 6 chambers) until the bullet fires. Let X be the random variable that represents the number of shots until the game ends. Clearly, this number includes all shots, including the one in which bullet fires. 
a) What values can X take? 
b) Find the probability for each of the following: 
   X = 1, 3, 5, 7, 9
 A: There are two types of Russian Roulette. Type (i): We always randomize (twirl the chamber) between shots, and Type (ii):  We do not randomize. 
It is not clear what type the questioner has in mind, so we analyze each type.  
Type (i): This version  is exactly like tossing a fair die until we get, say, a $5$. It is a version of sampling with replacement.
The random variable $X$ can, in principle, take on any positive integer value. 
The probability that $X=1$ is $\frac{1}{6}$. 
The event $X=3$ occurs precisely if we survive the first two "games," and do not survive the third. The probability of this is $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$. 
Essentially the same argument shows that the probability that $X=5$ is $\left(\frac{5}{6}\right)^4\frac{1}{6}$. And you can quickly derive a general formula for $\Pr(X=n)$. 
Type (ii) If we always go to the next chamber, then the only possibilities for $X$ are $1,2,3,4,5$ and $6$. In particular, $\Pr(X=7)=\Pr(X=9)=0$. 
Again, $\Pr(X=1)=\frac{1}{6}$. For the event  $X=2$ to occur, we must survive the first round, but not the second. The probability of this is $\frac{5}{6}\cdot \frac{1}{5}$. 
For the event  $X=3$ to occur, we must survive first and second round, but not the third. This has probability $\frac{5}{6}\cdot\frac{4}{5}\cdot \frac{1}{4}$.  
If you calculate the numbers we have obtained so far, you will note they each simplify to $\frac{1}{6}$. If we think about it, it is clear that $\Pr(X=n)$ is $\frac{1}{6}$ for each of $n=1,2,3,4,5,6$.  For the bullet is equally likely to be in any of the six chambers. 
A: Just want to add some further calculations based on Nicolas' logic.
1) In the first case, chambers was shuffled, 
   $$
E(X)=\frac{1}{6}(1\times (\frac{5}{6})^0+2\times (\frac{5}{6})^1+ 3\times (\frac{5}{6})^2+4\times (\frac{5}{6})^3+5\times (\frac{5}{6})^4+…)
   $$
   the sum of the above infinite series is 6.
2) In the second case, without turning the chambers randomly, 
  $$
  E(X)=\frac{1}{6}(1+2+3+4+5+6)=\frac{7}{2}.
  $$
Because in the second case, the game won't last any longer than the sixth round.
