<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] The answer = 1/2
The game has to end by either A winning or B winning
Let's say A wins. He is just as likely to roll a 1 or a 2 on the last roll. Therefore in a game that A wins, probability of an even roll ending the game is 1/2, as 1(odd) and 2(even) are equally likely.
Let's say B wins. He is just as likely to roll a 3/4/5/6 on the last roll. Therefore in a game that B wins, probability of an even roll ending the game is 1/2, as 4 and 6 are favourable outcomes.
P.S. I have assumed that "ending on an even roll" as written in the title means the die outputs an even number. I agree that while the body of the question seems to suggest an even turn, this seems like the correct interpretation to me.
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