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 $\frac{1}{13}$?

C wins when:
1) A loses, B loses, C wins
or
2) A loses, B loses, C loses, A loses, B loses, C wins
or
3) A loses, B loses, C loses, A loses, B loses, C loses, A loses, B loses, C wins
or
..........
There are infinitely many possibilities, what leads to infinite series
$$
\sum_{i\ge 0}\left(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}\right)^i\cdot\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{1}{6}
$$
where $\frac{3}{6}=P(A\ loses)$, $\frac{4}{6}=P(B\ loses)$, $\frac{5}{6}=P(C\ loses)$, $\frac{1}{6}=P(C\ wins)$.
The final answer is $\frac{18}{13}\cdot\frac{1}{18}=\frac{1}{13}$.
