Decision Theory Game Theory

Decision Theory &
Game Theory
Game Theory

Elements of a game
Players: intelligent opponents (competitors
or enemies)
 Strategies: choices of what to do to defeat
the opponents
 Outcomes = Payoffs: functions of the
different strategies for each player
 Payoff Matrix: Tableau (of the gain of the
Row Player)
 Rule: how to allocate the payoffs to the
players


The Game: Example






Two Players: Player A (Row) and Player B
(Column)
Tossing a balanced coin
Possible outcomes: Head (H) and Tail (T)
Rule:
 If

the outcome matches (player A selects H and the
outcome is also H or player A select T and the
outcomes is also T), Player A gets $1 from Player B;
 Otherwise Player A loses $1 to Player B

The Game: Payoff Matrix
Player B
Player A
(Row Player)

H

T

H
1
–1

T
–1

1

Strategies of each player : H or T
An Optimum Solution is said to be reached if
neither player finds it beneficial to alter his
strategy

Optimum Solution


Optimum Solution can be achieved if

players choose to apply:



Pure strategy (e.g. select H or T)
A Mixture of pure strategies = Mixed Strategy

Two-Person Zero-Sum Game








A game with two players
A gain of one player equals a loss to the other
The focused player = the row player (Player A)
Player A maximize his minimum gain (why?)

Player B minimize her maximum loss (why?)
Optimum Solution is gained by Minimax-Maximin
criterion
Optimum Solution reflects that the game is
stable or in the a state of equilibrium

Two-Person Zero-Sum Game with
Saddle Point
1

Player B
2
3

4

Row
Min

1

8

2

9

5

2

Player A 2

6

5

7

18

5

3

7

3

–4

10

–4

Column Max 8

5

9

18

Minimax
Value

Maximin
Value

Two-Person Zero-Sum Game with
Saddle Point
Player A (Row): Maximin Value = Lower
Value of the game
 Player B (Column): Minimax Value =
Upper Value of the game
 Maximin value = Minimax value  Saddle
point = Value of the game


Two-Person Zero-Sum Game with

Saddle Point
Saddle point leads to Optimum Solution
 Saddle point indicates a stable game
 Players apply Pure Strategy


In general


To maintain the “Optimality” of a game:
maximin value  value of the game  minimax value
OR
lower value  value of the game  upper value

Mixed Strategies
Used for solving a game that does not
have a saddle point
 Used for solving a game that does not
have a saddle point
 Optimum Solution is gained using:






Graphical Solution for (2  N) and (M  2)
Simplex for (M  N) game

Unstable Game without Saddle Point
Player B

Player A

Column
Max

2
3
–10 9

Row

1

1
5

4
0

Min
–10

2

6

7

8

2

2

3

8

5

4

15

4 Maximin

4

7

4

–1

3

–1

8

7

9

15

Value

Minimax
Value

Minimax value = 7 > Maximin value = 4  sub-optimal

2  N game


2  N game:



Player A has 2 strategies
Player B has N ( 2) strategies
B

A

y1

y2

… yn

a11

a12

… a1n

x2 = 1 – x1 a21

a22

… a2n

x1

2  N game
B’s Pure Strategy A’s Expected Payoff
1

(a11 – a21)x1 + a21

2

(a12 – a22)x1 + a22

…

…

n

(a1n – a2n)x1 + a2n

2  N game: Example
B

A

y1

y2

y3

y4

x1

2

2

3

–1

x2

4

3

2

6

2  N game
B’s Pure Strategy

A’s Expected Payoff

1

– 2x1 + 4

2

– x1 + 3

3

x1 + 2

4

–7 x1 + 6
Optimum Solution: Graphical Solution

Graphical Solution x1 = 0 and x1 = 1 = x2
6
5

Maximin
1

4
3
3

2
2

1
4

x1 = 1

x1 = 0

-1

x*1 =1/2

Optimum Solution for Player A
Intercept between lines (2), (3) and (4)
(x1* = ½, x2*= ½)
(2) – x1 + 3 = – ½ + 3 = 5/2
v*
(3) x1 + 2 = ½ + 2 = 5/2
(4) –7 x1 + 6 = – 7/2 + 6 = 5/2
 Player B can mix all the 3 strategies
 Player A’s gains = 5/2


Optimum Solution for Player B
Combination (2), (3) and (4):
 (2,3)  y1 and y4 = 0, y3 = y2 –1 (y2* = y3*)
 (2,4)  y1 and y3 = 0, y4 = y2 –1 (y2* = y4*)
 (3,4)  y1 and y2 = 0, y4 = y3 –1 (y3* = y4*)


Optimum Solution for Player B
(2,3)  y1 and y4 = 0, y3 = y2 –1 (y2* = y3*)

A’s Pure
Strategy
2

B’s Expected Payoff

3

y2 + 2

– y2 + 3 = y2 + 2
– 2 y2 = – 1
y2* = 1/2 dan y3* = 1/2

– y2 + 3

B’s loss = 5/2

Optimum Solution for Player B
(2,4)  y1 and y3 = 0, y4 = y2 –1 (y2* = y4*)

A’s Pure
Strategy
2

B’s Expected Payoff

4

– 7y2 + 6

– y2 + 3 = –7y2 + 6
6 y2 = 3
y2* = 1/2 dan y4* = 1/2

– y2 + 3

B’s loss = 5/2

Optimum Solution for Player B
(3,4)  y1 and y2 = 0, y4 = y3 –1 (y3* = y4*)

