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GAMES AND DECISIONS Lecture 3 Mixed extension of a matrix game Matrix game, two players, intelligent, finite strategy spaces and zero-sum payoffs. Two players {1, 2} Strategy spaces {X, Y} note: simple notation X and Y Payoff functions Z1(x, y) Z2(x, y) = - Z1(x, y) Find Nash equilibrium Player 2 1 Player 1 1 2 2 1 -1 1 -1 No Nash equilibrium, no solution in pure strategies Stone, Paper, Scissors. Scissors cuts paper. Paper covers stone. Stone breaks scissors. The winner will get 1 Czech Koruna. Scissors Player 2 Paper Stone Scissors 0 -1 1 Paper 1 0 -1 Stone -1 1 0 No Nash equilibrium (no solution in pure strategies) But we all know how to play this game! x*T = (1/3, 1/3, 1/3) and y*T = (1/3, 1/3, 1/3) (!!) Player 1 knows the strategy of Player 2 (and vice versa), but he or she cannot use this information against Player 2. Basic Theorem on Matrix Games (John von Neumann, 1928) For any matrix A, there exist vectors of mixed strategies x* and y* such that xTA y* ≤ x*TA y* ≤ x*TA y for all mixed strategies x and y. Any matrix game has a solution in the mixed strategies. Calculation of mixed strategies A. Graphical solution (game 2x2) B. Simple formula (game 2x2) based on row and column differences C. Linear Programming 1. If there is a negative element in the payoff matrix, make all elements of the matrix positive by adding the same positive number to all elements of the matrix. You should know that games A and A + cE are strategically equivalent. E is the matrix with all elements equal to 1; c is the real number. 2. Solve linear programming problem: maximize p1 + p2 + … + pn (objective function) (n is the number of strategies of Player 2 = number of columns) a11p1 + a12p2 + … + a1npn ≤ 1 …………………………….. am1p1 + am2p2 + … + amnpn ≤ 1 p1 0; p2 0; …pn 0. (constraints) (m is the number of rows) (sign restrictions) 3. Divide the primal and dual solutions by the optimal value of the objective function. The primal solution is the strategy of Player 2. The dual solution is the strategy of Player 1. EXAMPLE 1 Player 1 1 0 1 2 Player 2 2 1.5 4 3 -1 2 1. Negative element, let c = 2 1 Player 1 1 2 3 2 Player 2 2 3.5 6 3 1 4 2. Solve linear programming problem maximize p1 + p2 + p3 (objective function) (n is the number of strategies of Player 2 = number of columns) 3p1 + 3.5p2 + 1p3 ≤ 1 2p1 + 6p2 + 4p3 ≤ 1 (m is the number of rows) p1 0; p2 0; p3 0. Solution: Objective function is 0.4 The primal solution pT = (0.3, 0, 0.1) The dual solution qT = (0.2, 0.2) 3. Divide the primal and dual solutions by the optimal value of the objective function. The primal solution is the strategy of Player 2. The dual solution is the strategy of Player 1. x*T = q T /0.4 = (0.5, 0.5) y* T = p T /0.4 =(0.75, 0, 0.25) Value of the game is x*TA y* or (1/objective function). Value of the original game is (1/objective function - c) 1/0.4 – 2 = 2.5 – 2 = 0.5