LECTURE_2_GAME_THEORY.pptx
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Game Theory : How to solve Sums
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LECTURE_2_GAME_THEORY.pptx
- 1. Prof. Sanjay Christian JGIBA (Assistant Professor) M.Phil (Statistics), M.Sc (Statistics) & B.Sc (Statistics) GAME THEORY LECTURE 2
- 2. The Expected payoff to a player in a game with payoff matrix 𝒂𝒊𝒋 of Order 𝒎 × 𝒏 is given by, 𝑬 𝒑. 𝒒 = 𝒊=𝟏 𝒎 𝒋=𝟏 𝒏 𝒑𝒊 𝒂𝒊𝒋𝒒𝒋 𝒑𝟏, 𝒑𝟐, … … 𝒑𝒎 mixed strategy for player A 𝒒𝟏, 𝒒𝟐, … … 𝒒𝒏 mixed strategy for player B The value of the game using mixed strategies represents least payoff which player A can expect to win and least payoff which player B can loose. Optimal strategy mixture for each player may be obtained, By assigning probabilities to each strategy of being chosen. Mixed Strategies: Games without Saddle Point When there is no saddle point, both players must obtain an optimal mixture of strategies to find a saddle point.
- 4. If 01 & 02 points are not satisfied then take average of two or more pure strategies and compare with the remaining pure strategies & delete if it is inferior. If all the elements of a column (kth column) are more than or equal to the elements of the other column (rth column), then kth column is dominated by rth column. 01 If all the elements of a row (kth row) are less than or equal to the elements of the other row (rth row), then kth row is dominated by rth row. 02 03 Dominance Property Dominance Property is used when the payoff matrix is a profit matrix for the player A and a loss matrix to player B. Otherwise the principle gets revered.
- 5. ALGEBRAIC METHOD Is used to determine probability value by using different strategies by player A and B Consider 2 × 2 two person zero sum game without saddle point having the payoff matrix for player A 𝑎11 𝑎12 𝑎21 𝑎22 𝐵1 𝐵2 𝐴1 𝐴2 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐 𝒑𝟏 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 𝒑𝟐 = 𝟏 − 𝒑𝟏 𝒒𝟐 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 𝒒𝟏 = 𝟏 − 𝒒𝟐 𝑽𝒂𝒍𝒖𝒆 𝒐𝒇 𝒕𝒉𝒆 𝑮𝒂𝒎𝒆: 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐
- 6. Solve the following game. Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 𝐁𝟒 Player A 𝐀𝟏 5 1 7 4 𝐀𝟐 9 4 15 6 𝐀𝟑 0 5 3 9 𝑨𝟒 5 0 10 4 Answer: 3 Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 𝐁𝟒 Player A 𝐀𝟏 5 1 7 4 𝐀𝟐 9 4 15 6 𝐀𝟑 0 5 3 9 𝑨𝟒 5 0 10 4 Row Minimum 1 4 0 0 Column Maximum 9 5 15 9 Here, From Row Minimum, we find Maximum Value i.e., Maxi-Min=4 and From Column Maximum, we find Minimum Value i.e., Mini-Max=5 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist. As Saddle Point does not exist, the player will use Mixed Strategies.
