jmc

Kevin hops $1/3$ of the remaining distance with every hop. His first hop takes $1/3$ closer. For his second hop, he has $2/3$ left to travel, so he hops forward $(2/3)(1/3)$. For his third hop, he has $(2/3{)}^2$ left to travel, so he hops forward $(2/3{)}^2(1/3)$. In general, Kevin hops forward $(2/3{)}^{k-1}(1/3)$ on his $k$th hop. We want to find how far he has hopped after five hops. This is a finite geometric series with first term $1/3$, common ratio $2/3$, and five terms. Thus, Kevin has hopped $\frac{\frac{1}{3}\left(1-{\left(\frac{2}{3}\right)}^5\right)}{1-\frac{2}{3}}=\boxed{\frac{211}{243}}$.