In a tournament, team X plays with each of the 6 other teams once. For each match the
probability of a win, draw and loss are equal. If the probability that the team X, finishes with more wins than losses is  $\frac{p}{q}$ where p and q are co-prime then the sum of digits of $|p-q|$ is.

The correct option is B 294
          Total number of results$=3^6=729$  
          Required number of ways
$=\frac{1}{2}(729-\left(\begin{array}{l}Numberofwaysinwhich\\ teamTfinishewithequalnumber\\ ofwinsandlosses\end{array}\right))$
Now, we shall consider following cases:
Case 1: 0 draw, 3 wins and 3 losses
WWWLLL
Number of ways$=\frac{6!}{3!3!}={}^{6}C_3=20$ 
Case II: 1 win, 1loss, 4draws: 
WWLLDD
Number of ways $=\frac{6!}{4!}=30$
Case III: 2 wins, 2losses, 2draws: WWLLDD
Number of ways $P\left(A\right)=\frac{999}{1000}\times \frac{998}{999}\times \mathrm{..........}\times \frac{1000-k+1}{1000-k+2}\times \frac{1}{1000-k+1}$
Case IC: no win and no loss, 6draws: DDDDDD
 Number of ways=1
Total ways with equal wins and draws$=30+30+90+4=141$ 
So, required number of ways $=\frac{1}{2}(729-141)=\frac{588}{2}=294$

$\frac{p}{q}=\frac{98}{293}$