# Graphical Method to solve 2 by n game

## Graphical method to solve $2\times n$ game

A game where one player has only two course of action and the other player has more than two (say $n$) course of action is called $2\times n$ game.

Graphical method can be used to solve a $2\times n$ or $m\times 2$ game or a game reducible to either $2\times n$ or $m\times 2$ after applying a dominance property.

In this tutorial we will discuss about the graphical method to solve a $2\times n$ game or a game reducible to $2\times n$ after applying dominance property.

## Step by Step Procedure to solve 2 by n game

Consider the $2\times n$ game. In this game player $A$ has $2$ strategies and player $B$ has $n$ strategies. Assume that the game does not have a saddle point.

Player A \ Player B $B_1$ $B_2$ $\cdots$ $B_n$
$A_1$ $a_{11}$ $a_{12}$ $\cdots$ $a_{1n}$
$A_2$ $a_{21}$ $a_{22}$ $\cdots$ $a_{2n}$

Following are the steps to solve $2\times n$ game using graphical method.

#### Step 1

Apply maximin-minimax principle to check whether saddle point exists or not. If saddle point exists, then stop the method and calculate the value of the game, otherwise go to next step.

#### Step 2

Construct two vertical axis, axis 1 and axis 2, using appropriate scales.

Axis 1 represent the payoff values of $A_2$ strategy for player $A$ and axis 2 represent payoff values of $A_1$ strategy for player $A$.

(Graph is just a sample.)

#### Step 3

Joint the point representing $a_{1j}$ on axis 2 to the point representing $a_{2j}$ on axis 1 for all $j = 1, 2,\cdots,n$. (Graph is just a sample.)

#### Step 4

Mark the lowest boundary (called Lower envelope) of the lines by thick line segment. The highest point on the lower envelope gives the maximin point $P$ and it identifies the two critical moves (strategies) of player $B$.

Suppose the lower envelope is created by $B_1$ and $B_n$ strategy. The highest point on the lower envelope gives the maximin point. This shows that the two critical moves for player $B$ are $B_1$ and $B_n$.

Using the above selection the reduced game becomes

Player A \ Player B $B_1$ $B_n$
$A_1$ $a_{11}$ $a_{1n}$
$A_2$ $a_{21}$ $a_{2n}$

#### Step 5

Solve the reduced game (i.e. $2\times 2$) using maximin criterion. (if saddle point exists then find the value of the game otherwise use mixed strategy technique to find the value of the game.)

## 2 by n game by graphical method Example 1

Solve the game with the following payoff for player A.

Player A \ Player B $B_1$ $B_2$ $B_3$ $B_4$
$A_1$ 1 3 -3 5
$A_2$ 2 5 4 -4

#### Solution

Given game is $2\times 4$. That is for Player $A$ has two strategies and player $B$ has four strategies. Let us check using maximin-minimax principle whether the game has a saddle point or not.

B1 B2 B3 B4
A1 1 3 -3 5
A2 2 5 4 -4

#### Step 1

Apply maximin-minimax principle to check whether saddle point exists or not.

B1 B2 B3 B4 RowMin
A1 1 3 -3 5 -3
A2 2 5 4 -4 -4
ColMax 2 5 4 5

Thus $Max(min) = Max(-3, -4)=-3$ and $Min(max)=Min(2, 5, 4, 5)=2$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.

#### Step 2

Construct two vertical axis, axis 1 and axis 2.

Axis 1 represent the payoff values of $A_2$ strategy for player $A$ and axis 2 represent payoff values of $A_1$ strategy for player $A$.

Let the player $A$ play the strategy $A_1$ with probability $p$ and strategy $A_2$ with probability $1-p$.

Expected payoff for player $A$ for any pure strategy of player $B$ would be given by

$B$'s Strategy Expected Payoff for $A$
$B_1$ $E_1= p+2(1-p) = -p+2$
$B_2$ $E_2= 3p+5(1-p)=-2p+5$
$B_3$ $E_3= -3p+4(1-p)=-7p+4$
$B_4$ $E_4= 5p-4(1-p)= 9p-4$

#### Step 3

Joint the point representing $a_{1j}$ on axis 2 to the point representing $a_{2j}$ on axis 1 for all $j = 1, 2,3$.

#### Step 4

Mark the lowest boundary (called Lower envelope) of the lines by thick line segment. The highest point on the lower envelope gives the maximin point $P$ and it identifies the two critical moves (strategies) of player $B$.

The two critical moves for player $B$ are $B_3$ and $B_4$.

The highest point $P$ on the lower envelope gives the maximin point.

#### Step 5

The reduced game (i.e. $2\times 2$) is

Player A \ Player B $B_3$ $B_4$
$A_1$ -3 5
$A_2$ 4 -4

Apply maximin criterion.

B3 B4
A1 -3 5
A2 4 -4
• Select the minimum element of each row of the payoff matrix,

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

B3 B4 RowMin
A1 -3 5 -3
A2 4 -4 -4
• For each column of the payoff matrix, select the maximum element and call it ColMax.

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

B3 B4 RowMin
A1 -3 5 -3
A2 4 -4 -4
ColMax 4 5
• From each RowMin, obtain the maximum value, i.e., $Max(RowMin)$.

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

Thus $Max(min) = Max(-3, -4)=-3$

• For each ColMax, obtain the minimum value, i.e. $Min(ColMax)$.

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

Thus $Min(max)=Min(4, 5)=4$.

