lec 24 game applications

Lec 24 Game Applications

Chapter 29

Nash Equilibrium In any Nash equilibrium (NE) each player

chooses a “best” response to the choices made by all of the other players.

A game may have more than one NE. How can we locate every one of a

game’s Nash equilibria? If there is more than one NE, can we

argue that one is more likely to occur than another?

Best Responses Think of a 2×2 game; i.e., a game with

two players, A and B, each with two actions.

A can choose between actions aA1 and aA

2. B can choose between actions aB

1 and aB2.

There are 4 possible action pairs;(aA

1, aB1), (aA

1, aB2), (aA

2, aB1), (aA

2, aB2).

Each action pair will usually cause different payoffs for the players.

Best Responses Suppose that A’s and B’s payoffs when

the chosen actions are aA1 and aB

1 areUA(aA

1, aB1) = 6 and UB(aA

1, aB1) = 4.

Similarly, suppose thatUA(aA


1, aB2) = 5

UA(aA2, aB

1) = 4 and UB(aA2, aB

1) = 3UA(aA


2, aB2) = 7.

Best Responses UA(aA


1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB


2) = 7.



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB


2) = 7. If B chooses action aB

1 then A’s best response is ??



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB



1 then A’s best response is action aA

1 (because 6 > 4).



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB




1 (because 6 > 4). If B chooses action aB

2 then A’s best response is ??



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB




1 (because 6 > 4). If B chooses action aB


2 (because 5 > 3).

Best Responses

If B chooses aB1 then A chooses aA

1. If B chooses aB

2 then A chooses aA2.

A’s best-response “curve” is therefore

A’s bestresponse

aA1

aA2

aB2aB

1 B’s action

+

+



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB


2) = 7.



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB


2) = 7. If A chooses action aA

1 then B’s best response is ??



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB



1 then B’s best response is action aB

2 (because 5 > 4).



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB




2 (because 5 > 4). If A chooses action aA

2 then B’s best response is ??.



1, aB1) = 4

UA(aA1, aB


2) = 5UA(aA


2, aB1) = 3

UA(aA2, aB




2 (because 5 > 4). If A chooses action aA


2 (because 7 > 3).

Best Responses If A chooses aA

1 then B chooses aB2.

If A chooses aA2 then B chooses aB

2. B’s best-response “curve” is

therefore

A’s action

aA1

aA2

aB2aB

1 B’s best response

Best Responses If A chooses aA

1 then B chooses aB2.

If A chooses aA2 then B chooses aB

2. B’s best-response “curve” is

therefore

A’s action

aA1

aA2

aB2aB

1 B’s best response

Notice that aB2 is a

strictly dominantaction for B.

Best Responses & Nash Equilibria

A’s response

aA1

aA2

aB2aB

1

aA1

aA2

aB2aB

1

+

+

A’s choice

B’s choice B’s response

How can the players’ best-response curves beused to locate the game’s Nash equilibria?

B A


A’s response

aA1

aA2

aB2aB

1

aA1

aA2

aB2aB

1

+

+

A’s choice


How can the players’ best-response curves beused to locate the game’s Nash equilibria? Put one curve on top of the other.B A

How can the players’ best-response curves beused to locate the game’s Nash equilibria? Put one curve on top of the other.


A’s response

aA1

aA2

aB2aB

1

aA1

aA2

aB2aB

1

+

+

A’s choice


B A



A’s response

aA1

aA2

aB2aB

1

+

+

B’s response

Is there a Nash equilibrium?



A’s response

aA1

aA2

aB2aB

1

+

+

Is there a Nash equilibrium?Yes, (aA

2, aB2). Why?

B’s response



A’s response

aA1

aA2

aB2aB

1

+

+

Is there a Nash equilibrium?Yes, (aA

2, aB2). Why?

aA2 is a best response to aB

2.aB

2 is a best response to aA2.

B’s response


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

aA2 is the only best response to aB

2. aB

2 is the only best response to aA2.

Here is the strategicform of the game.


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A



2. aB


Is there a 2nd Nasheqm.?


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

Is there a 2nd Nasheqm.? No, becauseaB

2 is a strictlydominant actionfor Player B.


