Expected value of time taken on random-walk on a cube

Question

$A$ and $B$ are opposite vertices of a cube. A spider sits on $A$ and takes steps randomly along any edge connected to the vertex it is on at that moment. This situation ends when the spider makes it to $B$. Compute the expected value of the number of moves the spider makes.

Answer 1

I'm not certain what the standard notation for this is, but here's my proprietary method for expected value questions: Call the expected value for time taken from the starting vertex $v_3$ (this is what we want), the expected value for vertices with Manhattan distance $2$ $v_2$, the expected value for vertices with Manhattan distance $1$ $v_1$, and the expected value for the ending vertex $v_0$. From $v_3$, there is a $\frac{3}{3}=1$ probability of ending up at $v_2$. I write this as $v_3=v_2+1$. From $v_2$, there is a $\frac{2}{3}$ probability of ending up at $v_1$ and a $\frac{1}{3}$ probability of ending up at $v_3$. I write this as $v_2=\frac{2}{3}v_1+\frac{1}{3}v_3+1$. From $v_1$, there is a $\frac{1}{3}$ probability of ending up at $v_0$ and a $\frac{2}{3}$ probability of ending up at $v_2$. I write this as $v_1=\frac{1}{3}v_0+\frac{2}{3}v_2+1$. But the expected time taken from $v_0=0$, so we can leave that out, getting $v_1=\frac{2}{3}v_2+1$. We now have a system of equations and want to solve for $v_3$:

$$v_3=v_2+1$$ $$v_2=\frac{2}{3}v_1+\frac{1}{3}v_3+1$$ $$v_1=\frac{2}{3}v_2+1$$

So $v_2=\frac{2}{3}v_1+\frac{1}{3}(v_2+1)+1$, or $\frac{2}{3}v_2=\frac{2}{3}v_1+\frac{4}{3}$, or $v_2=v_1+2$. But we also know that $v_1=\frac{2}{3}v_2+1$, so $v_2=(\frac{2}{3}v_2+1)+2=\frac{2}{3}v_2+3$, or $\frac{1}{3}v_2=3$, or $v_2=9$. Then $\boxed{v_3=9+1=10}$.

Honestly, this method is probably just a stripped-down version of a more formal way of writing down the same method of solving- I don't remember- but it's always worked for me in the past, and that's actually why I suspect it's equivalent to known methods.