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

Hence the optimal strategy for the reduced game can be obtained by algebraic method.

 \begin{aligned} p_1 &= \frac{d-c}{(a+d)-(b+c)}\\ &=\frac{(-4)-4}{(-3-4)-(5+4))}\\ &=\frac{-8}{-16}\\ &=\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{(-4)-5}{(-3-4)-(5+4)}\\ &=\frac{-9}{-16}\\ &=\frac{9}{16}\\ q_2 &= 1-q_1\\ &=\frac{7}{16}. \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 & B_3 & B_4\\ 0 & 0 & q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2 & B_3 & B_4\\ 0 & 0 & \frac{9}{16} & \frac{7}{16} \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{(-3)(-4)-(5)(4)}{(-3-4)-(5+4)} \\ &=\frac{-8}{-16}\\ &=\frac{1}{2}. \end{aligned}

Thus, we conclude that player $A$ may choose strategy $A_1$ with probability $\dfrac{1}{2}$ and strategy $A_2$ with probability $\dfrac{1}{2}$. Player $B$ may choose strategy $B_3$ with probability $\dfrac{9}{16}$ and strategy $B_4$ with probability $\dfrac{7}{16}$. And value of the game for player A is $\dfrac{1}{2}$ and for player B is $-\dfrac{1}{2}$.

## 2 by n game by graphical method Example 2

Solve the game with the following payoff for player A.

Player A \ Player B $B_1$ $B_2$ $B_3$
$A_1$ 1 3 10
$A_2$ 8 6 2

#### Solution

Given game is $2\times 3$. That is for Player $A$ has two strategies and player $B$ has three strategies. Let us check using maximin-minimax principle whether the game has a saddle point or not.

B1 B2 B3
A1 1 3 10
A2 8 6 2

#### Step 1

Apply maximin-minimax principle to check whether saddle point exists or not.

B1 B2 B3 RowMin
A1 1 3 10 1
A2 8 6 2 2
ColMax 8 6 10

Thus $Max(min) = Max(1, 2, )=2$ and $Min(max)=Min(8, 6, 10)=6$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.

#### Step 2

Construct two vertical axis, axis 1 and axis 2.
Axis 1 represent the payoff values of $A_2$ strategies for player $A$ and axis 2 represent payoff values of $A_1$ strategies for player $A$.

#### Step 3

Joint the point representing $a_{1j}$ on axis 2 to the point representing $a_{2j}$ on axis 1 for all $j = 1, 2,\cdots,n$.

Let the player $A$ play the strategy $A_1$ with probability $p$ and strategy $A_2$ with probability $1-p$.

Expected payoff for player $A$ for any pure strategy of player $B$ would be given by

$B$'s Strategy Expected Payoff for $A$
$B_1$ $E_1= p+8(1-p) = -7p+8$
$B_2$ $E_2= 3p+6(1-p)=-3p+6$
$B_3$ $E_3= 10p+2(1-p)=8p+2$

#### Step 4

Mark the lowest boundary (called Lower envelope) of the lines by thick line segment. The highest point on the lower envelope gives the maximin point $P$ and it identifies the two critical moves (strategies) of player $B$.

The two critical moves for player $B$ are $B_2$ and $B_3$.

The highest point $P$ on the lower envelope gives the maximin point.

#### Step 5

The reduced game (i.e. $2\times 2$) is

Player A \ Player B $B_2$ $B_3$
$A_1$ 3 10
$A_2$ 6 2

Apply maximin criterion.

B2 B3
A1 3 10
A2 6 2
• Select the minimum element of each row of the payoff matrix,

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

B2 B3 RowMin
A1 3 10 3
A2 6 2 2
• For each column of the payoff matrix, select the maximum element and call it ColMax.

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

B2 B3 RowMin
A1 3 10 3
A2 6 2 2
ColMax 6 10
• From each RowMin, obtain the maximum value, i.e., $Max(RowMin)$.

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

Thus $Max(min) = Max(3, 2)=3$

• For each ColMax, obtain the minimum value, i.e. $Min(ColMax)$.

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

Thus $Min(max)=Min(6, 10)=6$.

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

Hence the optimal strategy for the reduced game can be obtained by algebraic method.

 \begin{aligned} p_1 &= \frac{d-c}{(a+d)-(b+c)}\\ &=\frac{2-6}{(3+2)-(10+6)}\\ &=\frac{-4}{-11}\\ &=\frac{4}{11}\\ p_2 &= 1-p_1\\ &=\frac{7}{11} \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{2-10}{(3+2)-(10+6)}\\ &=\frac{-8}{-11}\\ &=\frac{8}{11}\\ q_2 &= 1-q_1\\ &=\frac{3}{11}. \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{4}{11} &\frac{7}{11} \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 & B_3\\ 0 & q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2 & B_3\\ 0 & \frac{8}{11} & \frac{3}{11} \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{(3)(2)-(10)(6)}{(3+2)-(10+6)} \\ &=\frac{-54}{-11}\\ &=4.9091. \end{aligned}

Thus, we conclude that player $A$ may choose strategy $A_1$ with probability $\dfrac{5}{12}$ and strategy $A_5$ with probability $\dfrac{7}{12}$. Player $B$ may choose strategy $B_1$ with probability $\dfrac{8}{11}$ and strategy $B_2$ with probability $\dfrac{3}{11}$. And value of the game for player A is $4.9091$ and for player B is $-4.9091$.

## Endnote

In this tutorial, you learned about graphical method to solve $2\times n$ game and how to use graphical method to solve $2\times n$ game with illustrated examples.

Let me know in the comments if you have any questions on Graphical method to solve $2\times n$ game and your thought on this article. 