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Graphical Method to solve m by 2 game

Graphical method to solve $m\times 2$ game

A game where one player has more than $m$ course of action and the other player has only two course of action is called $m\times 2$ 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 $m\times 2$ game or a game reducible to $m\times 2$ after applying dominance property.

Step by Step Procedure to solve m by 2 game

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

Player A \ Player B $B_1$ $B_2$
$A_1$ $a_{11}$ $a_{12}$
$A_2$ $a_{21}$ $a_{22}$
$\vdots$ $\vdots$ $\vdots$
$A_m$ $a_{m1}$ $a_{m2}$

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

Step 1

Construct two vertical axis, axis 1 and axis 2, by taking suitable scale. (Graph is just a sample)

m by 2 game
m by 2 game

Step 2

Axis 1 represent the payoff values of $B_2$ strategy for player $B$ and axis 2 represent payoff values of $B_1$ strategy for player $B$.

Step 3

Joint the point representing $a_{i1}$ on axis 2 to the point representing $a_{i2}$ on axis 1 for all $i = 1, 2,\cdots,m$. (Graph is just a sample)

m by 2 game
m by 2 game

Step 4

Mark the upper boundary (called upper envelope) of the lines by thick line segment. The lowest point on the upper envelope gives the minimax point $P$ and it identifies the two critical moves (strategies) of player $A$.

Suppose the upper envelope is created by $A_1$ and $A_m$ strategy. The lowest point on the upper envelope gives the minimax point. This shows that the two critical moves for player $A$ are $A_1$ and $A_m$.

Using the above selection the reduced game becomes

Player A \ Player B $B_1$ $B_2$
$A_1$ $a_{11}$ $a_{12}$
$A_m$ $a_{m1}$ $a_{m2}$
m by 2 game
m by 2 game

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.)

m by 2 game by graphical method Example 1

Solve the game with the following payoff for player A.

Player A / Player B $B_1$ $B_2$
$A_1$ -2 5
$A_2$ -5 3
$A_3$ 0 -2
$A_4$ -3 0
$A_5$ 1 -4

Solution

Given game is $5\times 2$. That is for Player $A$ has five strategies and player $B$ has two strategies. Let us check using maximin-minimax principle whether the game has a saddle point or not.

B1 B2
A1 -2 5
A2 -5 3
A3 0 -2
A4 -3 0
A5 -1 -4

Step 1

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

B1 B2 RowMin
A1 -2 5 -2
A2 -5 3 -5
A3 0 -2 -2
A4 -3 0 -3
A5 -1 -4 -4
ColMax 0 5

Thus $Max(min) = Max(-2, -5, -2, -3, -4, )=-5$ and $Min(max)=Min(0, 5)=0$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.

Step 2

Draw two vertical axis, axis 1 and axis 2. Mark appropriate scale on each axis. The two axis represents strategies for player B. Axis 1 represent the payoff values of $B_2$ strategy for player $B$ and axis 2 represent payoff values of $B_1$ strategy for player $B$.

m by 2 game
m by 2 game

Let the player B play the strategy $B_1$ with probability $q$ and strategy $B_2$ with probability $1-q$.

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

A’s Strategy Expected Payoff for B
$A_1$ $E_1=-2q+5(1-q) = 7q+5$
$A_2$ $E_2=-5q+3(1-q)=-8q+3$
$A_3$ $E_3 = 0q+(-2)(1-q)=2q-2$
$A_4$ $E_4=-3q+0(1-q)= -3q$
$A_5$ $E_5 = -1q+(-4)(1-q)=3q-4$

Step 3

Draw the graph of expected payoff $E_1,E_2,E_3, E_4, E_5$ against $q$

Joint the point representing $a_{i1}$ on axis 2 to the point representing $a_{i2}$ on axis 1 for all $i = 1, 2,\cdots,m$.

m by 2 game
m by 2 game

Step 4

Mark the upper boundary (called upper envelope) of the lines by thick line segment. The lowest point on the upper envelope gives the minimax point $P$.

m by 2 game
m by 2 game

The upper envelope is created by $A_1$ and $A_5$ strategy.

The two critical moves (strategies) of player $A$ are $A_1$ and $A_5$.

The lowest point $P$ on the upper envelope gives the minimax point.

Step 5

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

Player A / Player B $B_1$ $B_2$
$A_1$ -2 5
$A_5$ 1 -4

Apply maximin-minimax principle on the reduced $2\times 2$ game.

B1 B2
A1 -2 5
A5 1 -4

For each row of the payoff matrix, select the minimum element and call it RowMin.

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

B1 B2 RowMin
A1 -2 5 -2
A5 1 -4 -4

For each column of the payoff matrix, select the maximum element and call it ColMax.

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

B1 B2 RowMin
A1 -2 5 -2
A5 1 -4 -4
ColMax 1 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 $Max(min) = Max(-2, -4)=-2$

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 $Min(max)=Min(1, 5)= 1$.

