Lone ant on a cube

The problem can be modelled as a Markov chain . Let be $S=\{1,2,3,4\}$ the state space, in which $4$ is the marked point, $3$ are the three points directly connected to the marked one, $1$ is the farthest point from the marked point and $2$ are the three points directly connected to $1$ . Let be $\textbf{P}=p_{ij}$ the transition matrix in which each element represent the probability of passing from state $i$ to $j$ for $i,j \in \{1,2,3,4\}$

$\displaystyle \textbf{P}=\begin{pmatrix} 0 & 1 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 \\ 0 & \frac{2}{3} & 0 & \frac{1}{3} \\ 0 & 0 & 0 & 1 \end{pmatrix}$

Notice that $p_{4j}$ is an absorbing state. From Markov process theory, $\textbf{P}$ is already written in the canonical form, where $\textbf{Q}$ is

$\displaystyle \textbf{Q}=\begin{pmatrix} 0 & 1 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} \\ 0 & \frac{2}{3} & 0 \end{pmatrix} \hspace{5pt} \Longrightarrow \hspace{5pt} \textbf{I}-\textbf{Q}=\begin{pmatrix} 1 & -1 & 0 \\ -\frac{1}{3} & 1 & -\frac{2}{3} \\ 0 & -\frac{2}{3} & 1 \end{pmatrix}$

Hence,

$\displaystyle \textbf{N}=(\textbf{I}-\textbf{Q})^{-1}=\begin{pmatrix} \frac{5}{2} & \frac{9}{2} & 3 \\ \frac{3}{2} & \frac{9}{2} & 3 \\ 1 & 3 & 3 \end{pmatrix}$ .

So, $\displaystyle E_{i}[4]=\sum_{j=1}^{4} n_{ij}$ is the expected value of reaching $4$ starting from $i$ . If we sum up, we'll get $E_1[4]=10$ , $E_2[4]=9$ , $E_3[4]=7$ . Since the starting state is given by the stochastic vector $\displaystyle \textbf{s}_0=\begin{pmatrix} \frac{1}{7} & \frac{3}{7} & \frac{3}{7} & 0 \end{pmatrix}$ , where $s_i$ is the probability of starting the process at $i$ , the expected value of reaching $4$ , $E[4]$ , is

$E[4]=s_1 \cdot E_1[4]+s_2 \cdot E_2[4]+s_3 \cdot E_3[4]+s_4 \cdot E_4[4]=\displaystyle \frac{58}{7}=\frac{a}{b}$

Eventually

$\displaystyle \frac{a+5}{b}=\boxed{9}$