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