I thought using quantum theory to achieve that goal, namely that the distinction between players is artificial and in fact each player is a superposition of players given in the data, since they normally have some properties in common.
This means that the matrix of payoffs with probabilities of strategies were :
$$\begin{array}[cccc] & A & & p_3& p_4\\& B & q_3&q_4\\p_1&q_1&(1,-1)&(-1,1)\\p_2&q_2&(-1,1)&(1,-1)\end{array}$$
The conditions are $$\sum_{i=1}^4 p_i=\sum_{i=1}^4 q_i=1$$ And $$\sum_{i=n}^{n+1}p_i+q_i=1,n=1,3$$
Those equations are dependent leaving only 3 left.
Let say player A corresponds to mixing coefficients p and B to q. The problem arises when computing gains. Is it meaningful to say player A is a superposition with probability p1+p2 of 1 and so :
$$G_A=(p_1+p_2)G_{1,A}+(p_3+p_4)G_{2,A}$$
With $$G_{1,A}=-G_{2,A}$$ ?
In fact it's a question of interpretation of the players 1/2 becoming A/B which are a superposition of the formers. It's like this could happen in some sense if the we shall consider the 2 players as a whole. But this makes no sense in practice.
A pure competition game is when players have exactly opposite interests. This means there can be only $2$ players as it is not possible to have exactly opposite interests if the number of players is greater than $2$.
So, this is either a zero-sum game or a constant-sum game with $2$ players.
If it is a constant sum game, you can have a win-win. Example: two competitors fighting for declaring their technology as a standard. This is a constant-sum game and the competitors both gain if they agree on a standard and their winnings is greater than any other opposing strategy they could have individually followed.
In a zero-sum game, one of them wins and the other loses or both of them lose.