2x2 Matrix Game Formula

Problem

Let the payoff matrix of a 2 x 2 game be characterized by the matrix

$\mathbf{A} = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$.

All entries are positive real numbers.

Part 1: Use the simplex method to show that the value of the matrix game is $v = \frac{\det{(A)}}{S}$, where $S = (a+d) - (b+c)$ and $\det{(A)} = ad - bc$.

Part 2: Determine the strategy of the row player and the column player.

Solution

Part 1

Construct the initial tableau.

Tableau 1

	x	y	s1	s2	Z
s1	$a$	$b$	$1$	$0$	$0$	$1$
s2	$c$	$d$	$0$	$1$	$0$	$1$
Z	$-1$	$-1$	$0$	$0$	$1$	$0$

Leaving variable is $x$.

Entering variable is $s_1$.

Divide row 1 by $a$: ${\frac {1}{a}}R_{1}$.

	x	y	s1	s2	Z
s1	$1$	$b/a$	$1 / a$	$0$	$0$	$1 / a$
s2	$c$	$d$	$0$	$1$	$0$	$1$
Z	$-1$	$-1$	$0$	$0$	$1$	$0$

Perform the following row operations to get tableau 2: $R_{3}+R_{1}$ and $R_{2}-cR_{1}$.

Tableau 2

	x	y	s1	s2	Z
x	$1$	$\frac{b}{a}$	$\frac{1}{a}$	$0$	$0$	$\frac{1}{a}$
s2	$0$	${\frac {ad-bc}{a}}$	${\frac {-c}{a}}$	$1$	$0$	$1-{\frac {c}{a}}$
Z	$0$	${\frac {b}{a}}-1$	$\frac{1}{a}$	$0$	$1$	$\frac{1}{a}$

Leaving variable is $y$.

Entering variable is $s_2$.

Divide row 1 by ${\frac {ad-bc}{a}}$: $\left({\frac {a}{ad-bc}}\right)R_{1}$.

	x	y	s1	s2	Z
x	$1$	$\frac{b}{a}$	$\frac{1}{a}$	$0$	$0$	$\frac{1}{a}$
s2	$0$	$1$	${\frac {-c}{ad-bc}}$	${\frac {a}{ad-bc}}$	$0$	${\frac {a-c}{ad-bc}}$
Z	$0$	${\frac {b}{a}}-1$	$\frac{1}{a}$	$0$	$1$	$\frac{1}{a}$

Perform the following row operations to get tableau 3: $R_{1}-{\frac {b}{a}}R_{2}$ and $R_{3}-\left({\frac {b}{a}}-1\right)R2$.

	x	y	s1	s2	Z
x	$1$	$0$	${\frac {ad}{ad-bc}}$	${\frac {-b}{ad-bc}}$	$0$	${\frac {d-b}{ad-bc}}$
y	$0$	$1$	${\frac {-c}{ad-bc}}$	${\frac {a}{ad-bc}}$	$0$	${\frac {a-c}{ad-bc}}$
Z	$0$	$0$	${\frac {d-c}{ad-bc}}$	${\frac {a-b}{ad-bc}}$	$1$	${\frac {(a+d)-(b+c)}{ad-bc}}$

The value of the game is the reciprocal of $Z$ (the entry in the lower right corner). Therefore,

$v={\frac {ad-bc}{(a+d)-(b+c)}}$

$v = \frac{\det{(A)}}{S}$.

Part 2

To obtain the strategy for the row player, take the entries of the bottom row under $s1$ and $s2$ and divide each by the value of $Z$.

$R={\begin{bmatrix}r_{1}&r_{2}\\\end{bmatrix}}$

where $r_{1}={\frac {d-c}{S}}$ and $r_{2}={\frac {a-b}{S}}$.

To obtain the strategy for the column player, take the entries of the right-most column on row $x$ and $y$ and divide each by the value of $Z$.

$C={\begin{bmatrix}c_{1}\\c_{2}\\\end{bmatrix}}$

where $c_{1}={\frac {d-b}{S}}$ and $c_{2}={\frac {a-c}{S}}$.