Three players A,B, and C take turns to roll a dice; they do this in the order ABCABC.... Show that the probability that the first 6 to appear is thrown by A,the second 6 to appear is thrown by B, and the third 6 to appear is thrown by C is 46656/753571.
I'm new to this problem,the best clue I could find is a similar problem which calculating the probability A throws 6 first,B second and C third,but this problem is more complicate when A could throws 6 first and cann't throws multiple 6s until B did.I don't know how to tackle this case.Can you suggested me the way to do this problem.much appreciated
In order for A to roll the first $6$, he must either roll a $6$ on his first turn (probability $\frac16$) or all $3$ players must miss on their first turn, and A roll $6$ on his second turn (probability $\frac16\left(\frac56\right)^3$), or all $3$ players must miss on their first two turns, and A roll $6$ on his third turn (probability $\frac16\left(\frac56\right)^6$) and so on.
The probability that A rolls the first $6$ is $$\frac16\sum_{k=0}^\infty\left(\frac56\right)^{3k}=\frac{1/6}{1-(5/6)^3}=\frac{36}{91}$$
Now once A rolls the first $6$, the players continue rolling in the order B,C,A and we want B to roll the first $6$. Once again, the probability that this happens is $\frac{36}{91}$. Once B rolls his $6$ the players are rolling in the order C,A,B and the probability that C rolls the first $6$ is $\frac{36}{91}$.
The probability of the event is $$\left(\frac{36}{91}\right)^3$$ which agrees with the answer given.