I have a proposition, maybe you can find some weak points in it:
It's based on Elo. First, all the players get 1000 rating points ($R$). Let's focus on 1 vs 1, in the end I will show how to deal with 2 vs 1 and 2 vs 2. Player's A expected score $E_A$ is calculated as follows:
$$ E_A = \frac{1}{1+10^{(R_B - R_A)/500}} \in (0, 1)$$
Where $R_A$ is a point rating for player A. Player's B expected rating is $E_B = 1 - E_A$.
New rating for player A:
$$R_A' = R_A + 100(S_A - E_A)$$
Where $S_A$ is the actual score. In case of winning $10:i$ $(i \in {0, 1, ..., 9}$) it is1:
$$S_A = \frac{10}{10+i}$$
In case of losing $i:10$ it is:
$$S_A = 1-\frac{10}{10+i}$$
Example:
Player A with $R_A = 1000$ points won 10:6 with player B with $R_B = 1000$ points. In such case, expected score $E_A = E_B = 0.5$. Actual score $S_A$ for player A is $\frac{10}{10 + 6}=0.625$. So, $R_A' = 1000 + 100(0.625 - 0.5) = 1000 + 12.5 = 1012.5$, $R_B' = 1000 - 12.5 = 987.5$.
What to do with 2 players A and B in one team playing versus player C? We can treat player A and B as one player with rating $\frac{R_A + R_B}{2}$. So, $E_A = E_B = \frac{1}{1+10^{(R_C - \frac{R_A + R_B}{2})/500}}$. New rating will be calculated individually:
$$R_A' = R_A + 100(S_{AB} - E_A), R_B' = R_B + 100(S_{AB} - E_B)$$
$S_{AB}$ is the actual game score for player A and B. Since $E_A = E_B$, rating change for players in the same team is the same.
2 vs 2 case analogically.
1Why $S_A = \frac{10}{10+i}$?
Consider a winning $10 : i$. Let $p$ be the probability of the winning team scoring a goal. This will be our "actual score (performance)". Let's assume that this probability is constant for the whole game. After seeing the actual outcome, $10 : i$, we can ask: what's the most likely $p$? One can use maximum likelihood estimation. What's the probability that it was $10:i$, given that the probability of the winning team scoring a goal was $p$? It is:
$P(10:i | p) = \binom{9+i}{9}p^{10} (1-p)^i$
There were $10 + i$ goals in total. The last one belong to the winning team, it was their 10th goal, so this one is pinned. $\binom{9+i}{9}$ is the number of 9 winners goals and $i$ losers goals combinations. Multiplied by the probability of winners scoring 10 goals, $p^{10}$.
Multiplied by the probability of losers scoring $i$ goals, $(1-p)^i$. Their goals positions are already determined, so no combinations here.
Now we just need to find which $p$ maximizes $P(10:i | p)$. One could just take the first derivative and set it to zero. However, the maximum is the same for $\log P(10:i | p)$ and it will be easier to find maximum for it, for log-likelihood:
$\log P(10:i | p) = \log \binom{9+i}{9} + \log p^{10} + \log (1-p)^i = \log \binom{9+i}{9} + 10 \log p + i \log (1-p)$
Set first derivative of $P$ with respect to $p$ to 0:
$\log P(10:i | p)' = \frac{10}{p} + \frac{i}{p-1} = 0$
$\frac{10(p-1) + ip}{p(p-1)} = 0$
$10p-10 + ip = 0$
$p(10+i) = 10$
$p = \frac{10}{10+i}$
Best Answer
https://www.microsoft.com/en-us/research/project/trueskill-ranking-system/
The answer would be game points do not matter. The model is only based on wins and losses. Won matches over time would assist the model in getting a more accurate picture. See FAQ in the above link for team games.