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.

$$ \begin{aligned} p_1 &=\frac{d-c}{(a+d)-(b+c)}\\ &=\frac{(-4)-1}{(-2)+(-4)-(5+1)}\\ &=\frac{5}{12}\\ p_2 &= 1-p_1\\ &=\frac{7}{12} \end{aligned} $$

and the optimal strategies for player B can be determined by

$$ \begin{aligned} q_1 &= \frac{d-b}{(a+d)-(b+c)}\\ &=\frac{(-4)-5}{(-2) +(-4) -(5+1)}\\ &=\frac{9}{12}=\frac{3}{4}\\ q_2 &= 1-q_1\\ &=\frac{1}{4}. \end{aligned} $$

The optimal strategies for player A can be written as

$$ \begin{aligned} S_{A} &= \begin{bmatrix} A_1 & A_2 & A_3 & A_4 & A_5\\ p_1 & 0 & 0 & 0 & p_2 \end{bmatrix}\\ &=\begin{bmatrix} A_1 & A_2 & A_3 & A_4 & A_5\\ \frac{5}{12} & 0 & 0 & 0 & \frac{7}{12} \end{bmatrix} \end{aligned} $$

and the optimal strategies for player B can be written as

$$ \begin{aligned} S_{B} &= \begin{bmatrix} B_1 & B_2\\ q_1 & q_2 \end{bmatrix}\\ &= \begin{bmatrix} B_1 & B_2\\ \frac{3}{4} & \frac{1}{4} \end{bmatrix} \end{aligned} $$

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

$$ \begin{aligned} V &= \frac{ad-bc}{(a+d)-(b+c)}\\ &=\frac{(-2)(-4)-(5)(1)}{(-2)+(-4)-(5+1)}\\ &=\frac{3}{12}=\frac{1}{4}. \end{aligned} $$

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 $\dfrac{3}{4}$ and strategy $B_2$ with probability $\dfrac{1}{4}$. And value of the game for player A is $\dfrac{1}{4}$ and for player B is $-\dfrac{1}{4}$.

m by 2 game by graphical method Example 2

Solve the game with the following payoff for player A.

Player A / Player B $B_1$ $B_2$
$A_1$ 2 4
$A_2$ 2 3
$A_3$ 3 2
$A_4$ -2 6

Solution

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

B1 B2
A1 2 4
A2 2 3
A3 3 2
A4 -2 6

Step 1

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

B1 B2 RowMin
A1 2 4 2
A2 2 3 2
A3 3 2 2
A4 -2 6 -2
ColMax 3 6

Thus $Max(min) = Max(2, 2, 2, -2, )=2$ and $Min(max)=Min(3, 6)=3$. Since the $Max(min)\neq Min(max)$ for the game, the game has no saddle point.

Step 2

Draw two vertical axis, axis 1 and axis 2. Mark appropriate scale on each axis. The two axis represents strategies for player B. Axis 1 represent the payoff values of $B_2$ strategy for player $B$ and axis 2 represent payoff values of $B_1$ strategy for player $B$.

m by 2 game
m by 2 game

Let the player B play the strategy $B_1$ with probability $q$ and strategy $B_2$ with probability $1-q$.

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

A’s Strategy Expected Payoff for B
$A_1$ $E_1= 2q+4(1-q) = -2q+4$
$A_2$ $E_2= 2q+3(1-q)=-q+3$
$A_3$ $E_3= 3q+2(1-q)=q+2$
$A_4$ $E_4=-2q+6(1-q)= -8q+6$

Step 3

Draw the graph of expected payoff $E_1,E_2,E_3, E_4$ against $q$

Joint the point representing $a_{i1}$ on axis 2 to the point representing $a_{i2}$ on axis 1 for all $i = 1, 2,\cdots,m$.

m by 2 game
m by 2 game

Step 4

Mark the upper boundary (called upper envelope) of the lines by thick line segment. The lowest point on the upper envelope gives the minimax point $P$.

m by 2 game
m by 2 game

The upper envelope is created by $A_1$ and $A_3$ strategy.

The two critical moves (strategies) of player $A$ are $A_1$ and $A_3$.

The lowest point $P$ on the upper envelope gives the minimax point.

Step 5

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

Player A / Player B $B_1$ $B_2$
$A_1$ 2 4
$A_3$ 3 2

Apply maximin-minimax principle on the reduced $2\times 2$ game.

B1 B2
A1 2 4
A3 3 2

For each row of the payoff matrix, select the minimum element and call it RowMin.

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

B1 B2 RowMin
A1 2 4 2
A3 3 2 2

For each column of the payoff matrix, select the maximum element and call it ColMax.

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

B1 B2 RowMin
A1 2 4 2
A3 3 2 2
ColMax 3 4

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

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

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.

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

and the optimal strategies for player B can be determined by

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

The optimal strategies for player A can be written as

$$ \begin{aligned} S_{A} &= \begin{bmatrix} A_1 & A_2 & A_3 & A_4 \\ p_1 & 0 & p_2 & 0 \end{bmatrix}\\ &=\begin{bmatrix} A_1 & A_2 & A_3 & A_4 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 \end{bmatrix} \end{aligned} $$

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

$$ \begin{aligned} V &= \frac{ad-bc}{(a+d)-(b+c)}\\ &=\frac{2*2-4*3}{(2+2)-(4+3)}\\ &=\frac{-8}{-3}\\ &=\frac{8}{3}. \end{aligned} $$

Thus, we conclude that player $A$ may choose strategy $A_1$ with probability $\dfrac{1}{3}$ and strategy $A_3$ with probability $\dfrac{2}{3}$. Player $B$ may choose strategy $B_1$ with probability $\dfrac{2}{3}$ and strategy $B_2$ with probability $\dfrac{1}{3}$. And value of the game for player A is $\dfrac{8}{3}$ and for player B is $-\dfrac{8}{3}$.

Endnote

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

To learn more about different methods to solve a game please refer to the following tutorials:

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

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