Ant walking cube This is a modification of the ant crossing to the opposite diagonal problem:
An ant traverse from a cube vertex and does so on the edge. It has equal probability to choose all the edge including the original one. What is the expected number of edge it will traverse before returning to the original vertex?
My intuition is: there are 8 vertices, so if the ant just walking non-stop randomly to infinity, it would spend 1/8 of the time hitting each vertex. So intuitively, my answer is the average edge it would walk between each visit to the original vertex is 8 edges. But I don't know how to set it up in mathematically rigorous way.
 A: This is the approach Ross Millikan gave in the comments, but with more detail.
Let the vertices of the cube be $\{0,1\}^3$, and let the ant start at $(0,0,0)$.
For each vertex $v \in \{0,1\}^3$ (aside from $(0,0,0)$), let $E(v)$ denote the expected number of edges the ant will traverse before hitting $(0,0,0)$ if it starts from $v$.
By symmetry, $E(1,0,0) = E(0,1,0) = E(0,0,1) = A$ and $E(1,1,0) = E(1,0,1) = E(0,1,1) = B$. Also, let $E(1,1,1) = C$.
If the ant is currently at $(1,0,0)$, then it will take one step to get to its next vertex, which will be $(0,0,0)$ or $(1,1,0)$ or $(1,0,1)$ with equal probability. If the ant moves to $(0,0,0)$, the ant is done. If the ant moves to $(1,1,0)$, then the expected value of the number of additional steps it needs to reach $(0,0,0)$ is $E(1,1,0)$ (by definition). Similarly, if the ant moves to $(1,0,1)$, then the expected value of the number of additional steps it needs to reach $(0,0,0)$ is $E(1,0,1)$. Hence $$E(1,0,0) = \dfrac{1}{3} \cdot 1 + \dfrac{1}{3} \cdot (E(1,1,0)+1) + \dfrac{1}{3} \cdot (E(1,0,1)+1)$$ $$A = \dfrac{2}{3}B+1$$
Using the same logic, you can work out that $$E(1,1,0) = \dfrac{1}{3} \cdot (E(1,0,0)+1)+\dfrac{1}{3} \cdot (E(0,1,0)+1) + \dfrac{1}{3}\cdot (E(1,1,1)+1)$$ $$B = \dfrac{2}{3}A+\dfrac{1}{3}C+1$$ and that $$E(1,1,1) = \dfrac{1}{3} \cdot (E(1,1,0)+1) + \dfrac{1}{3}\cdot(E(1,0,1)+1) + \dfrac{1}{3}\cdot(E(0,1,1)+1)$$ $$C = B+1$$
You can easily solve the system of equations for $A,B,C$. Finally, since the ant starts at $(0,0,0)$, its first step takes it to either $(1,0,0)$ or $(0,1,0)$ or $(0,0,1)$ with equal probability. So the answer to the question is $$\dfrac{1}{3}\cdot(E(1,0,0)+1)+\dfrac{1}{3}\cdot(E(0,1,0)+1)+\dfrac{1}{3}\cdot(E(0,0,1)+1) = A+1.$$
