I'm searching for a function $C:D\to[0,1]$ (where $D=[0,1]^2\setminus\{(0,0),(1,1)\}$) that can describe the outcome of a match between two players, $p_1$ and $p_2$ with skills of $s_1$ and $s_2$. the function would determine the chance p1 would win, and will be written as $C(s_1,s_2)$. therefore, it must have the following properties:
- $C(1,x)=1 \space \forall x\in[0,1)$ ,since the perfect payer always wins (such player doesn't exist, of course)
- $C(x,y)+C(y,x)=1 \space\forall (x,y)\in D$, because if you have a chance of $p$, the other person has a chance of $1-p$.
while those are the two essential rules, I've derived some rules that would make the function behave nicer.
- $C(x,C(x,y))=y\space\forall (x,y)\in (0,1)\times[0,1]$
- $C(1-x,1-y)=C(y,x)\space\forall (x,y)\in D$
- $C$ is continuous and differentiable.
simple rules can be derived from those rules but it lead me nowhere close to figuring out a numeric description of such function, so if you can help me find a function that satisfies all 5 rules, preferably simple, I would very much appriciate it. also note, that is assuming a player's skill can be described at a range from 0 to 1, which is very much not always the case. however, please ignore the logic of the calculation, and focus on finding an answer. thank you.
Here it is:
$$C(x,y) = \frac{x(1-y)}{x(1-y)+y(1-x)}=\frac{x-xy}{x+y-2xy}$$
It satisfies all requirements from $1$ to $5$.
The function comes thinking about probabilities proportional to the skill, which would lead to $\frac{x}{x+y}$ and then adding the perfect player requirement.