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Home > Operation Research calculators > Game Theory >> 2Xn Games example
8. 2Xn Games example ( Enter your problem )
( Enter your problem )
  1. Method & Example-1
  2. Example-2
 		Other related methods
  1. Saddle point
  2. Dominance method
  3. Oddment method
  4. Algebraic method
  5. Calculus method
  6. Arithmetic method
  7. Matrix method
  8. 2Xn Games
  9. Graphical method
  10. Linear programming method
  11. Bimatrix method
7. Matrix method
(Previous method)		2. Example-2
(Next example)

1. Method & Example-1


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 BB1B2
A1-34
A2-11
A37-2


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

This material is intended as a summary. Use your textbook for detail explanation.
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7. Matrix method
(Previous method)		2. Example-2
(Next example)


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