- 7. 5 1 7 4 9 4 15 6 0 5 3 9 5 0 10 4 Now we reduce the size of the matrix by using Dominance Property. Step : 1 Elements of 1st Row is less than 2nd Row and hence 1st Row is dominated by 2nd Row. 9 4 15 6 0 5 3 9 5 0 10 4 Step : 2 Elements of 3rd Column is more than 1st Column and hence 3rd Column is dominated by 1st Column 9 4 6 0 5 9 5 0 4 Step : 3 Elements of 3rd Row is less than 1st Row and hence 3rd Row is dominated by 1st Row. 9 4 6 0 5 9 Step : 4 Elements of 3rd Column is more than 2nd Column and hence 3rd Column is dominated by 2nd Column 9 4 0 5 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐
- 8. 𝒂𝟏𝟏 = 𝟗 𝒂𝟏𝟐 = 𝟒 𝒂𝟐𝟏 = 𝟎 𝒂𝟐𝟐 = 𝟓 𝒑𝟐 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟓 − 𝟎 𝟗 + 𝟓 − 𝟎 + 𝟒 = 𝟓 𝟏𝟒 − 𝟒 = 𝟓 𝟏𝟎 𝒑𝟐 = 𝟏 𝟐 𝒑𝟑 = 𝟏 − 𝒑𝟏 − 𝒑𝟐 − 𝒑𝟒 = 𝟏 − 𝟎 − 𝟏 𝟐 − 𝟎 = 𝟏 𝟐 𝒒𝟏 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟓 − 𝟒 𝟗 + 𝟓 − 𝟎 + 𝟒 = 𝟏 𝟏𝟒 − 𝟒 𝒒𝟏 = 𝟏 𝟏𝟎 𝒒𝟐 = 𝟏 − 𝒒𝟏 − 𝒒𝟑 − 𝒒𝟒 = 𝟏 − 𝟏 𝟏𝟎 − 𝟎 − 𝟎 = 𝟗 𝟏𝟎 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐, 𝒑𝟑, 𝒑𝟒 = 𝟎, 𝟏 𝟐 , 𝟏 𝟐 , 𝟎 𝒑𝟏 = 𝟎 & 𝒑𝟒= 𝟎 𝒑𝟏 + 𝒑𝟐 + 𝒑𝟑 + 𝒑𝟒 = 𝟏 5 1 7 4 9 4 15 6 0 5 3 9 5 0 10 4 𝒒𝟑 = 𝟎 & 𝒒𝟒= 𝟎 𝒒𝟏 + 𝒒𝟐 + 𝒒𝟑 + 𝒒𝟒 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐, 𝒒𝟑, 𝒒𝟒 = 𝟏 𝟏𝟎 , 𝟗 𝟏𝟎 , 𝟎, 𝟎
- 9. 𝒂𝟏𝟏 = 𝟗 𝒂𝟏𝟐 = 𝟒 𝒂𝟐𝟏 = 𝟎 𝒂𝟐𝟐 = 𝟓 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟗 𝟓 − 𝟎 𝟒 𝟗 + 𝟓 − 𝟎 + 𝟒 = 𝟒𝟓 − 𝟎 𝟏𝟒 − 𝟒 = 𝟒𝟓 𝟏𝟎 = 𝟒. 𝟓
- 10. Two companies are competing for business under the conditions so that one company’s gain is another company’s loss. The following payoff in lakhs is derived by company A for different strategies used for advertisements. 4 Company A Company B No advertising Medium advertising Heavy advertising No advertising 5 1 -7 Medium advertising 8 7 8 Heavy advertising 11 9 5 Suggest optimal strategies for the two companies. Answer: Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 Player A 𝐀𝟏 5 1 -7 𝐀𝟐 8 7 8 𝐀𝟑 11 9 5 Here, From Row Minimum we find Maximum Value i.e., Maxi-Min=7 and From Column Maximum we find Minimum Value i.e., Mini-Max=8 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist, Row Minimum -7 7 5 Column Maximum 11 9 8 As Saddle Point does not exist, the player will use Mixed Strategies.
- 11. 5 1 -7 8 7 8 11 9 5 Now we reduce the size of the matrix by using Dominance Property. Step : 1 Elements of 1st Row is less than 2nd Row and hence 1st Row is dominated by 2nd Row. Step : 2 Elements of 1st Column is more than 2nd Column and hence 1st Column is dominated by 2nd Column 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐 7 8 9 5 8 7 8 11 9 5