2. aB




Now allow both players to randomize (i.e., mix)over their actions.

6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


Now allow both players to randomize (i.e., mix)over their actions.

6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.

Given B1, what

value of A1 is best

for A?


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

EVA(aA1) = 6B

1 + 3(1 - B1) = 3 + 3B

1.

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.

Given B1, what

value of A1 is best

for A?


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.

Given B1, what

value of A1 is best

for A?EVA(aA1) = 6B

1 + 3(1 - B1) = 3 + 3B

1.EVA(aA

2) = 4B1 + 5(1 - B

1) = 5 - B1.


A1 is the prob. A chooses action aA

1.B

1 is the prob. B chooses action aB1.

Given B1, what value of A

1 is best for A?EVA(aA

1) = 3 + 3B1.

EVA(aA2) = 5 - B

1.3 + 3B

1 5 - B1 as B

1 ??>=<

>=<



1.B




1) = 3 + 3B1.

EVA(aA2) = 5 - B

1.3 + 3B

1 5 - B1 as B

1 ½.>=<

>=<



1.B




1) = 3 + 3B1.

EVA(aA2) = 5 - B

1.3 + 3B

1 5 - B1 as B

1 ½. A’s best response is:

aA1 if B

1 > ½aA

2 if B1 < ½

aA1 or aA

2 if B1 = ½

>=<>=<



1.B




1) = 3 + 3B1.

EVA(aA2) = 5 - B

1.3 + 3B

1 5 - B1 as B

1 ½. A’s best response is:

aA1 (i.e. A

1 = 1) if B1 > ½

aA2 (i.e. A

1 = 0) if B1 < ½

aA1 or aA

2 (i.e. 0 A1 1) if B

1 = ½

>=<>=<


0

A1

1 B10

A’s best response

½

A’s best response is: aA1 (i.e. A

1 = 1) if B1

> ½ aA

2 (i.e. A1 = 0) if B

1 < ½ aA

1 or aA2 (i.e. 0 A

1 1) if B

1 = ½1


0

A1

1 B10

A’s best response

½

1


1 = 1) if B1

> ½ aA

2 (i.e. A1 = 0) if B

1 < ½ aA

1 or aA2 (i.e. 0 A

1 1) if B

1 = ½


0

A1

1 B10

A’s best response

½

1 This is A’s best responsecurve when players areallowed to mix over theiractions.


1 = 1) if B1

> ½ aA

2 (i.e. A1 = 0) if B

1 < ½ aA

1 or aA2 (i.e. 0 A

1 1) if B

1 = ½


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.

Given A1, what

value of B1 is best

for B?


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.

Given A1, what

value of B1 is best

for B?EVB(aB1) = 4A

1 + 3(1 - A1) = 3 + A

1.


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. A

chooses action aA1.

B1 is the prob. B

chooses action aB1.

Given A1, what

value of B1 is best

for B?EVB(aB1) = 4A

1 + 3(1 - A1) = 3 + A

1.EVB(aB

2) = 5A1 + 7(1 - A

1) = 7 - 2A1.



1.B


Given A1, what value of B

1 is best for B?EVB(aB

1) = 3 + A1.

EVB(aB2) = 7 - 2A

1.3 + A

1 7 - 2A1 as A

1 ??>=<

>=<



1.B


Given A1, what value of B

1 is best for B?EVB(aB

1) = 3 + A1.

EVB(aB2) = 7 - 2A

1.3 + A

1 < 7 - 2A1 for all 0 A

1 1.



1.B



1 is best for A?EVB(aB

1) = 3 + A1.

EVB(aB2) = 7 - 2A

1.3 + A

1 < 7 - 2A1 for all 0 A

1 1.B’s best response is:

aB2 always (i.e. B

1 = 0 always).


0

A1

1 B10

B’s best response

½

B’s best response is aB2 always (i.e. B

1 = 0 always).

1 This is B’s best responsecurve when players areallowed to mix over theiractions.


0

A1

1 B10

B’s best response

½

1

0

A1

1 B10

A’s best response

½

1

B A


0

A1

1 B10

B’s best response

½

1

A’s best response



0

A1

1 B10

B’s best response

½

1

A’s best response

Is there a Nash equilibrium? Yes. Just one.(A

1, B1) = (0,0);

i.e. A chooses aA2 only

& B chooses aB2 only.


