<s>[INST] Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game stops.
What is the probability that the game ends on an even turn when $A$ rolls first?
Now the book gives the answer as $\frac{4}{7}$, however, when try to calculate I end up with $\frac{2}{11}$.
Below is my work:
To calculate this probability, we decompose the event into two disjoint events, (a) the event where $A$ wins on an even roll, and (b) the event where $B$ wins on an even roll.
(a) Now, the probability $A$ wins can be calculated as follows
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{1}{3}\\
= \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{4}{81}\biggr)^k = \frac{2}{27}\cdot \frac{1}{1- \frac{4}{81}} = \frac{6}{77}.
\end{align*}
(b) Similarly we calculate the probability $B$ wins on an even roll as
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot \frac{2}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot\frac{2}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{4}{9}\\
= \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{4}{81}\biggr)^k = \frac{8}{81}\cdot \frac{1}{1- \frac{4}{81}} = \frac{8}{77}.
\end{align*}
Therefore, it follows that the probability of the game ending on an even number of rolls is
\begin{equation*}
\frac{6}{77} + \frac{8}{77} = \frac{2}{11}.
\end{equation*}
Am I missing something?
 [/INST] Working under the assumption that the intended interpretation of the question was merely asking the probability that $B$ wins (i.e. distinguishing between the term "rounds" as iterating whenever A has a turn and "turns" iterating whenever either A or B has a turn) two other approaches have already been written.  Here I will include yet another approach:
Consider the final round, that is a roll of $A$ followed by a roll of $B$, where we allow $B$ to roll even in the event that $A$ has already won despite the roll not influencing the final result of the game.
Ordinarily there are $6\times 6 = 36$ equally likely results for a round.  Here, we condition on the fact that it is the last round, implying that it was not the case that both players missed their respective targets.  This gives $6\times 6 - 4\times 2 = 28$ equally likely possible final rounds.
Of these, $4\times 4 = 16$ of them end with $A$ missing their target and $B$ hitting theirs.
The probability of $B$ winning the game is then: $$\dfrac{16}{28} = \dfrac{4}{7}$$
</s>