- 12. 𝒂𝟏𝟏 = 𝟕 𝒂𝟏𝟐 = 𝟖 𝒂𝟐𝟏 = 𝟗 𝒂𝟐𝟐 = 𝟓 𝒑𝟐 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟓 − 𝟗 𝟕 + 𝟓 − 𝟗 + 𝟖) = −𝟒 𝟏𝟐 − 𝟏𝟕 = −𝟒 −𝟓 𝒑𝟐 = 𝟒 𝟓 𝒑𝟑 = 𝟏 − 𝒑𝟏 − 𝒑𝟐 = 𝟏 − 𝟎 − 𝟒 𝟓 = 𝟏 𝟓 𝒒𝟐 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟓 − 𝟖 𝟕 + 𝟓 − 𝟗 + 𝟖 = −𝟑 𝟏𝟐 − 𝟏𝟕 = −𝟑 −𝟓 𝒒𝟑 = 𝟏 − 𝒒𝟏 − 𝒒𝟐 = 𝟏 − 𝟎 − 𝟑 𝟓 = 𝟐 𝟓 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐, 𝑨𝟑 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐, 𝒑𝟑 = 𝟎, 𝟒 𝟓 , 𝟏 𝟓 𝒑𝟏 = 𝟎 𝒑𝟏 + 𝒑𝟐 + 𝒑𝟑 = 𝟏 5 1 -7 8 7 8 11 9 5 𝒒𝟏 = 𝟎 𝒒𝟏 + 𝒒𝟐 + 𝒒𝟑 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐, 𝑩𝟑 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐, 𝒒𝟑 = 𝟎, 𝟑 𝟓 , 𝟐 𝟓 𝒒𝟐 = 𝟑 𝟓
- 13. 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟕 𝟓 − 𝟗 𝟖 𝟕 + 𝟓 − 𝟗 + 𝟖 = 𝟑𝟓 − 𝟕𝟐 𝟏𝟐 − 𝟏𝟕 = −𝟑𝟕 −𝟓 = 𝟕. 𝟒 𝒂𝟏𝟏 = 𝟕 𝒂𝟏𝟐 = 𝟖 𝒂𝟐𝟏 = 𝟗 𝒂𝟐𝟐 = 𝟓
- 14. Solve the following Game. 5 Answer: Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 Player A 𝐀𝟏 9 8 -7 𝐀𝟐 3 -6 4 𝐀𝟑 6 7 -7 Here, From Row Minimum we find Maximum Value i.e., Maxi-Min=-6 and From Column Maximum we find Minimum Value i.e., Mini-Max=4 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist, Row Minimum -7 -6 -7 Column Maximum 9 8 4 As Saddle Point does not exist, the player will use Mixed Strategies. 𝟗 𝟖 −𝟕 𝟑 −𝟔 𝟒 𝟔 𝟕 −𝟕
- 15. 9 8 -7 3 -6 4 6 7 -7 Now we reduce the size of the matrix by using Dominance Property. Step : 1 Elements of 3rd Row is less than 1st Row and hence 3rd Row is dominated by 1st Row. Step : 2 Elements of 1st Column is more than 2nd Column and hence 1st Column is dominated by 2nd Column 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐 8 -7 -6 4 9 8 -7 3 -6 4
- 16. 𝒂𝟏𝟏 = 𝟖 𝒂𝟏𝟐 = −𝟕 𝒂𝟐𝟏 = −𝟔 𝒂𝟐𝟐 = 𝟒 𝒑𝟏 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟒 − (−𝟔) 𝟖 + 𝟒 − −𝟔 + (−𝟕) = 𝟏𝟎 𝟏𝟐 − −𝟏𝟑 = 𝟏𝟎 𝟐𝟓 𝒑𝟏 = 𝟐 𝟓 𝒑𝟐 = 𝟏 − 𝒑𝟏 − 𝒑𝟑 = 𝟏 − 𝟐 𝟓 − 𝟎 = 𝟑 𝟓 𝒒𝟐 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟒 − (−𝟕) 𝟖 + 𝟒 − −𝟔 + (−𝟕) = 𝟏𝟏 𝟏𝟐 − −𝟏𝟑 = 𝟏𝟏 𝟐𝟓 𝒒𝟑 = 𝟏 − 𝒒𝟏 − 𝒒𝟐 = 𝟏 − 𝟎 − 𝟏𝟏 𝟐𝟓 = 𝟏𝟒 𝟐𝟓 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐, 𝑨𝟑 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐, 𝒑𝟑 = 𝟐 𝟓 , 𝟑 𝟓 , 𝟎 𝒑𝟑 = 𝟎 𝒑𝟏 + 𝒑𝟐 + 𝒑𝟑 = 𝟏 9 8 -7 3 -6 4 6 7 -7 𝒒𝟏 = 𝟎 𝒒𝟏 + 𝒒𝟐 + 𝒒𝟑 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐, 𝑩𝟑 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐, 𝒒𝟑 = 𝟎, 𝟏𝟏 𝟐𝟓 , 𝟏𝟒 𝟐𝟓 𝒒𝟐 = 𝟏𝟏 𝟐𝟓
