Find Solution of game theory problem using matrix method
Player A\Player B | B1 | B2 | B3 |
A1 | 10 | 5 | -2 |
A2 | 13 | 12 | 15 |
A3 | 16 | 14 | 10 |
Solution:1. Saddle point testing
Players
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | | |
Player `A` | `A_1` | | 10 | 5 | -2 | |
`A_2` | | 13 | 12 | 15 | |
`A_3` | | 16 | 14 | 10 | |
We apply the maximin (minimax) principle to analyze the game.
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | | Row Minimum |
Player `A` | `A_1` | | 10 | 5 | -2 | | `-2` |
`A_2` | | 13 | [12] | 15 | | `[12]` |
`A_3` | | 16 | (14) | 10 | | `10` |
| Column Maximum | | `16` | `(14)` | `15` | | |
Select minimum from the maximum of columns
Column MiniMax = (14)
Select maximum from the minimum of rows
Row MaxiMin = [12]
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` | `B_3` | | |
Player `A` | `A_1` | | 10 | 5 | -2 | |
`A_2` | | 13 | 12 | 15 | |
`A_3` | | 16 | 14 | 10 | |
row-1 `<=` row-3, so remove row-1
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | | |
Player `A` | `A_2` | | 13 | 12 | 15 | |
`A_3` | | 16 | 14 | 10 | |
column-1 `>=` column-2, so remove column-1
| | | Player `B` | | |
| | | `B_2` | `B_3` | | |
Player `A` | `A_2` | | 12 | 15 | |
`A_3` | | 14 | 10 | |
reduced matrix
| | | Player `B` | | |
| | | `B_2` | `B_3` | | |
Player `A` | `A_2` | | 12 | 15 | |
`A_3` | | 14 | 10 | |
For this reduced matrix, calculate `P_(Adj)` and `P_(Cof)`
`P_(Adj) = [[10,-15],[-14,12]]`
and `P_(Cof) = [[10,-14],[-15,12]]`
Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]])`
`=([[1,1]][[10,-15],[-14,12]])/([[1,1]][[10,-15],[-14,12]][[1],[1]])`
`=([[-4,-3]])/(-7)`
`=[[4/7,3/7]]`
`p_1=4/7` and `p_2=3/7`, where `p_1` and `p_2` represent the probabilities of player A's, using his strategies `A_2` and `A_3` respectively.
Similarly,
Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]])`
`=([[1,1]][[10,-14],[-15,12]])/([[1,1]][[10,-15],[-14,12]][[1],[1]])`
`=([[-5,-2]])/(-7)`
`=[[5/7,2/7]]`
`q_1=5/7` and `q_2=2/7`, where `q_1` and `q_2` represent the probabilities of player B's, using his strategies `B_2` and `B_3` respectively.
Hence, Value of the game `V` = (Player A's optimal strategies) `xx` (Payoff matrix `P_(ij)`) `xx` (Player B's optimal strategies)
`V=[[4/7,3/7]][[12,15],[14,10]][[5/7],[2/7]]=90/7`
This material is intended as a summary. Use your textbook for detail explanation.
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