, 2018

Solution:

Let "a 3" mean a move in which Crisp moves from $x$ to $x+3$, and "a 7" mean a move in which Crisp moves from $x$ to $x+7$. Note that Crisp stops precisely the first time his number of 3's and number of 7's differs by a multiple of $5$, and that he'll stop on a dime if they differ by $0$, and stop on a nickel if they differ by $5$. This fact will be used without justification.

We split into two cases:

a. Crisp begins with a 3. Rather than consider the integer Crisp is on, we'll count the difference, $n$, between his number of 3's and his number of 7's. Each 3 increases $n$ by $1$, and each 7 decreases $n$ by $1$. Currently, $n$ is $1$. The probability of stopping on a dime, then, is the probability $n$ reaches $0$ before $n$ reaches $5$, where $n$ starts at $1$. Let $a_{i}$ be the probability $n$ reaches $0$ first, given a current position of $i$, for $i=1,2,3,4$. We desire $a_{1}$. We have the system of linear equations
$$\begin{aligned} a_{1} & =\frac{2}{3} a_{2}+\frac{1}{3} \cdot 1 \\ a_{2} & =\frac{2}{3} a_{3}+\frac{1}{3} a_{1} \\ a_{3} & =\frac{2}{3} a_{4}+\frac{1}{3} a_{2} \\ a_{4} & =\frac{2}{3} \cdot 0+\frac{1}{3} a_{3} \end{aligned}$$
From which we determine that $a_{1}=\frac{15}{31}$.

b. Crisp begins with a 7. Now, let $m$ be the difference between his number of 7's and his number of 3's. Let $b_{i}$ denote his probability of stopping on a dime, given his current position of $m=i$. We desire $b_{1}$. We have the system of linear equations
$$\begin{aligned} b_{1} & =\frac{1}{3} b_{2}+\frac{2}{3} \cdot 1 \\ b_{2} & =\frac{1}{3} b_{3}+\frac{2}{3} b_{1} \\ b_{3} & =\frac{1}{3} b_{4}+\frac{2}{3} b_{2} \\ b_{4} & =\frac{1}{3} \cdot 0+\frac{2}{3} b_{3} \end{aligned}$$
From which we determine that $b_{1}=\frac{30}{31}$.

We conclude that the answer is $\frac{2}{3} a_{1}+\frac{1}{3} b_{1}=\frac{2}{3} \cdot \frac{15}{31}+\frac{1}{3} \cdot \frac{30}{31}=\frac{20}{31}$.