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GT_Week_5 - Game Theory Solutions to Exercises The...

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Game Theory Solutions to Exercises The Strategic Form of a Game, Matrix Games & Domination Jan-Jaap Oosterwijk Fall 2007
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2 Matrix Games – Domination Contrary to Ferguson’s notation, we prefer to use the horizontal notation ( · , · , . . . , · ) to represent a row vector and hence ( · , · , . . . , · ) T to denote a column vector. 2.1 Solve the game with matrix - 1 - 3 - 2 2 , that is find the value and an optimal (mixed) strategy for both players. Solution: This matrix does not have a saddle point. Hence, the value of the game is V = - 1 · 2 - ( - 3)( - 2) - 1 - ( - 3)+2 - ( - 2) = - 8 6 = - 4 3 = - 1 1 3 , the probability with which Player I should choose the first row is p = 2 - ( - 2) 6 = 2 3 , and the probability with which Player II should choose the first column is q = 2 - ( - 3) 6 = 5 6 . So the optimal mixed strategy for Player I is ( 2 3 , 1 3 ) and for Player II ( 5 6 , 1 6 ) . 2.2 Solve the game with matrix 0 2 t 1 for an arbitrary real number t . Draw the graph of v ( t ) , the value of the game, as a function of t , for -∞ < t < . Solution: If t < 0, then 0 is a saddle point and hence the value of the game would be 0 and the optimal strategies are pure: Player I always chooses the first row and Player II always the first column. If instead 0 < t 1, then t is a saddle point and hence the value of the game would be t and the optimal strategies are again pure: Player I always chooses the second row and Player II always the first column. If, however, t = 0, then both zero-elements in the first column are saddle points, the value of the game is 0, and Player II still keeps its pure optimal strategy of choosing the first column but Player I can now use an arbitrary mixed strategy of choosing between the first and second row to be sure of a non-negative average payoff. He might, however, prefer the pure strategy of choosing the first row, since it dominates the second one. Finally, if t > 1, there is no saddle point. The value of the game would be V ( t ) = 0 · 1 - 2 t 0 - 2+1 - t = - 2 t - 1 - t = 2 t t +1 = 1 + t - 1 t +1 , the probability with which Player I should choose the first row is p = 1 - t - 1 - t = t - 1 t +1 , and the probability with which Player II should choose the first column is q = 1 - 2 - 1 - t = 1 t +1 . So an optimal mixed strategy for Player I is then t - 1 t +1 , 2 t +1 · and for Player II 1 t +1 , t t +1 · . Finally, note that lim t →∞ V ( t ) = lim t →∞ 2 1+ 1 t = 2. 2
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1 2 3 4 5 t 0.5 1 1.5 2 v LParen1 t RParen1 Value of the game 2.3 Show that if a game with m × n matrix has two saddle points, then they have equal values. Solution: Let A be the m × n matrix of a game with two saddle points. Then there exist 1 i 1 , i 2 m and 1 j 1 , j 2 n such that a i 1 j 1 = min 1 j n a i 1 j = max 1 i m a ij 1 and a i 2 j 2 = min 1 j n a i 2 j = max 1 i m a ij 2 . Hence, for the elements a i 1 j 2 and a i 2 j 1 that share a row with one and a column with the other saddle point (which, in the most extreme case, can also be one of the saddle points itself if the two saddle points are in the same row or column), we have that a i 1 j 1 a i 1 j 2 a i 2 j 2 and a i 2 j 2 a i 2 j 1 a i 1 j 1 . So both a i 1 j 1 a i 2 j 2
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