Probability that team $A$ has more points than team $B$ 
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

I got that since, Team $B$ already has one loss, it doesnt matter how many games team $B$ wins.
We must find the probability that team $A$ wins the rest $5$ games. 
Since it says: "outcome of games is independent, I am confused." 
My first approach was:
$$P(\text{A Wins 5}) = \frac{1}{32} \implies m + n = 33$$
This was wrong.
Second approach.
Suppose $A$ has a match with Team $C$.
$$P(\text{Team A wins, Team C loses}) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}$$
But then overall: $\frac{1}{4^5} > 1000$ too big of an answer ($m + n < 1000$ requirement).
HINTS ONLY PLEASE!!
EDIT:
I did some casework and the work is very messy and I don't think I got the right answer anyway. I have to find:
$$\binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1.$$
Let $A = x$
$$P(B=1, x \ge 1) + P(B = 2, x\ge 2) + P(B=3, x\ge 3) + ... + P(B=5, x = 5)$$
$$P(B=1, x \ge 1) = \binom{5}{1}(0.5)^{5} \cdot \bigg(\binom{5}{1} (0.5)^5 + \binom{5}{2} (0.5)^5 + ... + (0.5)^5\binom{5}{1}  \bigg) = \frac{5}{1024} \cdot \bigg(31\bigg) = \frac{155}{1024} $$
$$P_2 = \frac{10}{1024} \bigg(\binom{5}{2} + ... + \binom{5}{5}\bigg) = \frac{260}{1024}$$
$$P_3 = \frac{10}{1024} \bigg(\binom{5}{3} + ... + \binom{5}{5} \bigg) = \frac{160}{1024}$$
$$P_4 = \frac{5}{1024} \bigg( \binom{5}{4} + \binom{5}{5}\bigg) = \frac{60}{1024}$$
$$P_5 = \frac{1}{1024} \bigg( 1\bigg) = \frac{1}{1024}$$
$$P(\text{Total}) = \frac{636}{1024} = \frac{318}{512} = \text{wrong}$$
What is wrong with this method? 

$A$ and $B$ have played one game against each pther, and $B$ lpst.
Bpth teams have each to play against the remaining five teams (independently).   Let $X$ be the count of those games $A$ wins (ie: the points), and let $Y$ be the count of games $B$ wins.   Since $A$ has at least one point from the game between the two teams, you wish to find: $\;\mathsf P(X+1>Y)\;$.
Independence means that how many and which teams $A$ wins against has no influence on the probabilities of $B$ winning against any team.
Hint: The random variables $X$, $Y$ have iid binomial distributions, parameters $p=0.5, n=5$.  $$X,Y\mathop{\sim}^{iid}\mathcal{Bin}(5, 0.5)$$ 
$$\begin{align}
m/n & = \mathsf P(X+1> Y) 
\\[1ex] & = \sum_{x=0}^5 \mathsf P(X=x \cap Y< x+1) 
\\[1ex]
 & = \sum_{x=0}^5 \mathsf P(X=x) \sum_{y=0}^{x}\mathsf P(Y=y)
\end{align}$$
The rest is left to you.
