Let $X_n$ be the simple random walk on $\Z_2^3$, $A=\{(1,1,1)\}$. Compute for all $x\in\Z_2^n$: $\E^xH_A$.
Lable the nodes:
$$
0=(0,0,0),
1=(1,0,0),
2=(0,1,0),
3=(0,0,1),
4=(1,0,1),
5=(0,1,1),
6=(1,1,0),
7=(1,1,1)~.
$$
P=matrix(QQ,[[0,1,1,1,0,0,0,0],[1,0,0,0,1,0,1,0],[1,0,0,0,0,1,1,0],[1,0,0,0,1,1,0,0],[0,1,0,1,0,0,0,1],[0,0,1,1,0,0,0,1],[0,1,1,0,0,0,0,1]])
P=(1/3)*P
I=matrix(QQ,[1,0,0,0,0,0,0,0],[0,1,0,0,0,0,0,0],[0,0,1,0,0,0,0,0],[0,0,0,1,0,0,0,0],[0,0,0,0,1,0,0,0],[0,0,0,0,0,1,0,0],[0,0,0,0,0,0,1,0])
L=I-P
a=vector(QQ,[1,1,1,1,1,1,1])
La=L.augment(a,subdivide=false)
eLa=La.echelon_form()
print(eLa.column(8))
This will give you the solution: $10,9,9,9,7,7,7$. In particular, the expected number of steps to go from $(0,0,0)$ to $(1,1,1)$ is $10$.