Go to the Right, Counter!

For integers $0 \leq n \leq 6,$ let $P_{n}$ be the probability that if the counter were on square $n,$ I win the game. (We let the green square be square $6$ and the red square to be square $0$ .) From the description given in the problem, if the counter were on square $n,$ it has a $\frac{n}{6}$ chance of moving left to square $n - 1$ and a $\frac{6 - n}{6}$ chance of moving right to square $n + 1.$ If the counter moves right, then the probability that I win would then become $P_{n+1},$ by our definition. Similarly, if the counter moves left, the probability that I win is $P_{n-1}.$ The combination of both of these probabilities gives $P_{n},$ so we get

$P_{n} = \left ( \frac{n}{6} \right ) P_{n-1} +\left ( \frac{6 - n}{6} \right ) P_{n + 1}.$

Noticing that $P_{0} = 0$ (I lose the game) and $P_{6} = 1$ (I win the game), we get the system of equations

$\begin{aligned} \frac{5}{6} P_{2} &= P_{1} \\ \frac{1}{3} P_{1} + \frac{2}{3} P_{3} &= P_{2} \\ \frac{1}{2} P_{2} + \frac{1}{2} P_{4} &= P_{3} \\ \frac{2}{3} P_{3} + \frac{1}{3} P_{5} &= P_{4} \\ \frac{5}{6} P_{4} + \frac{1}{6} & = P_{5}. \\ \end{aligned}$

Solving this for $P_{1}$ yields $P_{1} = \frac{5}{13},$ so our desired answer is $13 - 5 = \boxed{8}.$