Frogs jump on dodecahedron vertex's

We can make use of the dodecahedron's symmetries to help with this problem. All that matters is how far away the frog's starting vertex is from vertex $1$ . In the diagram below (a dodecahedral graph; vertices and edges on the graph correspond to vertices and edges on the dodecahedron), I've coloured vertex $1$ white; vertices one edge away from vertex $1$ are green, $2$ edges away blue, and so on:

If a frog starts on a red vertex, and makes one hop, then (with equal probability) it will end up on either a new red vertex, a blue one, or a yellow one.

Using $R$ for the expected number of steps to reach vertex $1$ from a red vertex, $B$ for the expected number of steps to reach vertex $1$ from a blue vertex, and so on, this can be written as $R=1+\frac13 (R+B+Y)$

Continuing this way, we find $\begin{aligned} R &=1+\frac13 (R+B+Y) \\ B &=1+\frac13 (R+B+G) \\ G &=1+\frac23 B \\ Y &=1+\frac13 (2R+P) \\ P &=1+Y \\ W &=1+G \end{aligned}$

where the last of these represents the fact that the frog starting on vertex $1$ has to hop away first.

This is a system of six linear equations in six unknowns; solving, we find $(R,B,G,Y,P,W)=(32,27,19,34,35,20)$

Counting the vertices of each colour and totalling, the required answer comes to $\boxed{568}$ .