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

Let’s change the game.

3,1


6,4 3,5

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

Here is a new2×2 game.

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

Here is a new2×2 game. Againlet A

1 be the prob.that A chooses aA

1and let B

1 be theprob. that B choosesaB

1. What are the NEof this game?

Notice that Player B no longer has a strictly dominant action.

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A

A1 is the prob. that A

chooses aA1.

B1 is the prob. that B

chooses aB1.

EVA(aA1) = ??

EVA(aA2) = ??

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVA(aA1) = 6B

1 + 3(1 - B1) = 3 + 3B

1. EVA(aA

2) = ??

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVA(aA1) = 6B

1 + 3(1 - B1) = 3 + 3B

1. EVA(aA

2) = 4B1 + 5(1 - B

1) = 5 - B1.

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVA(aA1) = 6B

1 + 3(1 - B1) = 3 + 3B

1. EVA(aA

2) = 4B1 + 5(1 - B

1) = 5 - B1.

3 + 3B1 5 - B

1 as B1 ½.>=<

>=<


EVA(aA1) = 6B

1 + 3(1 - B1) = 3 + 3B

1. EVA(aA

2) = 4B1 + 5(1 - B

1) = 5 - B1.

3 + 3B1 5 - B

1 as B1 ½.>=<

>=<

0

A1

1 B10

A’s best response

½

1

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVB(aB1) = ??

EVB(aB2) = ??

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVB(aB1) = 4A

1 + 3(1 - A1) = 3 + A

1. EVB(aB

2) = ??

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVB(aB1) = 4A

1 + 3(1 - A1) = 4 + A

1. EVB(aB

2) = A1 + 7(1 - A

1) = 7 - 6A1.

3,1


6,4

5,74,3

aA1

aA2

aB1 aB

2

Player B

Player A


chooses aA1.


chooses aB1.

EVB(aB1) = 4A

1 + 3(1 - A1) = 3 + A

1. EVB(aB

2) = A1 + 7(1 - A

1) = 7 - 6A1.

3 + A1 7 - 6A

1 as A1 .

>=<>=<

4 7/


EVB(aB1) = 4A

1 + 3(1 - A1) = 3 + A

1. EVB(aB

2) = A1 + 7(1 - A

1) = 7 - 6A1.

3 + A1 7 - 6A

1 as A1 .

>=<>=<

4 7/

0

A1

1 B10

1

4 7/

B’s best response


0

A1

1 B10

B’s best response

1

0

A1

1 B10

A’s best response

½

1

B A

4 7/


0

A1

1 B10

B’s best response

1

B A

4 7/

0

A1

1 B10

A’s best response

½

1



0

A1

1 B10

B’s best response

1

4 7/

A’s best response

½



0

A1

1 B10

B’s best response

1

4 7/

A’s best response

½

Is there a Nash equilibrium? Yes. 3 of them.


0

A1

1 B10

B’s best response

1

4 7/

A’s best response

½

Is there a Nash equilibrium? Yes. 3 of them.(A

1, B1) = (0,0)


0

A1

1 B10

B’s best response

1

4 7/

A’s best response

½


1, B1) = (0,0)

(A1, B

1) = (1,1)



0

A1

1 B10

B’s best response

1

4 7/

A’s best response

½


1, B1) = (0,0)

(A1, B

1) = (1,1)(A

1, B1) = ( , )½4 7/


Some Important Types of Games

Games of coordination Games of competition Games of coexistence Games of commitment Bargaining games

Coordination Games Simultaneous play games in which the

payoffs to the players are largest when they coordinate their actions. Famous examples are: The Battle of the Sexes Game The Prisoner’s Dilemma Game Assurance Games Chicken

Coordination Games; The Battle of the Sexes

Sissy prefers watching ballet to watching mud wrestling.

Jock prefers watching mud wrestling to watching ballet.

Both prefer watching something together to being apart.


1,28,4

4,82,1

B

MW

B MWJock

Sissy

SB is the prob. that

Sissy chooses ballet.J

B is the prob. thatJock chooses ballet.


