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)

• SB = {C, D}

SA = {C, D}

Prisoner’s Dilemma

Prisoner’s Dilemma:

• Alice Deny

Alice Confess Bob Deny (-1,-1) (-9,0) Bob Confess (0,-9) (-6,-6)

• SB = {C, D}

SA = {C, D}

• uB(D, D) = −1

uA(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)

• SB = {C, D}

SA = {C, D}

• uB(D, D) = −1

uA(D, D) = −1

• uB(D, C) = −9

uA(D, C) = 0 uB(C, D) = 0 uA(C, D) = −9 uB(C, C) = −6 uA(C, C) = −6

What is a Non-zero Sum Game?

• The sum of each player’s gain or loss = what they begin with.
• ∃(s1, s2, ..., sn) ∈ S1 × S2 × ... × Sn, n

i=1 ui(s1, s2, ..., sn) = 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 a1 and

choose No.1 by a2.

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 a1 and

choose No.1 by a2.

• Then if Bob go to No.3, his pay-off will be 10a1 + 3a2. If he

go to No.1, then pay-off will be 7a2.

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 a1 and

choose No.1 by a2.

• Then if Bob go to No.3, his pay-off will be 10a1 + 3a2. If he

go to No.1, then pay-off will be 7a2.

• 10a1 + 3a2 must be equal to 7a2, otherwise Bob can decide

indeed which restaurant to go. And it will be circulation again.

What should they do?

10a1 + 3a2 = 7a2 a1 + a2 = 1 So a1 = 2

7, a2 = 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 b1 and

choose No.1 by b2.

• Based on the same method, we can get that b1 = 10

11, b2 = 1 11

• If both of them choose based on the probability, then it is a

equilibrium.

More Complicate situation.

Alice can choose from (x1, x2, ..., xi), and the probability vector will be − → x . Bob can choose from (y1, y2, ..., yj), 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: ∃− → v s.t. B(− → x , − → v ) > B(− → x , − → y )?

Nash Equilibrium

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

Brouwer’s fixed point theorem

• Examples:
• Take an ordinary map of a country, and suppose that that map

is laid out on a table inside that country. There will always be a ”You are Here” point on the map which represents that same point in the country.

Gain function

• Now we introduce the idea of Gain function:

GainBob(− → x , − → y , i) = max{B(− → ei , − → y ) − B(− → x , − → y ), 0} GainAlice(− → x , − → y , j) = max{A(− → x , − → ej ) − A(− → x , − → y ), 0}

Gain function

• Now we introduce the idea of Gain function:

GainBob(− → x , − → y , i) = max{B(− → ei , − → y ) − B(− → x , − → y ), 0} GainAlice(− → x , − → y , j) = max{A(− → x , − → ej ) − A(− → x , − → y ), 0}

• In other words, the Gain is equal to the increase in payoff for

a player if he were to switch to another strategy.

Gain function

• Now we introduce the idea of Gain function:

GainBob(− → x , − → y , i) = max{B(− → ei , − → y ) − B(− → x , − → y ), 0} GainAlice(− → x , − → y , j) = max{A(− → x , − → ej ) − A(− → x , − → y ), 0}

• In other words, the Gain is equal to the increase in payoff for

a player if he were to switch to another strategy.

• Obviously, the Gain for all players is 0 in Nash Equilibrium.

Proof of Nash’ Theorem

• Now we define a function as follows:

f (− → x , − → y , i) = xi + GainBob(− → x , − → y , i) 1 +

i GainBob(−

→ x , − → y , i) g(− → x , − → y , j) = yj + GainAlice(− → x , − → y , j) 1 +

j GainAlice(−

→ x , − → y , j)

Proof of Nash’ Theorem

• Now we define a function as follows:

f (− → x , − → y , i) = xi + GainBob(− → x , − → y , i) 1 +

i GainBob(−

→ x , − → y , i) g(− → x , − → y , j) = yj + GainAlice(− → x , − → y , j) 1 +

j GainAlice(−

→ x , − → y , j)

• In other words, function f and g tries to boost the probability

mass that player places on various pure strategies depending

• n the each one’s gains in payoff the player would get by

switching to these strategies.

Proof of Nash’ Theorem

• These function is a map from a 2-dimension space to a

2-dimension space. (− → x , − → y )− →( − → x′, − → y′)

Proof of Nash’ Theorem

• It is easy to see that this function is continuous. So we can

use Brouwer’s fixed point theorem, there is at least one fixed point of the function.

Proof of Nash’ Theorem

• It is easy to see that this function is continuous. So we can

use Brouwer’s fixed point theorem, there is at least one fixed point of the function.

• For any fixed point

GainBob(− → x , − → y , i) = 0, ∀i ∈ [n] GainAlice(− → x , − → y , j) = 0, ∀j ∈ [n] It can be proved by by contradiction.

Proof of Nash’ Theorem

• It is easy to see that this function is continuous. So we can

use Brouwer’s fixed point theorem, there is at least one fixed point of the function.

• For any fixed point

GainBob(− → x , − → y , i) = 0, ∀i ∈ [n] GainAlice(− → x , − → y , j) = 0, ∀j ∈ [n] It can be proved by by contradiction.

• Then we claim that any fixed point of this function is a Nash

equilibrium.

The smile of John Nash

The End