jmc

The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$. We let this probability be $p$; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$. Of these three cases ($|A|>|B|,|A|<|B|,|A|=|B|$), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). There are $\left(\genfrac{}{}{0ex}{}{5}{k}\right)$ ways to $A$ to have $k$ victories, and $\left(\genfrac{}{}{0ex}{}{5}{k}\right)$ ways for $B$ to have $k$ victories. Summing for all values of $k$, $1-2p=\frac{1}{2^5\times 2^5}\left({\sum }_{k=0}^5{\left(\genfrac{}{}{0ex}{}{5}{k}\right)}^2\right)=\frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024}=\frac{126}{512}.$ Thus $p=\frac{1}{2}\left(1-\frac{126}{512}\right)=\frac{193}{512}$. The desired probability is the sum of the cases when $|A|\ge |B|$, so the answer is $\frac{126}{512}+\frac{193}{512}=\frac{319}{512}$, and $m+n=\boxed{831}$.