- 17. 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟖 𝟒 − −𝟔 −𝟕 𝟖 + 𝟒 − −𝟔 + (−𝟕) = 𝟑𝟐 − 𝟒𝟐 𝟏𝟐 − −𝟏𝟑 = −𝟏𝟎 𝟐𝟓 𝒂𝟏𝟏 = 𝟖 𝒂𝟏𝟐 = −𝟕 𝒂𝟐𝟏 = −𝟔 𝒂𝟐𝟐 = 𝟒 = −𝟐 𝟓
- 18. Solve the following Game. (Extra Example for Practice). 6 Answer: Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 Player A 𝐀𝟏 1 7 2 𝐀𝟐 6 2 7 𝐀𝟑 6 1 6 Here, From Row Minimum we find Maximum Value i.e., Maxi-Min=2 and From Column Maximum we find Minimum Value i.e., Mini-Max=6 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist, Row Minimum 1 2 1 Column Maximum 6 7 7 As Saddle Point does not exist, the player will use Mixed Strategies. 𝟏 𝟕 𝟐 𝟔 𝟐 𝟕 𝟔 𝟏 𝟔
- 19. 1 7 2 6 2 7 6 1 6 Now we reduce the size of the matrix by using Dominance Property. Step : 1 Elements of 3rd Row is less than 2nd Row and hence 3rd Row is dominated by 2nd Row. Step : 2 Elements of 3rd Column is more than 1st Column and hence 3rd Column is dominated by 1st Column 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐 1 7 6 2 1 7 2 6 2 7
- 20. 𝒂𝟏𝟏 = 𝟏 𝒂𝟏𝟐 = 𝟕 𝒂𝟐𝟏 = 𝟔 𝒂𝟐𝟐 = 𝟐 𝒑𝟏 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟐 − 𝟔 𝟏 + 𝟐 − 𝟔 + 𝟕 = −𝟒 𝟑 − 𝟏𝟑 = −𝟒 −𝟏𝟎 𝒑𝟏 = 𝟐 𝟓 𝒑𝟐 = 𝟏 − 𝒑𝟏 − 𝒑𝟑 = 𝟏 − 𝟐 𝟓 − 𝟎 = 𝟑 𝟓 𝒒𝟏 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟐 − 𝟕 𝟏 + 𝟐 − 𝟔 + 𝟕 = −𝟓 𝟑 − 𝟏𝟑 = −𝟓 −𝟏𝟎 𝒒𝟐 = 𝟏 − 𝒒𝟏 − 𝒒𝟑 = 𝟏 − 𝟏 𝟐 − 𝟎 = 𝟏 𝟐 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐, 𝑨𝟑 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐, 𝒑𝟑 = 𝟐 𝟓 , 𝟑 𝟓 , 𝟎 𝒑𝟑 = 𝟎 𝒑𝟏 + 𝒑𝟐 + 𝒑𝟑 = 𝟏 1 7 2 6 2 7 6 1 6 𝒒𝟑 = 𝟎 𝒒𝟏 + 𝒒𝟐 + 𝒒𝟑 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐, 𝑩𝟑 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐, 𝒒𝟑 = 𝟏 𝟐 , 𝟏 𝟐 , 𝟎 𝒒𝟏 = 𝟏 𝟐
- 21. 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟏 𝟐 − 𝟔 𝟕 𝟏 + 𝟐 − 𝟔 + 𝟕 = 𝟐 − 𝟒𝟐 𝟑 − 𝟏𝟑 = −𝟒𝟎 −𝟏𝟎 = 𝟒 𝒂𝟏𝟏 = 𝟏 𝒂𝟏𝟐 = 𝟕 𝒂𝟐𝟏 = 𝟔 𝒂𝟐𝟐 = 𝟐
- 22. Solve the following game. (Extra Example for Practice) Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 𝐁𝟒 Player A 𝐀𝟏 1 0 2 -2 𝐀𝟐 1 2 0 2 𝐀𝟑 2 0 2 -2 𝑨𝟒 -2 2 -2 1 Answer: 7 Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 𝐁𝟒 Player A 𝐀𝟏 1 0 2 -2 𝐀𝟐 1 2 0 2 𝐀𝟑 2 0 2 -2 𝑨𝟒 -2 2 -2 1 Row Minimum -2 0 -2 -2 Column Maximum 2 2 2 2 Here, From Row Minimum, we find Maximum Value i.e., Maxi-Min=0 and From Column Maximum, we find Minimum Value i.e., Mini-Max=2 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist. As Saddle Point does not exist, the player will use Mixed Strategies.
