Random walk on a cube

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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 $h(s)=10$.

It is clear that the expectation is finite. It will be handy to have a cube to play with.

There are $4$ types of vertex: Type A, the vertex we start at; Type B, the ones at distance $1$ from the start; Type C, the ones at distance $2$ from the start; and finally Type D, the vertex opposite from the start.

Let $a$ be the expected number of steps from A to D (this is what we want). Let $b$ be the expected number of (additional) steps to get to D given that we are at a Type B vertex. And let $c$ be the expected number of additional steps given that we are at a Type C vertex.

If we are at A, then with probability $1$, we will be in one step at a Type B vertex, so $$a=b+1.$$

If we are at a tyoe B vertex, then with probability $\frac{1}{3}$ at the next stage we return to A, and with probability $\frac{2}{3}$ we go to a Type C vertex. It follows that $$b=\frac{1}{3}(a+1)+\frac{2}{3}(c+1).$$ Finally, if we are at a Type C vertex, with probability $\frac{2}{3}$ we next go to a Type B, and with probability $\frac{1}{3}$ we get to D. Thus $$c=\frac{2}{3}(b+1)+\frac{1}{3}.$$ We have three linear equations in $3$ unknowns. Solve for $a$.