# Algebraic Method to solve a game-examples

Algebraic Method to solve a game-examples

## Algebraic Method to solve a game

Algebraic method is used to solve the game under following situations:

• if the game is $2\times 2$ game and the game has no saddle point, or
• if the game is $m\times n$ game and the game has no saddle point and it can be reduced to $2\times 2$ game after applying dominance property.

## Algebraic method to solve game Example 1

Solve the following $2\times 2$ game. The payoff matrix is given for player A as follows:

PlayerA \ Player B $B_1$ $B_2$
$A_1$ 4 1
$A_2$ 2 3

#### Solution

We have two players A and B. Player A has two strategies and player B also has two strategies.

B1 B2
A1 4 1
A2 2 3

Since the game is $2 \times 2$, let us apply Maximin-Minimax principle to check if there exists a saddle point.

#### Step 1

Select the minimum element of each row of the payoff matrix,

 $$\begin{equation*} \text{i.e., } \min_{j} \{a_{ij}\}, i=1,2,\cdots, m. \end{equation*}$$

B1 B2 RowMin
A1 4 1 1
A2 2 3 2

#### Step 2

Select the maximum element of each column of the payoff matrix,

 $$\begin{equation*} \text{i.e., } \max_{i} \{a_{ij}\}, j=1,2,\cdots, n. \end{equation*}$$

B1 B2 RowMin
A1 4 1 1
A2 2 3 2
ColMax 4 3

#### Step 3

Obtain the maximum value of each row minimum,
 $$\begin{equation*} \text{i.e., } \max_{i}\min_{j} \{a_{ij}\}=\underline{v}. \end{equation*}$$

Thus $Max(min) = Max(1, 2)=2$

#### Step 4

Obtain the minimum value of each column maximum,

 $$\begin{equation*} \text{i.e., } \min_{j}\max_{i} \{a_{ij}\}=\overline{v}. \end{equation*}$$

Thus $Min(max)=Min(4, 3)=3$.

#### Step 5

$Max(min) =2$ and $Min(max)=3$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.

Hence the optimal strategies for the given game can be obtained by the algebraic method.

 \begin{aligned} p_1 &= \frac{d-c}{(a+d)-(b+c)}\\ &=\frac{3-2}{(4+3)-(1+2)}\\ &=\frac{1}{4}\\ p_2 &= 1-p_1\\ &=\frac{3}{4} \end{aligned}

and the optimal strategies for player B can be determined by

 \begin{aligned} q_1 &= \frac{d-b}{(a+d)-(b+c)}\\ &=\frac{3-1}{(4+3)-(1+2)}\\ &=\frac{2}{4}=\frac{1}{2}\\ q_2 &= 1-q_1\\ &=\frac{1}{2}. \end{aligned}

The optimal strategies for player A can be written as

 \begin{aligned} S_{A} &= \begin{bmatrix} A_1 & A_2\\ p_1 & p_2 \end{bmatrix}\\ &=\begin{bmatrix} A_1 & A_2\\ \frac{1}{4} & \frac{3}{4} \end{bmatrix} \end{aligned}

and the optimal strategies for player B can be written as

 \begin{aligned} S_{B} &= \begin{bmatrix} B_1 & B_2\\ q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \end{aligned}

And the value of the game for player A is given by

 \begin{aligned} V &= \frac{ad-bc}{(a+d)-(b+c)}\\ &=\frac{(4)(3)-(1)(2)}{(4+3)-(2+1)}\\ &=\frac{10}{4}=\frac{5}{2}. \end{aligned}

Thus, we conclude that player $A$ may choose strategy $A_1$ with probability $\dfrac{1}{4}$ and strategy $A_2$ with probability $\dfrac{3}{4}$. Player $B$ may choose strategy $B_1$ with probability $\dfrac{1}{2}$ and strategy $B_2$ with probability $\dfrac{1}{2}$. And value of the game for player A is $\dfrac{5}{2}$ and for player B is $-\dfrac{5}{2}$.

## Algebraic method to solve game Example 2

Solve the following $3\times 3$ game. The payoff matrix for player A is as follows:

PlayerA \ Player B $B_1$ $B_2$ $B_3$
$A_1$ -5 10 20
$A_2$ 5 -10 -10
$A_3$ 5 -20 -20

#### Solution

We have two players A and B. Player A has three strategies and player B also has three strategies.

B1 B2 B3
A1 -5 10 20
A2 5 -10 -10
A3 5 -20 -20

Since the game is $3 \times 3$, let us apply Maximin-Minimax principle to check if there exists a saddle point.