- 23. 1 0 2 -2 1 2 0 2 2 0 2 -2 -2 2 -2 1 Now we reduce the size of the matrix by using Dominance Property. Step : 1 Elements of 1st Row is less than 3rd Row and hence 1st Row is dominated by 3rd Row. 1 2 0 2 2 0 2 -2 -2 2 -2 1 Step : 2 Elements of 1st Column is more than 3rd Column and hence 1st Column is dominated by 3rd Column 2 0 2 0 2 -2 2 -2 1 Step : 3 Elements of 3rd Row is less than 1st Row and hence 3rd Row is dominated by 1st Row. 2 0 2 0 2 -2 Step : 4 Elements of 1st Column is more than 3rd Column and hence 1st Column is dominated by 3rd Column 0 2 2 -2 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐
- 24. 𝒂𝟏𝟏 = 𝟎 𝒂𝟏𝟐 = 𝟐 𝒂𝟐𝟏 = 𝟐 𝒂𝟐𝟐 = −𝟐 𝒑𝟐 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = −𝟐 − 𝟐 𝟎 + (−𝟐) − 𝟐 + 𝟐 = −𝟒 −𝟐 − 𝟒 = −𝟒 −𝟔 𝒑𝟐 = 𝟐 𝟑 𝒑𝟑 = 𝟏 − 𝒑𝟏 − 𝒑𝟐 − 𝒑𝟒 = 𝟏 − 𝟎 − 𝟐 𝟑 − 𝟎 = 𝟏 𝟑 𝒒𝟑 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = −𝟐 − 𝟐 𝟎 + (−𝟐) − 𝟐 + 𝟐 = −𝟒 −𝟐 − 𝟒 = −𝟒 −𝟔 𝒒𝟒 = 𝟏 − 𝒒𝟏 − 𝒒𝟐 − 𝒒𝟑 = 𝟏 − 𝟎 − 𝟎 − 𝟐 𝟑 = 𝟏 𝟑 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐, 𝒑𝟑, 𝒑𝟒 = 𝟎, 𝟐 𝟑 , 𝟏 𝟑 , 𝟎 𝒑𝟏 = 𝟎 & 𝒑𝟒= 𝟎 𝒑𝟏 + 𝒑𝟐 + 𝒑𝟑 + 𝒑𝟒 = 𝟏 1 0 2 -2 1 2 0 2 2 0 2 -2 -2 2 -2 1 𝒒𝟏 = 𝟎 & 𝒒𝟐= 𝟎 𝒒𝟏 + 𝒒𝟐 + 𝒒𝟑 + 𝒒𝟒 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐, 𝒒𝟑, 𝒒𝟒 = 𝟎, 𝟎, 𝟐 𝟑 , 𝟏 𝟑 𝒒𝟑 = 𝟐 𝟑
- 25. 𝒂𝟏𝟏 = 𝟎 𝒂𝟏𝟐 = 𝟐 𝒂𝟐𝟏 = 𝟐 𝒂𝟐𝟐 = −𝟐 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟎 −𝟐 − 𝟐 𝟐 𝟎 + (−𝟐) − 𝟐 + 𝟐 = 𝟎 − 𝟒 −𝟐 − 𝟒 = −𝟒 −𝟔 = 𝟐 𝟑
- 26. Solve the following game. (Extra Example for Practice) Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 𝐁𝟒 Player A 𝐀𝟏 6 8 -16 -10 𝐀𝟐 0 3 -4 2 𝐀𝟑 18 4 10 51 𝑨𝟒 10 2 5 16 Answer: 8 Player B 𝐁𝟏 𝐁𝟐 𝐁𝟑 𝐁𝟒 Player A 𝐀𝟏 6 8 -16 -10 𝐀𝟐 0 3 -4 2 𝐀𝟑 18 4 10 51 𝑨𝟒 10 2 5 16 Row Minimum -16 -4 4 2 Column Maximum 18 8 10 51 Here, From Row Minimum, we find Maximum Value i.e., Maxi-Min=4 and From Column Maximum, we find Minimum Value i.e., Mini-Max=8 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist. As Saddle Point does not exist, the player will use Mixed Strategies.
