# Three players A,B,C throw dice in turns

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

## 1 Answer

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.

• thanks very much for your solution,it's a very clear explanation to me ! Commented Jan 6, 2021 at 5:54
• It was my pleasure. Commented Jan 6, 2021 at 5:55