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: 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 $m+n=831.$
A: $A$ 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.
A: Let us compute some figures so that you can confirm answer !
Firstly, notice that when p = q = 1/2, the binomial distribution formula simplifies to P(X) = $\dfrac{n\choose X}{2^n}$
To simply computations, we can leave the division by $2^n$ till the end.
n(A wins) = $n[A = X]\cdot n[B\le X] =1\cdot1 + 5\cdot6 + 10\cdot16 + 10\cdot26 + 5\cdot31 + 1\cdot32$= 638
and P(A wins) = 638/1024 = 319/512
Finally, m+n = 831 
A: Hints:


*

*Team A has $5$ more games to play, and so does team B.

*The outcome of games is independent. Thus, for team A or B, the probability that the team wins exactly $k$ more games (after their game with each other) has binomial distribution, namely


$$P(\text{exactly $k$ more games won})={5 \choose k}\left(\frac 12\right)^5$$


*There are really four main ways that B can finish with more points than A: (a) A wins $0$ more games and B wins $2$ through $5$ more games; (b) A wins $1$ more game and B wins $3$ through $5$ more games; (c) A wins $2$ more games and B wins $4$ through $5$ more games; (d) A wins $3$ more games and B wins $5$ more games.

*The probability of (a) happening is 


$$P(\text{A wins $0$ more games}) \times [P(\text{B wins $2$ more games}) + P(\text{B wins $3$ more games}) + P(\text{B wins $4$ more games}) + P(\text{B wins $5$ more games})]$$


*Possibilities (a) through (d) are mutually exclusive, so you can just add their probabilities.


I'm sure you get the idea. The final calculation is made easier by the fact that $\left(\frac 12\right)^5$ factors out nicely. Calculating everything else has $6$ additions, $4$ multiplications, $3$ more additions, and $1$ more multiplication. This may be more calculating than you would like, but it does get the job done in a reasonable amount of time. Some calculators, like the TI-84 and TI-Nspire, can add ranges of binomial probabilities together all at once, which makes the calculations easier.
Of course, when you find the final probability, finding $m+n$ is easy.