- 27. 6 8 -16 -10 0 3 -4 2 18 4 10 51 10 2 5 16 Now we reduce the size of the matrix by using Dominance Property. Step : 1 Elements of 2nd Row is less than 3rd Row and hence 2nd Row is dominated by 3rd Row. 6 8 -16 -10 18 4 10 51 10 2 5 16 Step : 2 Elements of 1st Column is more than 3rd Column and hence 1st Column is dominated by 3rd Column 8 -16 -10 4 10 51 2 5 16 Step : 3 Elements of 3rd Row is less than 2nd Row and hence 3rd Row is dominated by 2nd Row. 8 -16 -10 4 10 51 Step : 4 Elements of 3rd Column is more than 2nd Column and hence 3rd Column is dominated by 2nd Column 8 -16 4 10 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐
- 28. 𝒂𝟏𝟏 = 𝟖 𝒂𝟏𝟐 = −𝟏𝟔 𝒂𝟐𝟏 = 𝟒 𝒂𝟐𝟐 = 𝟏𝟎 𝒑𝟏 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟏𝟎 − 𝟒 𝟖 + 𝟏𝟎 − 𝟒 + (−𝟏𝟔) = 𝟔 𝟏𝟖 − −𝟏𝟐 = 𝟔 𝟑𝟎 𝒑𝟏 = 𝟏 𝟓 𝒑𝟑 = 𝟏 − 𝒑𝟏 − 𝒑𝟐 − 𝒑𝟒 = 𝟏 − 𝟏 𝟓 − 𝟎 − 𝟎 = 𝟒 𝟓 𝒒𝟐 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟏𝟎 − (−𝟏𝟔) 𝟖 + 𝟏𝟎 − 𝟒 + (−𝟏𝟔) = 𝟐𝟔 𝟏𝟖 − −𝟏𝟐 = 𝟐𝟔 𝟑𝟎 𝒒𝟑 = 𝟏 − 𝒒𝟏 − 𝒒𝟐 − 𝒒𝟒 = 𝟏 − 𝟎 − 𝟏𝟑 𝟏𝟓 − 𝟎 = 𝟐 𝟏𝟓 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐, 𝒑𝟑, 𝒑𝟒 = 𝟏 𝟓 , 𝟒 𝟓 , 𝟎, 𝟎 𝒑𝟐 = 𝟎 & 𝒑𝟒= 𝟎 𝒑𝟏 + 𝒑𝟐 + 𝒑𝟑 + 𝒑𝟒 = 𝟏 6 8 -16 -10 0 3 -4 2 18 4 10 51 10 2 5 16 𝒒𝟏 = 𝟎 & 𝒒𝟒= 𝟎 𝒒𝟏 + 𝒒𝟐 + 𝒒𝟑 + 𝒒𝟒 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐, 𝒒𝟑, 𝒒𝟒 = 𝟎, 𝟏𝟑 𝟏𝟓 , 𝟐 𝟏𝟓 , 𝟎 𝒒𝟐 = 𝟏𝟑 𝟏𝟓
- 29. 𝒂𝟏𝟏 = 𝟖 𝒂𝟏𝟐 = −𝟏𝟔 𝒂𝟐𝟏 = 𝟒 𝒂𝟐𝟐 = 𝟏𝟎 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟖 𝟏𝟎 − 𝟒 −𝟏𝟔 𝟖 + 𝟏𝟎 − 𝟒 + (−𝟏𝟔) = 𝟖𝟎 + 𝟔𝟒 𝟏𝟖 − −𝟏𝟐 = 𝟏𝟒𝟒 𝟑𝟎 = 𝟐𝟒 𝟓 = 𝟒. 𝟖
