Answer: 37

Small Hint:

Use states for $0,$ $1,$ $2,$ $3,$ and $4$ consecutive heads and one separate state for a single trailing tail

Big Hint:

Express every head-run state’s success probability in terms of the trailing-tail state

Solution:

Let $q_i$ be the success probability with $i$ consecutive heads and no trailing tail, and let $t$ be the probability after one tail. Then $$\begin{aligned}q_i&=\frac{q_{i+1}+t}{2}\quad(0\leq i<4),\\q_4&=\frac{1+t}{2},\\t&=\frac{q_1}{2}.\end{aligned}$$ Working backward gives $q_1=\frac{1+15t}{16}.$ Since $q_1=2t,$ we get $t=\frac{1}{17}$ and $q_1=\frac{2}{17}.$ Therefore $$p=q_0=\frac{q_1+t}{2}=\frac3{34},$$ so $m+n=3+34=37.$