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? 

You can actually get away with computing only one binomial coefficient; I'll leave out one detail so this qualifies as a hint.
As @GrahamKemp observes, both $X$ and $Y$ are distributed as $\mathcal{Bin}(5,1/2)$; we want the probability that $X\ge Y$, i.e. that $Z:=X-Y\ge 0$.  But $Z+5$ is distributed as $\mathcal{Bin}(10,1/2)$ (why?)  From this you can compute $p=\textrm{Pr}(X=Y)=\textrm{Pr}(Z=0)$.  Finally, by symmetry, we have $\textrm{Pr}(X<Y)=\textrm{Pr}(X>Y)$, so both of these are equal to $(1-p)/2$.  It follows that
$$\textrm{Pr}(X\ge Y)=(1+p)/2.$$
Incidentally, I got $m+n=831.$
