Non-zero-sum Game and Nash Equilibarium Team nogg December 21, 2016 - PowerPoint PPT Presentation

Non-zero-sum Game and Nash Equilibarium Team nogg December 21, 2016

Overview

Prisoner’s Dilemma Prisoner’s Dilemma: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) • Bob Confess (0,-9) (-6,-6)

Prisoner’s Dilemma Prisoner’s Dilemma: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) • Bob Confess (0,-9) (-6,-6) • S B = { C , D } S A = { C , D }

Prisoner’s Dilemma Prisoner’s Dilemma: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) • Bob Confess (0,-9) (-6,-6) • S B = { C , D } S A = { C , D } • u B ( D , D ) = − 1 u A ( D , D ) = − 1

Prisoner’s Dilemma Prisoner’s Dilemma: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) • Bob Confess (0,-9) (-6,-6) • S B = { C , D } S A = { C , D } • u B ( D , D ) = − 1 u A ( D , D ) = − 1 • u B ( D , C ) = − 9 u A ( D , C ) = 0 u B ( C , D ) = 0 u A ( C , D ) = − 9 u B ( C , C ) = − 6 u A ( C , C ) = − 6

What is a Non-zero Sum Game? • The sum of each player’s gain or loss � = what they begin with. • ∃ ( s 1 , s 2 , ..., s n ) ∈ S 1 × S 2 × ... × S n , � n i =1 u i ( s 1 , s 2 , ..., s n ) � = 0

Strict Domination Assuming you are Bob, what should you do? Alice Deny Alice Confess • Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6)

Strict Domination Assuming you are Bob, what should you do? Alice Deny Alice Confess • Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) • What if Alice chooses to Deny ?

Strict Domination Assuming you are Bob, what should you do? Alice Deny Alice Confess • Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) • What if Alice chooses to Deny ? • Result: Bob is free and Alice will spend 9 years.

Strict Domination Assuming you are Bob, what should you do? Alice Deny Alice Confess • Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) • What if Alice chooses to Deny ? • Result: Bob is free and Alice will spend 9 years. • What if Alice chooses to Confess ?

Strict Domination Assuming you are Bob, what should you do? Alice Deny Alice Confess • Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) • What if Alice chooses to Deny ? • Result: Bob is free and Alice will spend 9 years. • What if Alice chooses to Confess ? • Result: Bob and Alice will spend 6 years together.

Strict Domination Assuming you are Bob, what should you do? Alice Deny Alice Confess • Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) • What if Alice chooses to Deny ? • Result: Bob is free and Alice will spend 9 years. • What if Alice chooses to Confess ? • Result: Bob and Alice will spend 6 years together. • In both cases, Bob will definitely choose to confess.

Strict Domination The process is like: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6)

Strict Domination The process is like: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) ⇓ Alice Deny Alice Confess Bob Confess (0,-9) (-6,-6)

Strict Domination The process is like: Alice Deny Alice Confess Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6) ⇓ Alice Deny Alice Confess Bob Confess (0,-9) (-6,-6) ⇓ Alice Confess Bob Confess (-6,-6)

Strict Domination • If one of the players strategies is never the right thing to do, no matter what the opponents do, it is Strictly Dominated .

Strict Domination • If one of the players strategies is never the right thing to do, no matter what the opponents do, it is Strictly Dominated . • Get rid of the strictly dominated strategies because they won’t happen .

Strict Domination • If one of the players strategies is never the right thing to do, no matter what the opponents do, it is Strictly Dominated . • Get rid of the strictly dominated strategies because they won’t happen . • This is called iterated elimination of dominated strategies .

Bob and Alice • Bob and Alice are students in some school. • Bob loves Alice but Alice dont like Bob. • The situation arises when they decide where to eat lunch.

Restaurant • Restaurant No.1’s food is awful. • Restaurant No.3’s food is better.

