, 2016

Solution:

Note that at any point in the game after the first roll, the probability that Allen wins depends only on the most recent roll, and not on any rolls before that one. So we may define $p$ as the probability that Allen wins at any point in the game, given that the last roll was a $1, 2$, or $3$, and $q$ as the probability that he wins given that the last roll was a $4, 5$, or $6$.

Suppose at some point, the last roll was $r_1 \in \{1,2,3\}$, and the next roll is $r_2 \in \{1,2,3,4,5,6\}$. By the definition of $p$, Allen wins with probability $p$. Furthermore, if $r_2 = r_1$, which happens with probability $\frac{1}{6}$, Allen wins. If $r_2 \in \{1,2,3\}$ but $r_2 \neq r_1$, which happens with probability $\frac{2}{6}$, neither Allen nor Brian wins, so they continue playing the game, now where the last roll was $r_2$. In this case, Allen wins with probability $p$. If $r_2 \in \{4,5,6\}$, which happens with probability $\frac{3}{6}$, neither Allen nor Brian wins, so they continue playing, now where the last roll was $r_2$. In this case, Allen wins with probability $q$. Hence, the probability that Allen wins in this case can be expressed as $\frac{1}{6} + \frac{2}{6} p + \frac{3}{6} q$, and thus
$$p = \frac{1}{6} + \frac{2}{6} p + \frac{3}{6} q$$
By a similar analysis for $q$, we find that
$$q = \frac{1}{6} \cdot 0 + \frac{2}{6} p + \frac{3}{6} q$$
Solving, we get $p = \frac{1}{2}$ and $q = \frac{1}{3}$. Allen wins with probability $p = \frac{1}{2}$ if the first roll is $1, 2$, or $3$, and he wins with probability $q = \frac{1}{3}$ if the first roll is $4, 5$, or $6$. We conclude that the overall probability that he wins the game is $\frac{1}{2} p + \frac{1}{2} q = \frac{5}{12}$.