<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] Thanks to the comment of @JMoravitz I realized my mistake. I was interpreting turns as the rolls $A$ AND $B$, as in $\{A_1,B_1\}, \{A_2,B_2\}, \dots$. In reality the question is merely asking what the probability of $B$ winning if $A$ rolls first.
The work is as follows:
We calculate the probability of $B$ winning. Denote the probability of $B$ winning on their $i$th roll as $S_i$. Now, the probabilities of $B$ winning on her first roll, second roll, third roll, etc., are as follows:
\begin{equation*}
 P(S_1) = \biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \quad P(S_2) = \biggr(\frac{2}{3}\biggr)\biggr(\frac{1}{3}\biggr)\biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \quad P(S_3) = \biggr(\biggr(\frac{2}{3}\biggr)\biggr(\frac{1}{3}\biggr)\biggr)^2\biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \dots
\end{equation*}
It then follows that in general that $\displaystyle P(S_i) = \biggr(\frac{2}{9}\biggr)^{i-1} \biggr(\frac{4}{9}\biggr).$ Thus, it follows that the probability of $B$ winning is calculated as
\begin{equation*}
 P(S) = P\biggr(\bigcup_{i=1}^\infty S_i\biggr) = \sum_{i=1}^\infty P(S_i) = \sum_{i=1}^\infty \biggr(\frac{2}{9}\biggr)^{i-1} \biggr(\frac{4}{9}\biggr) = \frac{4}{9} \sum_{i=1}^\infty \biggr(\frac{2}{9}\biggr)^{i-1} = \frac{4}{9} \cdot \frac{9}{7} = \frac{4}{7}.
\end{equation*}
</s>