- 30. Player A and B toss a dice. If even number occurs on both dice, player A gets 8 Rs, if odd number occurs on both dice, player A gets 6 Rs. If one gets odd and other gets even, player B gets 5 Rs. Determine the best strategy for each players. 9 Player B 𝑬𝒗𝒆𝒏 𝑶𝒅𝒅 Player A 𝑬𝒗𝒆𝒏 𝑶𝒅𝒅 8 6 -5 -5 Row Minimum -5 Column Maximum Here, From Row Minimum, we find Maximum Value i.e., Maxi-Min=-5 and From Column Maximum, we find Minimum Value i.e., Mini-Max=6 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 ≠ 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 Saddle point does not exist. As Saddle Point does not exist, the player will use Mixed Strategies. Player B 𝑬𝒗𝒆𝒏 𝑶𝒅𝒅 Player A 𝑬𝒗𝒆𝒏 8 -5 𝑶𝒅𝒅 -5 6 -5 8 6
- 31. 𝒂𝟏𝟏 = 𝟖 𝒂𝟏𝟐 = −𝟓 𝒂𝟐𝟏 = −𝟓 𝒂𝟐𝟐 = 𝟔 𝒑𝟏 = 𝒂𝟐𝟐 − 𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟔 − (−𝟓) 𝟖 + 𝟔 − −𝟓 + (−𝟓) = 𝟏𝟏 𝟏𝟒 − −𝟏𝟎 = 𝟏𝟏 𝟐𝟒 𝒑𝟏 = 𝟏𝟏 𝟐𝟒 𝒑𝟐 = 𝟏 − 𝒑𝟏 = 𝟏 − 𝟏𝟏 𝟐𝟒 = 𝟏𝟑 𝟐𝟒 𝒒𝟏 = 𝒂𝟐𝟐 − 𝒂𝟏𝟐 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟔 − (−𝟓) 𝟖 + 𝟔 − −𝟓 + (−𝟓) = 𝟏𝟏 𝟏𝟒 − −𝟏𝟎 = 𝟏𝟏 𝟐𝟒 𝒒𝟐 = 𝟏 − 𝒒𝟏 = 𝟏 − 𝟏𝟏 𝟐𝟒 = 𝟏𝟑 𝟐𝟒 Optimum mixed strategy for Player A is 𝑨𝟏, 𝑨𝟐 𝑺𝑨 = 𝒑𝟏, 𝒑𝟐 = 𝟏𝟏 𝟐𝟒 , 𝟏𝟑 𝟐𝟒 𝒒𝟏 + 𝒒𝟐 = 𝟏 Optimum mixed strategy for Player B is 𝑩𝟏, 𝑩𝟐 𝑺𝑩 = 𝒒𝟏, 𝒒𝟐 = 𝟏𝟏 𝟐𝟒 , 𝟏𝟑 𝟐𝟒 𝒒𝟏 = 𝟏𝟏 𝟐𝟒 𝒑𝟏 + 𝒑𝟐 = 𝟏
- 32. 𝒂𝟏𝟏 = 𝟖 𝒂𝟏𝟐 = −𝟓 𝒂𝟐𝟏 = −𝟓 𝒂𝟐𝟐 = 𝟔 𝑽 = 𝒂𝟏𝟏𝒂𝟐𝟐 − 𝒂𝟏𝟐𝒂𝟐𝟏 𝒂𝟏𝟏 + 𝒂𝟐𝟐 − 𝒂𝟐𝟏 + 𝒂𝟏𝟐 = 𝟖 𝟔 − −𝟓 −𝟓 𝟖 + 𝟔 − −𝟓 + (−𝟓) = 𝟒𝟖 − 𝟐𝟓 𝟏𝟒 − −𝟏𝟎 = 𝟐𝟑 𝟐𝟒
- 33. Find the range of the values of x and y which will make the element (2,2) a saddle point for the same. 10 Here, It is said that the Saddle Point is at (2,2) 𝑴𝒂𝒙𝒊. 𝒎𝒊𝒏 = 𝑴𝒊𝒏𝒊. 𝒎𝒂𝒙 = 𝟕 Player B 𝑩𝟏 𝑩𝟐 𝑩𝟑 Player A 𝑨𝟏 2 4 5 𝑨𝟐 10 7 y 𝑨𝟑 4 x 6 2 4 5 10 7 𝑦 4 𝑥 6 𝐴1 𝐴2 𝐴3 𝐵1 𝐵2 𝐵3 Player B Player A Row Minimum 2 Column Maximum 7 10 7 𝒚 ≥ 𝟕 ( ∵ 𝒊𝒇 𝒚 𝒊𝒔 𝒎𝒐𝒓𝒆 𝒕𝒉𝒂𝒏 𝟕 𝒕𝒉𝒆𝒏 𝒐𝒏𝒍𝒚 𝟕 𝒐𝒄𝒄𝒖𝒓𝒔 𝒊𝒏 𝑹𝒐𝒘 𝑴𝒊𝒏𝒊𝒎𝒖𝒎) 𝒙 ≤ 𝟕 ( ∵ 𝒊𝒇 𝒙 𝒊𝒔 𝒍𝒆𝒔𝒔 𝒕𝒉𝒂𝒏 𝟕 𝒕𝒉𝒆𝒏 𝒐𝒏𝒍𝒚 𝟕 𝒐𝒄𝒄𝒖𝒓𝒔 𝒊𝒏 𝑪𝒐𝒍𝒖𝒎𝒏 𝑴𝒂𝒙𝒊𝒎𝒖𝒎) 𝑶𝑹 𝑯𝒆𝒏𝒄𝒆 𝒙 ≤ 𝟕 & 𝒚 ≥ 𝟕