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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

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up vote 4 down vote accepted

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.

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Thank you very much. It is much clearer now. – Matthew Dec 29 '12 at 22:37

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.

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