A’s Pure
Strategy
3

B’s Expected Payoff

4

– 7y3 + 6

y3 + 2 = –7y3 + 6
8 y3 = 4
y3* = 1/2 dan y4* = 1/2

y3 + 2

B’s loss = 5/2

M  2 game


M 2 game:



Player A has M ( 2) strategies
Player B has 2 strategies
B
x1
A
x2
…

y1
a11
a21
…

y2= 1 – y1
a12
a22
…

xm

am1

am2

M  2 game
A’s Pure Strategy B’s Expected Payoff
1

(a11 – a12)y1 + a12

2

(a21 – a22)y1 + a22

…

…

m

(am1 – am2)y1 + am2

M 2 game: Example
B

A

y1

y2

x1

2

4

x2

3

2

x3

–2

6

M  2 game
A’s Pure Strategy

B’s Expected Payoff

1

– 2 y1 + 4

2

y1 + 2

3

– 8 y1 + 6
Optimum Solution: Graphical Solution

Graphical Solution y1 = 0 and y1 = 1 = y2
6

Minimax
5
3

4
2

3

1

2
y1 = 1

1

-1
-2

y1* = y3* = 1/3

Optimum Solution for Player B


Intercept between lines (1) and (3)
(y1* = 1/3, y3*= 1/3)
(1) – 2y1 + 4 = – 2/3 + 4 = 10/3
v*
(3) – 8y1 + 6 = – 8/3 + 6 = 10/3

Player B can mix all the 2 strategies
 Player B’s loss = 10/3


Optimum Solution for Player A
Combination (1) and (3):
 (1,3)  x2 and x4 = 0, x3 = x1 –1 (x1* = x3*)


Optimum Solution for Player A
(1,3)  x2 and x4 = 0, x3 = x1 –1 (x1* = x3*)
B’s Pure Strategy

A’s Expected Payoff

1

– 2x1 + 4

3

– 8x1 +6

–2x1 + 4 = – 8x1 +6
6 x1 = 2
x1* = 1/3 dan x3* = 1/3

A’s gain = 10/3

M  N Games: Simplex
Focus on Row (Player A)
 Duality Problem
 Objective Function: maximize
w = Y1 + Y2 + . . . Yn


M  N Games: Simplex
Subject to (Constraints):
a11 Y1 + a12 Y2 + . . . + a1nYn  1
a21 Y1 + a22 Y2 + . . . + a2nYn  1
…
…
…
am1 Y1 + am2 Y2 + . . . + amnYn  1
Y1, Y2, . . . , Yn  0
 w = 1/v 
v* = 1/w
 Yj = Yi /v,
j = 1,2,. . . , n


M  N Games: Simplex
Ensure the tableau does not contain any
zero and negative value
 Use K (a constant value) to make sure that
the tableau does not contain any zero and
negative value
 K > negative of the maximin value
 K > negatif of the most negative value


M  N Games: Simplex
If K is used in the tableau,
v* = 1/w – K
z=w
 X1* = X1/z, X2* = X2/z, . . . , Xm* = Xm/z


M  N Games: Example
B
A

Column

Row

1

2

3

Min

1

3

–1

–3

–3

2

–3

3

–1

–3

3

–4

–3

3

–4

Max

3

3

3

K=5

B
A

Column

Row

1

2

3

Min

1

8

4

2

2

2

2

8

4

2

3

1

2

8

1

Max

8

8

8

Objective Function:
Maximize: w = Y1 + Y2 + Y3

B
A

Column

Row

1

2

3

Min

1

8

4

2

2

2

2

8

4

2

3

1

2

8

1

Max

8

8

8

Subject to :

8Y1 + 4Y2 + 2Y3  1
2Y1 + 8Y2 + 4Y3  1
1Y1 + 2Y2 + 8Y3  1
Y1, Y2,Y3  0

Subject to :
8Y1 + 4Y2 + 2Y3  1  8Y1 + 4Y2 + 2Y3 + S1 = 1
2Y1 + 8Y2 + 4Y3  1  2Y1 + 8Y2 + 4Y3 + S2 = 1
1Y1 + 2Y2 + 8Y3  1 1Y1 + 2Y2 + 8Y3 + S3 = 1
Y1, Y2,Y3  0
Objective Function:
Maximize: w = Y1 + Y2 + Y3 + S1+S2+S3

Basic Y1 Y2 Y3 S1 S2

S3 Solution

w

-1

-1

-1

0

0

0

0

S1

8

4

2

1

0

0

1

S2
S3

2
1

8
2

4
8

0
0

1
0

0
1

1
1

Final Optimal Tableau
Basic Y1 Y2 Y3

S1

S2

S3

w

0

0

0

5/49 11/196 1/14

Y1

1

0

0

1/7

Y2
Y3

0
0

1
0

0
1

-1/14

0

-3/98 31/196 -1/14
-1/98 -3/98
1/7

Solution
45/196
1/14
11/96
5/49

Solution for B
w = 45/196
 v* = 1/w – K = 196/45 – 225/45 = –29/45
 y1* = Y1/w = (1/14)/(45/196) = 14/45
 y2* = Y2/w = (11/196)/(45/196) = 11/45
 y3* = Y3/w = (5/49)/(45/196) = 20/45


Solution for A
z = w = 45/196
 X1 = 5/49
 X2 = 11/196
 X3 = 1/14
 x1* = X1/z = (5/49)/(45/196) = 20/45
 x2* = X2/z = (11/196)/(45/196) = 11/45
 x3* = X3/z = (1/14)/(45/196) = 14/45


The End


This is the end of Chapter 8B