$A$ and $B$ are two players. The probability of $A$ winning a particular game against $B$ is $1/3$ and the probability of $B$ winning the game is $2/3$. They play a series in which the rules are such that $A$ wins the series if he wins $2$ consecutive games, while $B$ wins the series if he wins $4$ consecutive games. They start the first game and play until one of them wins the series. Find the probability that $B$ wins the series.
The options that go with the problem are - $$\begin{align} &1)\dfrac{64}{243}\\ &2)\dfrac{16}{81}\\ &3)\dfrac{64}{129}\\ &4) \ \text{None of the above}\end{align}$$
This problem appeared in a test I gave today, but I was not able to solve it. What I did was right down the ways in which $B$ can win the series, for eg, $$WWWW, LWWWW,WLWWWW,WLWLWWWW,....$$ where $W$ means $B$ wins the game and $L$ means he looses the game. But I am not able to find a pattern. Please help me out.
Notice that we must end with $WWWW$. The possible "building blocks" before we arrive at $B$'s victory are $WWWL$, $WWL$ and $WL$. The individual probabilities of each "block" are $\frac{8}{81}, \frac{4}{27}, \frac{2}{9}$ respectively. They sum up to $\frac{38}{81}$.
To construct at sequence ending in $B$'s victory, we pick an arbitrary number of building blocks (notice the sequence can go on arbitrarily long) and attach $WWWW$ at the end. This then corresponds to an infinite geometric progression. The total probability for this part will be the sum to infinity $\displaystyle \frac{\frac{16}{81}}{1-\frac{38}{81}} = \frac{16}{43}$ (first term corresponding to just $WWWW$ by itself).
We then observe that, in addition to the sequences mentioned, we could also start the game with an $L$. Hence total probability = $\frac{16}{43} + \frac{1}{3} \cdot \frac{16}{43} = \frac{64}{129}$