Find Solution of game theory problem using graphical method

Player A\Player B	B1	B2	B3	B4
A1	2	2	3	-2
A2	4	3	2	6

Solution:
1. Saddle point testing
Players

			Player `B`
			`B_1`	`B_2`	`B_3`	`B_4`
Player `A`	`A_1`		2	2	3	-2
	`A_2`		4	3	2	6

We apply the maximin (minimax) principle to analyze the game.

			Player `B`
			`B_1`	`B_2`	`B_3`	`B_4`		Row Minimum
Player `A`	`A_1`		2	2	3	-2		`-2`
	`A_2`		4	(3)	[2]	6		`[2]`
	Column Maximum		`4`	`(3)`	`3`	`6`

Select minimum from the maximum of columns
Column MiniMax = (3)

Select maximum from the minimum of rows
Row MaxiMin = [2]

Here, Column MiniMax `!=` Row MaxiMin

`:.` This game has no saddle point.

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2. Dominance rule to reduce the size of the payoff matrix
Using dominance property

			Player `B`
			`B_1`	`B_2`	`B_3`	`B_4`
Player `A`	`A_1`		2	2	3	-2
	`A_2`		4	3	2	6

column-1 `>=` column-2, so remove column-1

			Player `B`
			`B_2`	`B_3`	`B_4`
Player `A`	`A_1`		2	3	-2
	`A_2`		3	2	6

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Solution using graphical method
First, we draw two parallel lines 1 unit distance apart and mark a scale on each.
The two parallel lines represent strategies of player `A`.

If player `B` selects strategy `B_1`, player `A` can win 2 or 3 units depending on A's selection of strategies.

The value 2 is plotted along the vertical axis under strategy `A_1` and the value 3 is plotted along the vertical axis under strategy `A_2`.

A straight line joining the two points is then drawn.
Similarly, we can plot strategies `B_2,B_3` also. The problem is graphed in the following figure.

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The highest point `V` in the shaded region indicates the value of game. From the above figure, the value of the game is `2.44` units.

1. The point of optimal solution occurs at the intersection of two lines
`E_2 = 3p_1 +2p_2`

`E_3 = -2p_1 +6p_2`

Comparing the above two equations, we have
`3p_1 +2p_2 = -2p_1 +6p_2`

Substituting `p_2 = 1 - p_1`

`3p_1 +2(1 - p_1) = -2p_1 +6(1 - p_1)`

Solving `p_1=4/9`

`p_2=1-p_1=1-4/9=5/9`

Substituting the values of `p_1` and `p_2` in equation `E_2`

`V = 3(4/9) +2(5/9)=22/9`


2. The point of optimal solution occurs at the intersection of two lines
`L_2 = 3q_1 -2q_2`

`L_3 = 2q_1 +6q_2`

Comparing the above two equations, we have
`3q_1 -2q_2 = 2q_1 +6q_2`

Substituting `q_2 = 1 - q_1`

`3q_1 -2(1 - q_1) = 2q_1 +6(1 - q_1)`

Solving `q_1=8/9`

`q_2=1-q_1=1-8/9=1/9`

Substituting the values of `q_1` and `q_2` in equation `L_2`

`V = 3(8/9) -2(1/9)=22/9`

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