Two Bugs on an Icosahedron

Define the following states:

• 0: The bugs meet on a vertex
• 1: The bugs are on adjacent vertices
• 2: The bugs aren't on the same vertex, adjacent vertices, or opposite vertices
• 3: The bugs are on opposite vertices
• 4: The bugs meet on an edge

Note: Except for state 4, the number of the state represents the Hamming distance between the bugs.

And let,

$E_n =$ Expected number of moves to meet from state $n$ .

For each possible state where they don't meet ( $n=1,2,3$ ) we can set up equations like this:

$E_n = 1 + \sum_{m=0}^{4} P(n \rightarrow m)E_m$

And clearly, $E_0 = E_4 = 0$

The $1$ accounts for the move going from $n$ to $m$ . And the sum account for all the possible outcomes after the move.

This gives the following set of linear equations:

• $E_0 = 0$
• $E_1 = 1 + \frac{2}{25}E_0 + \frac{12}{25}E_1 + \frac{8}{25}E_2 + \frac{2}{25}E_3 + \frac{1}{25}E_4$
• $E_2 = 1 + \frac{2}{25}E_0 + \frac{8}{25}E_1 + \frac{13}{25}E_2 + \frac{2}{25}E_3$
• $E_3 = 1 + \frac{2}{5}E_1 + \frac{2}{5}E_2 + \frac{1}{5}E_3$
• $E_4 = 0$

Solving, $E_1 = \frac{550}{51}$

$550+51 = \boxed{601}$