Symmetric random walk on a cube

Question

A particle is on a symmetric random walk on the vertices of a cube that looks like this:

The particle starts in $A$ has to jump to bordering vertex after one second. Every bordering vertex has a $1/3$ probability of being jumped to.

What is the probability that the particle reaches $H$ before it reaches $B$ and how long does it take the particle on average to reach one of the two?

I am not quite sure where to start, can someone help me?

Answer 1

Let $P_X$ be the probability of reaching $H$ before $B$, assuming the current vertex is $X$, and let $p=P_A$.

Our goal is to find the value of $p$.

It's immediate that $P_B=0$ and $P_H=1$.

Also, by symmetry, we have $$ \left\lbrace \begin{align*} &P_C=P_F=p\\[4pt] &P_E=P_G=P_D=1-p\\[4pt] \end{align*} \right. $$ Then starting from $A$, considering the possibilities for the first move, we get $$ P_A = \Bigl({\small{\frac{1}{3}}}\Bigr)P_D + \Bigl({\small{\frac{1}{3}}}\Bigr)P_E + \Bigl({\small{\frac{1}{3}}}\Bigr)P_B $$ hence $$ p = \Bigl({\small{\frac{1}{3}}}\Bigr)(1-p) + \Bigl({\small{\frac{1}{3}}}\Bigr)(1-p) + \Bigl({\small{\frac{1}{3}}}\Bigr)(0) $$ which yields $p={\large{\frac{2}{5}}}$.

Next, let $e$ be the expected number of moves to reach one of $B,H$, assuming the current vertex is $A$.

By symmetry, the expected number of moves to reach one of $B.H$, assuming the current vertex is any vertex other than $A,B,H$, is also equal to $e$, hence we get the recursion $$ e = \Bigl({\small{\frac{1}{3}}}\Bigr)(1) + \Bigl({\small{\frac{2}{3}}}\Bigr)(1+e) $$ which yields $e=3$.

Answer 2

$B$ and $H$ partition the cube, so $\text{Pr(B first)}=\frac32\text{Pr(H first)}$, and $\text{Pr(B first)}+\text{Pr(H first)}=1$ so $\text{Pr(B first)}=\frac35$.

Estimated number of turns is:

$$\sum_{k=1}^\infty \frac{k}3\left(\frac23\right)^{k-1}=3$$