Their Pay-off when eating Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) Bob go to No.1

Their Pay-off when eating Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1

Their Pay-off when eating Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10)

Their Pay-off when eating Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0)

How will they act? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) * (3,5) Bob go to No.1 (0,10) (7,0) • If Bob claims in WeChat that he will go to No.3 and Alice claims that she will go to No.3. • What will happen?

How will they act? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) * Bob go to No.1 (0,10) (7,0) • Alice will choose to go to No.3 restaurant.

How will they act? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0) * • Then Bob will choose to go to No.3 restaurant.

How will they act? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) * (7,0) • Then Alice will choose to go to No.3 restaurant.

How will they act? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) * (3,5) Bob go to No.1 (0,10) (7,0) • Then Bob will choose to go to No.3 restaurant. • It is a circulation!!

What should they do? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0) • When the two persons are stuck in this dilemma, a third person comes out and say, ”Why don’t you just choose the restaurant by probability? ” • That’s it!

What should they do? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0) • Supposed that Alice will choose No.3 by probability a 1 and choose No.1 by a 2 .

What should they do? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0) • Supposed that Alice will choose No.3 by probability a 1 and choose No.1 by a 2 . • Then if Bob go to No.3, his pay-off will be 10 a 1 + 3 a 2 . If he go to No.1, then pay-off will be 7 a 2 .

What should they do? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0) • Supposed that Alice will choose No.3 by probability a 1 and choose No.1 by a 2 . • Then if Bob go to No.3, his pay-off will be 10 a 1 + 3 a 2 . If he go to No.1, then pay-off will be 7 a 2 . • 10 a 1 + 3 a 2 must be equal to 7 a 2 , otherwise Bob can decide indeed which restaurant to go. And it will be circulation again.

What should they do? 10 a 1 + 3 a 2 = 7 a 2 a 1 + a 2 = 1 So a 1 = 2 7 , a 2 = 5 7

What should they do? Alice go to No.3 Alice go to No.1 Bob go to No.3 (10,4) (3,5) Bob go to No.1 (0,10) (7,0) • Supposed that Bob will choose No.3 by probability b 1 and choose No.1 by b 2 . • Based on the same method, we can get that b 1 = 10 11 , b 2 = 1 11 • If both of them choose based on the probability, then it is a equilibrium.

More Complicate situation. Alice can choose from ( x 1 , x 2 , ..., x i ), and the probability vector will be − → x . Bob can choose from ( y 1 , y 2 , ..., y j ), and the probability vector will be − → y . A( − → x , − → y ) means Alice’s paid-off. B( − → x , − → y ) means Bob’s paid-off.

More Complicate situation. Suppose Alice plays − → x . Can Bob do better than B( − → x , − → y )? That is: ∃− → s . t . B ( − → x , − → v ) > B ( − → x , − → v y )?

Nash Equilibrium DEF: − → x , − → y is a Nash equilibrium if ∀− → A ( − → x , − → y ) ≥ A ( − → u , − → u y ) ∀− → B ( − → x , − → y ) ≥ B ( − → x , − → v v )

Nash’s Theorem • Nash’s Theorem : Every game with a finite number of players and a finite number of actions available to each player has a Nash equilibrium.

Nash’s Theorem • Nash’s Theorem : Every game with a finite number of players and a finite number of actions available to each player has a Nash equilibrium. • As for Bob and Alice, there must be a point that they won’t change their strategies.

How to prove it? • Nash’s original proof of it used Kakutani’s fixed point theorem .

How to prove it? • Nash’s original proof of it used Kakutani’s fixed point theorem . • But a year later Nash simplified his proof to only use Brouwer’s fixed point theorem .

How to prove it? • Nash’s original proof of it used Kakutani’s fixed point theorem . • But a year later Nash simplified his proof to only use Brouwer’s fixed point theorem .

Brouwer’s fixed point theorem • Brouwer’s fixed point theorem : Let D be a convex, compact subset of the Euclidean space. If f : D − → D is continuous, then there exists x ∈ D such that f ( x ) = x .