Three desperados Three desperados A, B and C play Russian roulette in which they take turns pulling the trigger of a six-cylinder revolver loaded with one bullet. Each time the magazine is spun to randomly select a new cylinder to fire as long the deadly shot has not fallen. The desperados shoot according to the order A, B, C, A, B, C, . . .. Determine for each of the three desperados the probability that this desperado will be the one to shoot himself dead.
I have to calculate the probability that A dies at $i$th trial, so B and C never die. If at $1$th trial nobody dies we have $(\frac{5}{6})^3$. If at $i$th trial nobody dies we should have $(\frac{5}{6})^3+...+(\frac{5}{6})^3=\sum_{k=0}^{\infty}k(\frac{5}{6})^3$, because
$\mathbb{P}(A)=\mathbb{P}[(A\cap B_1)\cup...\cup(A\cap B_k)]=\mathbb{P}(A\cap B_1)+...+\mathbb{P}(A\cap B_k)=\sum_{k=0}^{\infty}\mathbb{P}(A\cap B_k)=\sum_{k=0}^{\infty}\mathbb{P}(B_k)\mathbb{P}(A|B_k)$
and since $A\perp B_k\Rightarrow \mathbb{P}(A)=\sum_{k=0}^{+\infty}\mathbb{P}(B_k)\mathbb{P}(A)=(\frac{5}{6})^2\frac{5}{6}+...+(\frac{5}{6})^2\frac{5}{6}=(\frac{5}{6})^3+...+(\frac{5}{6})^3=k(\frac{5}{6})^3$
where $A=[$A not die$]$. Thus A should die at $i+1$th trial with probability $\frac{1}{6}\sum_{k=0}^{\infty}k(\frac{5}{6})^3$.
Unfortunately the result is $\frac{1}{6}\sum_{k=0}^{\infty}(\frac{5}{6})^{3k}$. Why?
 A: Your notation is overly complicating the problem.  The probability that the gun first fires on the $i$-th pulling of the trigger is $(5/6)^{i- 1}\cdot(1/6)$... that is, the probability that the first $i-1$ trigger pulls don’t fire and then the $i$-th one does.  $A$ dies if this happens on the first, fourth, seventh, etc.... pulling of the trigger: if $i=3k+1$ for some natural $k$.  So
$$
P(A)=\frac{1}{6}\sum_{k=0}^{\infty}\left(\frac{5}{6}\right)^{3k}=\frac{36}{91}.
$$
For completeness, since $B$ and $C$ require $i=3k+2$ and $i=3k+3$ respectively,
$$
P(B)=\frac{5}{6}P(A)=\frac{30}{91}
$$
and
$$
P(C)=\frac{25}{36}P(A)=\frac{25}{91}.
$$
A: You can solve this problem without any fancy series summing. You know that:
$$ P(A) = \frac{1}{6} + \frac{5}{6} P(C)$$
One time out of six, $A$ will lose right away. In the other $\frac{5}{6}$ of cases, $A$ will be in the same situation $C$ is at the beginning of the game.
Next you have:
$$P(B) = \frac{5}{6} P(A)$$
If A doesn't lose at the beginning, $B$ will lose with probability $P(A)$.
Of course you also have:
$$P(A) + P(B) + P(C) = 1 \; ,$$
since someone has to lose. You have 3 equations and 3 unknowns.
