Lost on an Octahedron

Let us consider two states:

In state #1, Alice and Bob are on opposite vertices of the octahedron, i.e. they are separated by a distance of 2 edge lengths. In state #2, Alice and Bob are on adjacent vertices of the octahedron i.e. they are separated by a distance of 1 edge length.

Let $T_1$ denote the expected amount of time for Alice and Bob to meet each other given that they are in state #1, and define $T_2$ similarly for state #2. Our goal is to find the value of $T_1,$ since Alice and Bob start out on opposite vertices of the octahedron.

We can find that, after Alice and Bob choose a vertex, it takes $\dfrac{1}{2}$ minutes for them to meet up if they meet up in the middle of an edge, $1$ minute if they meet up at a vertex, $T_1 + 1$ minutes if they end up in state #1, and $T_2 + 1$ minutes if they end up in state #2.

When Alice and Bob are in state #1, there is a $\dfrac{1}{4}$ probability that they meet up at a vertex; a $\dfrac{1}{4}$ probability that they will end up back in state #1; and a $\dfrac{1}{2}$ probability that they will end up in state #2. Thus,

$T_1 = \dfrac{1}{4}(1) + \dfrac{1}{4}(T_1 + 1) + \dfrac{1}{2}(T_2 + 1).$

When Alice and Bob are in state #2, there is a $\dfrac{1}{16}$ probability that they meet up in the middle of an edge; a $\dfrac{1}{8}$ probability they will meet up at a vertex; a $\dfrac{11}{16}$ probability that they will end up back in state #2; and a $\dfrac{1}{8}$ probability that they will end up in state #1. Thus,

$T_2 = \dfrac{1}{16} \left (\dfrac{1}{2} \right) + \dfrac{1}{8} (1) + \dfrac{11}{16}(T_2 + 1) + \dfrac{1}{8}(T_1 + 1).$

This gives us a system of equations for $T_1$ and $T_2.$ Solving for $T_1$ yields $T_1 = \dfrac{51}{11},$ so $p+q = \boxed{62}.$