TITLE: Random walk on a cube
QUESTION [12 upvotes]: Start a random walk on a vertex of a cube, with equal probability going along the three edges that you can see (to another vertex). what is the expected number of steps to reach the opposite vertex that you start with?

REPLY [20 votes]: $\hskip1.75in$ 
A random  walk on the cube has  eight states, but by symmetry
we can reduce this to four states: start, nearest neighbors, further neighbors, and opposite.
The "boundary" is the single state {o}.

Define $h$ as the expected time to hit the boundary  starting
 at $x$, i.e., $h(x)=\mathbb{E}_x(T).$
First step analysis gives 
\begin{eqnarray*}
h(s)&=&1+h(n)\vphantom{1\over3}\\[3pt]
h(n)&=&1+{1\over3}\, h(s)+{2\over 3}\, h(f)\\[3pt]
h(f)&=&1+{1\over3}\, 0+{2\over 3}\, h(n).
\end{eqnarray*}
You can  work out that $h(s)=10$.