Expected Length of Walk on Truncated Icosahedron

The symmetry of the Markov chain associated with the random walk enables you to calculate the expected return time fairly easily. The chain has $\ 32\ $ states $\ h_1,h_2,\dots,h_{20},p_1,p_2,\dots,p_{12}\ ,$ where $\ h_i\ $ are the hexagonal faces and $\ p_i\ $ the pentagonal faces. If $\ \pi\ $ is the stationary distribution of the chain, then by symmetry we must have $$ \pi_{h_i}=\pi_{h_j}=\mathfrak{h}\ , $$ say, for all $\ i,j\in\{1,2,\dots,20\}\ ,$ and $$ \pi_{p_i}=\pi_{p_j}=\mathfrak{p}\ , $$ say, for all $\ i,j\in\{1,2,\dots,12\}\ ,$ with $$ 20\mathfrak{h}+12\mathfrak{p}=1 $$ and $$ \mathfrak{h}=\frac{\mathfrak{h}}{2}+\frac{3\mathfrak{p}}{5}\ , $$ because the entry to a hexagon from each of its three adjacent hexagons occurs with probability $\ \frac{1}{6}\ $, and the entry to it from each of its three adjacent pentagons occurs with probability $\ \frac{1}{5}\ $. These solution to these equations is $\ \mathfrak{p}=\frac{1}{36}, \mathfrak{h}=\frac{1}{30}\ .$ It now follows from one of the fundamental theorems of ergodic Markov chains that the expected first passage time from a hexagon to itself is $\ \frac{1}{\mathfrak{h}}=30\ .$