<s>[INST] What is the probability that Cathy wins? 
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
  die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
  he wins, and Cathy wins if she rolls a 6. They continue playing until
  a player wins. What is the probability (as a fraction) that Cathy
  wins?

This was I question I got when competing at a Mathematics competition earlier this week. I thought the answer was $\frac{1}{18}$ but it turns out the actual answer is $\frac{1}{13}$... 
This is my working out of why I thought it was $\frac{1}{18}$:
For Cathy to win:


*

*Alice needs to get a 4, 5 or 6 and there is a $\frac{3}{6}$ chance of that happening when she rolls the die

*Bob needs to get a 1, 2, 3 or 6 and there is a $\frac{4}{6}$ probability of that happening when he rolls the die

*Cathy needs to get 6 and there is a $\frac{1}{6}$ probability of that happening when she rolls the die


Those three need to happen for Cathy to win so:
$$\frac{3}{6} \times \frac{4}{6} \times \frac{1}{6} =  \frac{1}{18}$$
However, the answer was $\frac{1}{13}$ so something must be wrong! Could someone please demonstrate why it was $\frac{1}{13}$?
 [/INST] What you are leaving out is the possibility that nobody wins the first round. There is a $\frac{1}{18}$ chance that Cathy wins AND does so in the first round. Her overall odds will be $\sum_{n=1}^{\infty}(\frac{1}{2})^n(\frac{2}{3})^n(\frac{5}{6})^{n-1}(\frac{1}{6})=\frac{6}{5}\frac{1}{6}\sum_{n=1}^{\infty}(\frac{5}{18})^n=\frac{1}{5}\frac{5}{13}=\frac{1}{13}$
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