# Game with Logistic Chance of Winning

by tbes   Last Updated November 03, 2017 01:19 AM

Two players (player $A$ and $B$) play a game. Let $a$ denote $A's$ effort and $b$ denote $B's$ effort. Suppose that the chance of winning in this game is a logistic function of effort (i.e. is a function of the difference between the amount of effort exerted):

$$p(win_a)=\frac{e^{a}}{(e^{a}+e^{b})}$$

Suppose the winner receives some prize $P$. Effort costs $c$ per unit for both players. A will maximize

$$argmax_a\bigg(p\frac{e^{a}}{(e^{a}+e^{b})}-ca\bigg)$$

Taking the derivatives yields

$$p\frac{e^{a+b}}{(e^{a}+e^{b})^2}-c$$

What is A's choice of effort?

My attempt: Define $X:=e^a$ and $Y:=e^b$, then we have the condition $c=p\frac{XY}{(X+Y)^2}$.

Then we get that $$X=\frac{pY-2cY+Y\sqrt{p(p-4c)}}{2c},\:X=\frac{pY-2cY-Y\sqrt{p(p-4c)}}{2c}$$

However, this seems unsatisfactory because i) $p>4c$ for the radical to make sense ii) we get two different equations for $a$, and I'm not sure which to choose iii) $a$ seems to be strictly increasing in $Y$, meaning that the more effort $B$ exerts, the more effort $A$ wants to exert.

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