Scoring system for two-party game Suppose we have a game that operates with '$a~$' players for one party, and '$b~$' players on the opposing party. Each party has a long-run average win percentage for this game, with $A\%$ and $B\%~$(such that $(A + B) = 100~$) respectively for the first and second party. After $(a+b)$ players are willing to participate in the game, each will be assigned to a party (with equal chance of joining either party) 
Firstly, I was wondering if the expected long-run average win percentage for a random player joining this game, before his alignment has been assigned, can be calculated using the formula $\frac{aA + bB}{a + b}\% ~$, and if there are any improvements that can be made to this description.   
More particularly, this game is one of a system of games. The system aims to discourage players exclusively selecting games with a high $\frac{aA + bB}{a + b}\% ~$ percentage, or in other words, games in which a random player joining can more easily win (before his alignment has been assigned)
Furthermore, suppose the first party wins the game, and one were to allocate points to the party based on some scoring system. The maximum number of points allowed to be allocated to a specific party is 100 points. Would the following formula allocate the points in a fair manner, under the assumption that loss rate of party $\propto$ points deserved ? 

$\displaystyle \frac{(100 - A) + (100 - \frac{aA + bB}{a + b})}{2}~$ noting that $(100 -A) = B$ 

How could this model possibly be improved?  
Please comment if this description is too vague or unclear, and I apologize if this question is not suitable for this website.
Note: In the formula, I'm averaging functions of two values (long-run average win percentage before  and after alignment has been dictated) to determine the points deserved.
As an example, suppose a game has 9 players for the first party, and 1 player for the second. The long-run average win probabilities are 60% and 40% respectively for each party. By the first formula, the average win probability for a random player joining this game is $\frac{60\times9 + 40}{9 +1}\%=58\%$ The scoring system punishes joining games with a high win probability, hence, half the score will be determined by $(100 - 58) = 42~$,  as loss rate $\propto$ points deserved. The scoring system also rewards games won by a party with a high win probability less, hence if the first party wins, the second half of the score is determined by $(100 - 60) = 40~$, as loss rate of party $\propto$ points deserved. Therefore, the final score is the average, $\frac{40 + 42}{2} = 41~$.This is modeled by the above formulas.
Comment: Specifically, the variation in long-run average win percentages for the parties arises naturally as a result of specific methods, unique to each party, that can be exploited to win the game,  and thus, this may unintentionally favor a specific party (as reflected in win probabilities, if $A\ne B$)
 A: Your calculation of the probability of a win before assignment is correct.  One approach to make it fair is to make each game fair:  Assign $B$ points for the $a$ side winning and $A$ points for the $b$ side.  Then the expected win rate becomes $\frac {aAB+bAB}{a+b}=AB$ for each side.
A: $\frac{aA+bB}{a+b}$ is the long-term mean winning percentage of the $a+b$ members of the existing group of players; the expected winning percentage of a new player depends on how the assignment to parties is made. If it's done by coin toss, for instance, his expected winning percentage [EWP] is $\frac{A+B}{2}$. If it's weighted by the existing distribution, (i.e., if his probability of being assigned to the first party is $\frac{a}{a+b}$), his EWP is given by your formula. If it's reverse weighted by the existing distribution (i.e., if his probability of being assigned to the first party is $\frac{b}{a+b}$), his EWP is $\frac{bA+aB}{a+b}$. In your example with $a=9,b=1,A=60,B=40$ these three EWPs are respectively $50$%, $58$%, and $42$%.
I don't think that you solve the fairness problem until you decide how the assignment to sides is made, because a player's choice of which game to enter will depend not on $\frac{aA+bB}{a+b}$, but on $pA+(1-p)B$, where $p$ is the probability that he'll be assigned to the first party.
