Markov chain on an $n$-dimensional hypercube? I am looking at the following Cambridge Tripos question, which has left me rather stumped for quite a while.

A particle moves on the $2^n$ vertices of the hypercube $\{0,1\}^n$. At each step the particle is equally likely to move to each of the $n$ adjacent vertices, independently of its past motion. The initial vertex occupied by the particle is $(0,0,...,0)$. Calculate the expected number of steps until the particle
i) first returns to $0,0,...,0)$;
ii) first visits $(0,0,...,1)$;
iii) first visits $(0,0,...,1,1)$.

I have seen multiple answers on here regarding the 3D case, as well as a general $D$-dimensional case presented here, but I am quite stuck on generalising to other vertices, including the three presented here. Would be lovely to see some guidance here, thanks!
 A: 2018 Part IB Paper 2 Q20H if anybody is looking.  Also asked in 1991.  The introductory part to the question was

For a finite irreducible Markov chain, what is the relationship
between the invariant probability distribution and the mean recurrence
times of states?

which can be applied here to give the answer for (i) of $2^n$ given the symmetry of the hypercube
If we call the expectation when starting zero steps away from the end point $E_0=2^n$, when starting one step away $E_1$, and starting two steps away $E_2$ then:

*

*for (ii) we can say $E_0=1+\frac{n}{n}E_1$ as you start by stepping one step from the start, making $E_1 = 2^{n}-1$


*similarly for (iii) we can say $E_1=1+\frac1n0 +\frac{n-1}{n}E_2$ as you start by stepping one step either to the end point or (more likely) to somewhere two steps away  $E_2=\frac{n}{n-1}(2^n-2)$
The pattern continues: in general for $1 \lt k \lt n$ you have $E_k=1+\frac kn E_{k-1} +\frac{n-1}{n} E_{k+1}$ so can find $E_{k+1}$ from $E_{k}$ and $E_{k-1}$.  And finally for the diametrically opposite corner you will have $E_n=1+E_{n-1}$.
A: For (i), note that the Markov chain $(X_n)$ is clearly irreducible, so since the state space is finite, it is positive recurrent. So there is a stationary distribution, which by symmetry is just $\pi(v)=1/2^n$ for all vertices $v$. So the expected first return time is the inverse of this, i.e. $2^n$.
Now let $T_k$ be the mean time to hit the origin from a vertex of distance $k$ away, with $T_0=2^n$ as above.
For (ii), we want $T_1$. But clearly $T_0=1+T_1\implies T_1=2^n-1$.
For (iii), we want to find $T_2$. Just note that after two steps, either we are back at the origin (with probability $1/n$) , or we end up a distance $2$ away. So
$$T_0=2+\left(1-\frac{1}{n}\right)\cdot T_2\implies T_2=\frac{n\left(2^n-2\right)}{n-1}$$
A: Just a thought:
Let $w(x)$ denote the weight of $x\in\{0,1\}^n$, i.e., $w(x)=\sum x_j$. There is no one-step path between two $x$'s with the same weight; taking each step, the weight changes by $+1$ or $-1$. Take the weight as the state, so, there exist $n+1$ states associated with the weights. Let $p_{i}^+$ denote the probability of transition from state $i$ to state $i+1$, and $p_i^-$ denote the transition from state $i$ to $i-1$, then $p_i^-=\frac{i}{n}$, and $p_{i}^+=1-\frac{i}{n}$.
At each state, due to the symmetry of hyper-cube, members are equiprobable, i.e., the probability of visiting a specific member of $i^{th}$ state, given you are at the $i^{th}$ state, is$\dfrac{1}{{n\choose i}}$.
I suspect that it builds a simple Markov model to answer the question.
