There are 2 particles are positioned at the vertices of a cube. If at a given time $t$ the particles are in the same edge, then they remain in the same position up to time $t + 1$. Otherwise, one of them is chosen at random with equal probability, and at time $t + 1$ it is moved to one of the 3 neighboring vertices with equal probability.
If the particles start at opposite vertices of the same face, what is the expected value of the least time they are in the same edge?
It doesn't matter which particle is chosen to move since the situation is always symmetrical with respect to interchange of the particles. There are three states, corresponding to distances $1$, $2$ and $3$ of the particles. The initial state is $2$ and the final state is $1$. State $3$ always transitions to state $2$, and state $2$ transitions to state $1$ with probability $\frac23$ and to state $3$ with probability $\frac13$. Thus every pair of steps starting at $3$ can be seen as an attempt to reach state $1$ from state $3$ with success probability $\frac23$, so the expected time starting at state $3$ is $2\div\frac23=3$. Thus, since state $3$ is certain to transition to state $2$, the expected time starting at state $2$ is $3-1=2$.
