, 2019

Solution:

Note that we desire to compute the number of times James moves to the right before moving down to the line $y=0$. Note also that we can describe James's current state based on whether his $y$-coordinate is $0$ or $1$ and whether or not the other vertically adjacent point has been visited. Let $E(1, N)$ be the expected number of times James will go right before stopping if he starts at a point with $y$-coordinate $1$ and the other available point with the same $x$-coordinate has not been visited. Define $E(1, Y)$, $E(2, N)$, and $E(2, Y)$ similarly. Then we can construct equations relating the four variables:
$$E(1, N)=\frac{1}{3} E(2, Y)+\frac{1}{3}(E(1, N)+1)$$
as James can either go up, right, or down with probability $1/3$ each if he starts in the state $(1, N)$. Similarly, we have
$$E(2, N)=\frac{1}{2} E(1, Y)+\frac{1}{2}(E(2, N)+1), \quad E(1, Y)=\frac{1}{2}(E(1, N)+1)$$
and $E(2, Y)=E(2, N)+1$. Solving these equations, we get $E(1, N)=\frac{7}{3}$, which is our answer, as James starts in that state having gone left $0$ times.