Solution
= Solution
Let $h$ be the expected <hitting time> from any neighbour of the origin back to the origin. After the first step, the return-time calculation from part (i) gives
$$
2^n=1+h,
$$
so $h=2^n-1$. Translation by the target vertex is a <graph automorphism> of the hypercube that exchanges the origin and that adjacent vertex, so the reverse expected hitting time is the same. Hence
$$
\boxed{\mathbb E_0T_{(0,\ldots,0,1)}=2^n-1.}
$$