Nikhil the Ant

Let us analyze this two-state Markov chain. Our main goal is to find a closed formula for the probability that Nikhil is at vertex A after n minutes.

Let $p_n=\frac{1}{4}+d_n$ be the probability that Nikhil finds itself at vertex A after n minutes, where $\frac{1}{4}$ is the equilibrium state and $d_n$ is the deviation from the equilibrium. The probability that Nikhil isn't at A after n minutes is $q_n=\frac{3}{4}-d_n$ so that $p_{n+1}=\frac{q_n}{3}=\frac{1}{4}-\frac{d_n}{3}$ . (The probability is 1/3 that the ant travels from a point other than A to A.) Note that $d_{n+1}=-\frac{d_n}{3}$ . Starting from $p_1=0=\frac{1}{4}-\frac{1}{4}$ , we find that $d_n=\frac{(-1)^n}{4*3^{n-1}}$ and $p_n=\frac{1}{4}+d_n=\frac{1}{4}\left(1+\frac{(-1)^n}{3^{n-1}}\right)$ In particular, we have $p_7=\boxed{\frac{182}{729}}$ .

$\begin{aligned} A_n & = \text{\# of ways Nikhil will be at vertex A after n minutes.} \\ B_n & = \text{\# of ways Nikhil will be at any other vertex after n minutes.} \\ A_n & = B_{n-1} \\ B_n & = 3A_{n-1} + 2B_{n-1} \\ A_{n+1} & = 2A_{n} + 3A_{n-1}\\ A_1 & = 0 \\ A_2 & = 3 \\ A_3 & = 6 \\ A_4 & = 21 \\ A_5 & = 60 \\ A_6 & = 183 \\ A_7 & = 546 \end{aligned}$

Thus the probability is $\frac{546}{3^7} = \frac{182}{729} \implies m = \boxed{182}.$