# How to solve a bimatrix game?

Given a bimatrix game of $$\left(\begin{matrix}(0,-1) & (0,0)\\(-90,-6)&(10, -10)\end{matrix}\right)$$ Source

How to find the nash equilibrium strategy for both players?

• I assume you mean mixed strategy here, since there is no pure strategy equilibrium. – Arthur Nov 6 '17 at 9:03
• @Arthur Yessssss – Siwei Nov 6 '17 at 9:06

## 2 Answers

I assume that the entries $(a,b)$ in the payoff-matrix are interpreted as: $a$ is the row player's payoff and $b$ is the column player's payoff. To find a mixed strategy Nash equilibrium you use the fact that for a mixed strategy to be optimal for a player, the player must be indifferent between the pure strategies over which he or she mixes.

Denote by $x$ the probability that the row player chooses the upper row. If the column player chooses left, he or she gets $-x-6(1-x)$, if he or she chooses right, the payoff is $-10(1-x)$. Hence, it must hold that $-6+5x=-10+10x$, which yields $x=4/5$.

Denote by $y$ the probability that the column player chooses left. An argument analogous to the one above yields the condition $0=-90y+10(1-y)$, from which we obtain $y=1/10$.

I'll follow the method described in the text

$\qquad$Thomas$\,-\,$Games, Theory and Applications (1984)

on pages 59-61.

Suppose players $1$ and $2$ use mixed strategies $(x,1-x)$ and $(y,1-y)$, respectively, where

• The probability that player $1$ chooses row $1$ is $x$.$\\[2pt]$
• The probability that player $1$ chooses row $2$ is $1-x$.$\\[8pt]$
• The probability that player $2$ chooses row $1$ is $y$.$\\[2pt]$
• The probability that player $2$ chooses row $2$ is $1-y$.

Then the value of the game for player $1$ is \begin{align*} v_1(x,y) &= xy(0)+x(1-y)(0)+(1-x)y(-90)+(1-x)(1-y)(10)\\[4pt] &= (100y-10)x + (10-100 y)\\[4pt] \end{align*} and the value of the sgame for player $2$ is \begin{align*} v_2(x,y) &= xy(-1)+x(1-y)(0)+(1-x)y(-6)+(1-x)(1-y)(-10)\\[4pt] &= (4-5x)y + (10x-10) \end{align*}

Suppose $(x,y)$ yields a Nash equlibrium.

From the given payoffs, it's clear that there is no pure strategy equilbrium, hence we must have $0 < x < 1$, and $0 < y < 1$.

Since $0 < x < 1$, the equation $$v_1= (100y-10)x + (10-100 y)$$ implies $100y-10 = 0$, else player $1$ can change $x$ slightly, and do better.

Since $0 < y < 1$, the equation $$v_2= (4-5x)y + (10x-10)\qquad\;$$ implies $4-5x= 0$, else player $2$ can change $y$ slightly, and do better.

It follows that the unique Nash equilbrium has $x = {\large{\frac{4}{5}}}$, and $y = {\large{\frac{1}{10}}}$.

The corresponding values of the game are $v_1 = 0$ for player $1$, and $v_2 = -2$ for player $2$.