Roger is guaranteed to be able to take at least one step. Suppose he takes that step in a direction $u$. Let $e_{1}$ be the expectation of the number of additional steps Roger will be able to take after that first move. Notice that Roger is again guaranteed to be able to make a move, and that three types of steps are possible: (1) With probability $\frac{1}{4}$, Roger takes a step in the direction $-u$ and his path ends. (2) With probability $\frac{1}{4}$, Roger again takes a step in the direction $u$, after which he is expected to take another $e_{1}$ steps. (3) With probability $\frac{1}{2}$, Roger takes a step in a direction $w$ perpendicular to $u$, after which he is expected to take some other number $e_{2}$ of additional steps. If Roger makes a move of type (3), he is again guaranteed to be able to take a step. Here are the options: (1) With probability $\frac{1}{2}$, Roger takes a step in one of the directions $-u$ and $-w$ and his path ends. (2) With probability $\frac{1}{2}$, Roger takes a step in one of the directions $u$ and $w$, after which he is expected to take an additional $e_{2}$ steps. Using these rules, we can set up two simple linear equations to solve the problem. $$\begin{aligned} & e_{2}=\frac{1}{2} e_{2}+1 \Longrightarrow e_{2}=2 \\ & e_{1}=\frac{1}{2} e_{2}+\frac{1}{4} e_{1}+1=\frac{1}{4} e_{1}+2 \Longrightarrow e_{1}=\frac{8}{3} \end{aligned}$$ Since Roger takes one step before his expectation is $e_{1}$, the answer is $\frac{11}{3}$.
Roger the ant is traveling on a coordinate plane, starting at $(0,0)$. Every second, he moves from one lattice point to a different lattice point at distance 1, chosen with equal probability. He will continue to move until he reaches some point $P$ for which he could have reached $P$ more quickly had he taken a different route. For example, if he goes from $(0,0)$ to $(1,0)$ to $(1,1)$ to $(1,2)$ to $(0,2)$, he stops at $(0,2)$ because he could have gone from $(0,0)$ to $(0,1)$ to $(0,2)$ in only 2 seconds. The expected number of steps Roger takes before he stops can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$.