Find Solution of game theory problem using 2Xn Games

Player A\Player B	B1	B2
A1	1	-3
A2	3	5
A3	-1	6
A4	4	1

Solution:
1. Saddle point testing
Players

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		1	-3
	`A_2`		3	5
	`A_3`		-1	6
	`A_4`		4	1

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

			Player `B`
			`B_1`	`B_2`		Row Minimum
Player `A`	`A_1`		1	-3		`-3`
	`A_2`		[3]	5		`[3]`
	`A_3`		-1	6		`-1`
	`A_4`		(4)	1		`1`
	Column Maximum		`(4)`	`6`

Select minimum from the maximum of columns
Column MiniMax = (4)

Select maximum from the minimum of rows
Row MaxiMin = [3]

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`
Player `A`	`A_1`		1	-3
	`A_2`		3	5
	`A_3`		-1	6
	`A_4`		4	1

row-1 `<=` row-4, so remove row-1

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_2`		3	5
	`A_3`		-1	6
	`A_4`		4	1

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So, we consider each 2`xx`2 sub-game and obtain their values

Players

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_2`		3	5
	`A_3`		-1	6
	`A_4`		4	1

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2`xx`2 sub-game : (1) Row 1, 2

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_2`		3	5
	`A_3`		-1	6

This game has saddle point and value of the game, `V_1` is 3

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2`xx`2 sub-game : (2) Row 1, 3

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_2`		3	5
	`A_4`		4	1

This game has no saddle point, so we use the algebraic method.
Value of game, `V_2=(a * d - b * c)/((a + d) - (b + c))=((3 xx 1) - (5 xx 4))/((3 +1) - (5 +4))=(3 -20)/(4 -9)=17/5`

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2`xx`2 sub-game : (3) Row 2, 3

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_3`		-1	6
	`A_4`		4	1

This game has no saddle point, so we use the algebraic method.
Value of game, `V_3=(a * d - b * c)/((a + d) - (b + c))=((-1 xx 1) - (6 xx 4))/((-1 +1) - (6 +4))=(-1 -24)/(0 -10)=5/2`

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The 2`xx`2 sub-game with the lowest value is `5/2` in (3) and hence the solution to this game provides the solution to the larger game.

so now solution using algebraic method
1. Saddle point testing
Players

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_3`		-1	6
	`A_4`		4	1

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

			Player `B`
			`B_1`	`B_2`		Row Minimum
Player `A`	`A_3`		-1	6		`-1`
	`A_4`		(4)	[1]		`[1]`
	Column Maximum		`(4)`	`6`

Select minimum from the maximum of columns
Column MiniMax = (4)

Select maximum from the minimum of rows
Row MaxiMin = [1]

Here, Column MiniMax `!=` Row MaxiMin

`:.` This game has no saddle point.

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Matrix size is 2`xx`2, so dominance rule is not required.

Solution using algebraic method
Here `a=-1,b=6,c=4,d=1`

`p_1=(d - c)/((a + d) - (b + c))=(1 -4)/((-1 +1) - (6 +4))=(-3)/(0 -10)=3/10`

`p_2=1-p_1=1-3/10=7/10`

`q_1=(d - b)/((a + d) - (b + c))=(1 -6)/((-1 +1) - (6 +4))=(-5)/(0 -10)=1/2`

`q_2=1-q_1=1-1/2=1/2`

`V=(a * d - b * c)/((a + d) - (b + c))=((-1 xx 1) - (6 xx 4))/((-1 +1) - (6 +4))=(-1 -24)/(0 -10)=5/2`

A plays `(0,3/10,7/10)`

B plays `(1/2,1/2)`

Value of game is `5/2`

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