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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 $A\ne B$)

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If you know the exact win probability for each party you just have to give points proportional to the win probability to make a fair game. Its as simple as that (if I understood this question right). –  Listing Jul 10 '11 at 12:09
    
@Listing: Sorry that I didn't add this to the question (I will edit it now), but 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) –  j_z Jul 10 '11 at 12:16
    
the main question is: what do you meant allocating points in a fair manner? Do you mean that the expected point haul joining any single game should be the same? –  Willie Wong Jul 10 '11 at 13:08
    
@Willie: Basically, I intend for the points allocated to discourage players from exclusively selecting easier games, and reward players for winning harder games (based on win rates). I have mirrored this idea to correspond (wth equal weighting) to two relationships: (1) expected win rate before joining the game is inversely proportional to points earned, (2) win probability of alignment of player is inversely proportional to points earned. As for the validity of using these relationships to represent the initial idea, and if my methods correctly utilise the two relationships, I am unsure. –  j_z Jul 10 '11 at 13:18
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2 Answers 2

up vote 1 down vote accepted

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.

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I think this is the simplest approach. In other words, don't worry about the probability of a player being assigned to either one of the teams; just make sure that everyone on both teams gets the same expected number of points. However, this probably also needs to be normalized among the different games, so a player can't select games with a high expected number of points. –  joriki Jul 10 '11 at 21:40
    
@Ross, joriki: Just to clarify, for the formula $AB$, are we calculating the expected number of points on average for a random player joining the game (assuming that we are only assigning B and A points respectively), and why would this be a better indicator than expected win probability? Also, what would be a method to modify the score determination to punish joining easier games, and if such a score modification were implemented, wouldn't that directly influence your proposed formula for expected win points $\frac{aAB + bAB}{a + b}$? Sorry if I misunderstood. –  j_z Jul 11 '11 at 13:28
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Yes, $AB$ is the expected number of points to win. As joriki comments, if you have several games you need to scale the awards to keep the expected gain from each game the same. Maybe you decide a game should be worth $10$ on average. Then if game $1$ is $4:1$ for the first side, the first side gets $12.5$ for a win and the second gets $50$ for a win. If game $2$ is even, each side gets $20$ for a win. Then the players have no advantage picking any game. –  Ross Millikan Jul 11 '11 at 13:46
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$\frac{aA+bB}{a+b}$ 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.

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Although it's not entirely clear from the description, I took the question to imply that $a+b$ players join and all have the same probabilities $a/(a+b)$ and $b/(a+b)$, respectively, of being assigned to one of the sides. –  joriki Jul 10 '11 at 21:36
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