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Suzie flips a fair coin $6$ times. The probability that Suzie flips $3$ heads in a row but not $4$ heads in a row is given by $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. $m+n=$~\uline{~~~~~~~~~~}~.  $\textasciitilde$ 
Overseas Competition->Knowledge Point->Counting Modules
Consider the distribution of the three consecutive heads. We have four cases:  1110xx, 01110x, x01110, xx0111.  Since case 1 and case4, case 2 and case 3 are the same, we only need to think about case 1 and case 2.  The probability of case 1: $(\dfrac{1}{2})^{4}=\frac{1}{16}$.  The probability of case 2: $(\dfrac{1}{2})^{5}=\frac{1}{32}$.  The sum of the probabilities of 4 cases is $2\times(\frac{1}{16}+\frac{1}{32})=\frac{3}{16}$. 