Decision Theory Game Theory

Decision 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

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;

The Game: Payoff Matrix

Player A

(Row Player)

Player B

1

– 1

T

– 1

1 H

T H

Strategies of each player : H or T

Optimum Solution

 Optimum Solution can be achieved if

players choose to apply:

 _{Pure strategy (e.g. select H or T)}

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

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

Two-Person Zero-Sum Game with

Saddle Point

 Saddle point leads to Optimum Solution  Saddle point indicates a stable game

In general

 To maintain the “Optimality” of a game:

maximin value  value of the game  minimax value OR

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:

Unstable Game without Saddle Point 15 9 7 8 Column Max –1 3 –1 4 7 4 4 15 4 5 8 3 2 2 8 7 6 2 –10 0 9 –10 5 1 Player A Min 4 3 2 1 Row Player B Minimax Value Maximin Value

2



N game

 2  N game:

 Player A has 2 strategies

 Player B has N ( 2) strategies

a_2n …

a₂₂ a₂₁

x₂ = 1 – x₁

a_1n …

a₁₂ a₁₁

x₁ A

y_n …

y₂ y₁

2



N game

(a₁₂ – a₂₂)x₁ + a₂₂ 2

(a₁₁ – a₂₁)x₁ + a₂₁ 1

… …

(a_1n – a_2n)x₁ + a_2n n

2



N game: Example

B

6 2

3 4

x₂

–1 3

2 2

x₁ A

y₄ y₃

2



N game

– x₁ + 3 2

– 2x₁ + 4 1

x₁ + 2 3

–7 x₁ + 6 4

A’s Expected Payoff B’s Pure Strategy

Graphical Solution x₁ = 0 and x₁ = 1 = x₂

x₁ = 0 x₁ = 1

1

5

2 3

4 6

-1

4 2

x*₁ =1/2

3 1

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

Optimum Solution for Player B

 Combination (2), (3) and (4):

Optimum Solution for Player B

y₂ + 2 3

– y₂ + 3 2

B’s Expected Payoff A’s Pure

Strategy

(2,3)  y₁ and y₄ = 0, y₃ = y₂ –1 (y₂* = y₃*)

– y₂ + 3 = y₂ + 2 – 2 y₂ = – 1

y₂* = 1_{/2 dan y}

3* = 1/2

Optimum Solution for Player B

– 7y₂ + 6 4

– y₂ + 3 2

B’s Expected Payoff A’s Pure

Strategy

(2,4)  y₁ and y₃ = 0, y₄ = y₂ –1 (y₂* = y₄*)

– y₂ + 3 = –7y₂ + 6 6 y₂ = 3

y₂* = 1_{/2 dan y}

4* = 1/2

Optimum Solution for Player B

– 7y₃ + 6 4

y₃ + 2 3

B’s Expected Payoff A’s Pure

Strategy

(3,4)  y₁ and y₂ = 0, y₄ = y₃ –1 (y₃* = y₄*)

y₃ + 2 = –7y₃ + 6 8 y₃ = 4

y₃* = 1_{/2 dan y}

4* = 1/2

M



2 game

 M 2 game:

 _{Player A has M (} _{2) strategies}  _{Player B has 2 strategies}

… …

…

a₁₂ a₁₁

x₁ A

y₂₌ 1 – y₁ y₁

a_m2 a_m1

x_m

a₂₂ a₂₁

x₂

M



2 game

(a₂₁ – a₂₂)y₁ + a₂₂ 2

(a₁₁ – a₁₂)y₁ + a₁₂ 1

… …

(a_m1 – a_m2)y₁ + a_m2 m

M



2 game: Example

B

A x2 3 2

4 2

x₁

6 – 2

x₃

M



2 game

y₁ + 2 2

– 2 y₁ + 4 1

– 8 y₁ + 6 3

B’s Expected Payoff A’s Pure Strategy

Graphical Solution y₁ = 0 and y₁ = 1 = y₂

y₁ = 1

1

5

2 3 4 6

-1

2

1 3

-2

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

(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):

Optimum Solution for Player A

– 8x₁ +6 3

– 2x₁ + 4 1

A’s Expected Payoff B’s Pure Strategy

–2x₁ + 4 = – 8x₁ +6 6 x₁ = 2

x₁* = 1_{/3 dan x}

3* = 1/3

A’s gain = 10_/3

M



N Games: Simplex

 Focus on Row (Player A)  Duality Problem

 Objective Function: maximize

M



N Games: Simplex

 Subject to (Constraints):

a₁₁ Y₁ + a₁₂ Y₂ + . . . + a_1nY_n  1 a₂₁ Y₁ + a₂₂ Y₂ + . . . + a_2nY_n  1

… … …

a_m1 Y₁ + a_m2 Y₂ + . . . + a_mnY_n  1 Y₁, Y₂, . . . , Y_n  0

 w = 1/v  v* = 1/w

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

M



N Games: Simplex

 If K is used in the tableau,

v* = 1/w – K

 z = w

M



N Games: Example

3 3 3 Max Column –4 3 –3 –4 3 –3 –1 3 –3 2 –3 –3 –1 3 1 Min 3 2 1 Row B A

8 8 8 Max Column 1 8 2 1 3 2 4 8 2 2 2 2 4 8 1 Min 3 2 1 Row B A

8 8 8 Max Column 1 8 2 1 3 2 4 8 2 2 2 2 4 8 1 Min 3 2 1 Row B A

8Y₁ + 4Y₂ + 2Y₃  1 Subject to :

2Y₁ + 8Y₂ + 4Y₃  1 1Y₁ + 2Y₂ + 8Y₃  1

8Y₁ + 4Y₂ + 2Y₃  1  8Y₁ + 4Y₂ + 2Y₃ + S₁ = 1 Subject to :

2Y₁ + 8Y₂ + 4Y₃  1  2Y₁ + 8Y₂ + 4Y₃ + S₂ = 1

1Y₁ + 2Y₂ + 8Y₃  1 1Y₁ + 2Y₂ + 8Y₃ + S₃ = 1

Y₁, Y₂,Y₃  0

5/49 1/7 -3/98 -1/98 1 0 0 Y₃ 11/96 -1/14 31/196 -3/98 0 1 0 Y₂ 1/14 0 -1/14 1/7 0 0 1 Y₁ 45/196 1/14 11/196 5/49 0 0 0 w Solution S₃ S₂ S₁ Y₃ Y₂ Y₁ Basic

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

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

The End