Method

2Xn Games Steps (Rule)
Step-1:	If this game does not have a saddle point and not reducible by the dominance method, then first write all 2`xx`2 sub-games and find the value of each 2`xx`2 sub-games.
Step-2:	Find 2`xx`2 sub-game which has the lowest value and this sub-game provides the solution to the larger game. Solve this sub-game using algebraic method.

---

Example-1
1. Find Solution of game theory problem using 2Xn Games

Player A\Player B	B1	B2
A1	-3	4
A2	-1	1
A3	7	-2

Solution:
1. Saddle point testing
Players

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		-3	4
	`A_2`		-1	1
	`A_3`		7	-2

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

			Player `B`
			`B_1`	`B_2`		Row Minimum
Player `A`	`A_1`		-3	(4)		`-3`
	`A_2`		[-1]	1		`[-1]`
	`A_3`		7	-2		`-2`
	Column Maximum		`7`	`(4)`

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.

---

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

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		-3	4
	`A_2`		-1	1
	`A_3`		7	-2

Also, no course of action dominates the other

---

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

Players

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		-3	4
	`A_2`		-1	1
	`A_3`		7	-2

---

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

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		-3	4
	`A_2`		-1	1

This game has saddle point and value of the game, `V_1` is -1

---

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

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		-3	4
	`A_3`		7	-2

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 -2) - (4 xx 7))/((-3 -2) - (4 +7))=(6 -28)/(-5 -11)=11/8`

---

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

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_2`		-1	1
	`A_3`		7	-2

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 -2) - (1 xx 7))/((-1 -2) - (1 +7))=(2 -7)/(-3 -8)=5/11`

---

The 2`xx`2 sub-game with the lowest value is `-1` in (1) and hence the solution to this game provides the solution to the larger game.

Using algebraic method
1. Saddle point testing
Players

			Player `B`
			`B_1`	`B_2`
Player `A`	`A_1`		-3	4
	`A_2`		-1	1

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

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

Select minimum from the maximum of columns
Column MiniMax = (-1)

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

Here, Column MiniMax = Row MaxiMin = -1
`:.` This game has a saddle point and value of the game is -1

---

---