#### Step 1

Select the minimum element of each row of the payoff matrix,

 $$\begin{equation*} \text{i.e., } \min_{j} \{a_{ij}\}, i=1,2,\cdots, m. \end{equation*}$$

B1 B2 B3 RowMin
A1 -5 10 20 -5
A2 5 -10 -10 -10
A3 5 -20 -20 -20

#### Step 2

Select the maximum element of each column of the payoff matrix,

 $$\begin{equation*} \text{i.e., } \max_{i} \{a_{ij}\}, j=1,2,\cdots, n. \end{equation*}$$

B1 B2 B3 RowMin
A1 -5 10 20 -5
A2 5 -10 -10 -10
A3 5 -20 -20 -20
ColMax 5 10 20

#### Step 3

Obtain the maximum value of each row minimum,
 $$\begin{equation*} \text{i.e., } \max_{i}\min_{j} \{a_{ij}\}=\underline{v}. \end{equation*}$$

Thus $Max(min) = Max(-5, -10, -20)=-5$

#### Step 4

Obtain the minimum value of each column maximum,

 $$\begin{equation*} \text{i.e., } \min_{j}\max_{i} \{a_{ij}\}=\overline{v}. \end{equation*}$$

Thus $Min(max)=Min(5, 10, 20)=5$.

#### Step 5

$Max(min) =-5$ and $Min(max)=5$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.

Thus we reduce the size of pay-off matrix using dominance property.

As every element of $3^{rd}$ column ($B_3$) $\geq$ every element of $2^{nd}$ column ($B_2$), the $3^{rd}$ strategy of player B is dominated. (i.e. strategy $B_3$ is inferior to strategy $B_2$). Hence deleting $B_3$, the reduced game is

Player A \ Player B $B_1$ $B_2$ Min
$A_1$ -5 10 -5
$A_2$ 5 -10 -10
$A_3$ 5 -20 -20
Max 5 10

Apply the Maximin-minimax principle on the reduced game. We have $Max(min) = max(-5,-10,-20)=-5$ and $Min(max)=min(5,10)=5$. Since the $Max(min)\neq min(max)$ for the reduced game, the game has no saddle point.

As every element of $3^{rd}$ row ($A_3$) $\leq$ every element of $2^{nd}$ row ($A_2$), the $3^{rd}$ strategy of player A is dominated. (i.e strategy $A_3$ is inferior to $A_2$). Hence deleting $A_3$, the reduced game is

Player A \ Player B $B_1$ $B_2$ Min
$A_1$ -5 10 -5
$A_2$ 5 -10 -10
Max 5 10

Apply the Maximin-minimax principle on the reduced game. We have $Max(min) = max(-5,-10)=-5$ and $Min(max)=min(5,10)=5$. Since the $Max(min)\neq min(max)$ for the reduced game, the game has no saddle point.

Hence the optimal strategies can be obtained by the algebraic method.

Comparing the reduced game with

Player A \ Player B $B_1$ $B_2$
$A_1$ $a$ $b$
$A_2$ $c$ $d$

The optimal strategy for player $A$ can be determined as

 \begin{aligned} p_1 &=\frac{d-c}{(a+d)-(b+c)}\\ &=\frac{(-10)-5}{(-5)+(-10)-(10+5)}\\ &=\frac{1}{2}\\ p_2 &= 1-p_1\\ &=\frac{1}{2} \end{aligned}

and the optimal strategies for player B can be determined by

 \begin{aligned} q_1 &= \frac{d-b}{(a+d)-(b+c)}\\ &=\frac{(-10)-10}{(-5)+(-10)-(10+5)}\\ &=\frac{2}{3}\\ q_2 &= 1-q_1\\ &=\frac{1}{3}. \end{aligned}

The optimal strategies for player A can be written as

 \begin{aligned} S_{A} &= \begin{bmatrix} A_1 & A_2\\ p_1 & p_2 \end{bmatrix}\\ &=\begin{bmatrix} A_1 & A_2\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \end{aligned}

and the optimal strategies for player B can be written as

 \begin{aligned} S_{B} &= \begin{bmatrix} B_1 & B_2\\ q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2\\ \frac{2}{3} & \frac{1}{3} \end{bmatrix} \end{aligned}

And the value of the game for player A is given by

 \begin{aligned} V &= \frac{ad-bc}{(a+d)-(b+c)}\\ &=\frac{(-5)(-10)-(10)(5)}{((-5)+(-10))-(10+5)}\\ &=0. \end{aligned}

Thus, we conclude that player $A$ may choose strategy $A_1$ and $A_2$ with probability $\dfrac{1}{2}$ each and will never choose $A_3$. Player $B$ may choose strategy $B_1$ with probability $\dfrac{2}{3}$ and strategy $B_2$ with probability $\dfrac{1}{3}$, while he never choose strategy $B_3$. And since value of the game is zero, the game will end in a draw.

## Endnote

In this tutorial, you learned about how to solve 2 by 2 or m by n game by algebraic method.