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I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

$f(0,1)$ does not exist. All other points $(x,y) \in [0,1]\times [0,1]$ should be defined.

I do not believe any polynomial solution exists. I am wondering if a solution exists and if so how to find it. I believe that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where $a$, $b$, and $c$ are linear or quadratic functions of $x$ and $y$, and maybe quartic at worst.

The intent of this function is to make an iterative solver of the game binario based solely on probability. I am aware that this function would not be fully able to solve the game but I am still interested in its existence.

I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(1-x,1-y)=1-f(x,y)$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

$f(0,1)$ does not exist. All other points $(x,y) \in [0,1]\times [0,1]$ should be defined.

I do not believe any polynomial solution exists. I am wondering if a solution exists and if so how to find it. I believe that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where $a$, $b$, and $c$ are linear or quadratic functions of $x$ and $y$, and maybe quartic at worst.

The intent of this function is to make an iterative solver of the game binario based solely on probability. I am aware that this function would not be fully able to solve the game but I am still interested in its existence.

I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

I do not believe any polynomial solution exists. I am wondering if a solution exists and if so how to find it. I believe that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where $a$, $b$, and $c$ are linear or quadratic functions of $x$ and $y$, and maybe quartic at worst.

I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

$f(0,1)$ does not exist. All other points $(x,y) \in [0,1]\times [0,1]$ should be defined.

I do not believe any polynomial solution exists. I am wondering if a solution exists and if so how to find it. I believe that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where $a$, $b$, and $c$ are linear or quadratic functions of $x$ and $y$, and maybe quartic at worst.

The intent of this function is to make an iterative solver of the game binario based solely on probability. I am aware that this function would not be fully able to solve the game but I am still interested in its existence.

I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

I do not believe any polynomial solutions exist. I am wondering if a solution exists and if so how to find it. I belive that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where a, b, and c are linear or quadratic functions of x and y, and maybe quartic at worst.

I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

I do not believe any polynomial solution exists. I am wondering if a solution exists and if so how to find it. I believe that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where $a$, $b$, and $c$ are linear or quadratic functions of $x$ and $y$, and maybe quartic at worst.