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
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.