Solution:
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 $\frac{11}{3}$.