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