1,28,4

4,82,1

B

MW

B MWJock

Sissy



B is the prob. thatJock chooses ballet.What are the players’best-responsefunctions?


1,28,4

4,82,1

B

MW

B MWJock

Sissy




EVS(B) = 8JB + (1 - J

B) = 1 + 7JB.


1,28,4

4,82,1

B

MW

B MWJock

Sissy




EVS(B) = 8JB + (1 - J

B) = 1 + 7JB.

EVS(MW) = 2JB + 4(1 - J

B) = 4 - 2JB.

EVS(B) = 8JB + (1 - J

B) = 1 + 7JB.

EVS(MW) = 2JB + 4(1 - J

B) = 4 - 2JB.

1 + 7JB 4 - 2J

B as JB .1 3/


1,28,4

4,82,1

B

MW

B MWJock

Sissy




>=<>=<

EVS(B) = 8JB + (1 - J

B) = 1 + 7JB.

EVS(MW) = 2JB + 4(1 - J

B) = 4 - 2JB.

1 + 7JB 4 - 2J

B as JB .1 3/


1,28,4

4,82,1

B

MW

B MWJock

Sissy




>=<>=<

SB

JB1

1

00 1 3/

EVS(B) = 8JB + (1 - J

B) = 1 + 7JB.

EVS(MW) = 2JB + 4(1 - J

B) = 4 - 2JB.

1 + 7JB 4 - 2J

B as JB .1 3/


1,28,4

4,82,1

B

MW

B MWJock

Sissy




>=<>=<

SB

JB1

1

00 1 3/

Sissy


SB

JB1

1

00 1 3/

Sissy SB

JB1

1

00

1 3/

Jock


SB

JB1

1

00 1 3/

Sissy SB

JB1

1

00

1 3/

Jock

The game’s NE are ??


SB

JB1

1

00 1 3/

Sissy

Jock


1 3/


SB

JB1

1

00 1 3/

Sissy

Jock

The game’s NE are: (JB, S

B) = (0, 0); i.e., (MW, MW)

1 3/


SB

JB1

1

00 1 3/

Sissy

Jock


B) = (0, 0); i.e., (MW, MW) (J

B, SB) = (1, 1); i.e., (B, B)

1 3/


SB

JB1

1

00 1 3/

Sissy

Jock


B) = (0, 0); i.e., (MW, MW) (J

B, SB) = (1, 1); i.e., (B, B)

(JB, S

B) = ( , ); i.e., bothwatch the ballet with prob. 1/9, both watch the mudwrestling with prob. 4/9, and with prob. 4/9 they watch different events.

1 3/

1 3/1 3/


1,28,4

4,82,1

B

MW

B MWJock

Sissy




For Sissy the expected value of the NE (JB, S

B) = ( , ) is8× + 1× + 2× + 4× = < 4 and 8.1 9/ 2 9/ 2 9/ 4 9/ 10 3/

1 3/ 1 3/


1,28,4

4,82,1

B

MW

B MWJock

Sissy




For Sissy the expected value of the NE (JB, S

B) = ( , ) is8× + 1× + 2× + 4× = < 4 and 8.1 9/ 2 9/ 2 9/ 4 9/ 10 3/

1 3/ 1 3/

For Jock the expected value of the NE (JB, S

B) = ( , ) is4× + 2× + 1× + 8× = ; 4 < < 8.1 9/ 2 9/ 2 9/ 4 9/ 14 3/

1 3/1 3//14 3

Coordination Games; The Prisoner’s Dilemma

A simultaneous play game in which each player has a strictly dominant action.

The only NE, therefore, is the choice by each player of her strictly dominant action.

Yet both players can achieve strictly larger payoffs than in the NE by coordinating with each other on another pair of actions.


Tim and Tom are in police custody. Each can confess (C) to a crime or stay silent (S).

Confession by both results in 5 years each in jail.

Silence by both results in 2 years each in jail.

If Tim confesses and Tom stays silent then Tim gets no penalty and Tom gets 10 years in jail (and conversely).


-10,0-2,-2

-5,-50,-10Confess

SilentTom

TimSilent

Confess

For Tim, Confess strictly dominates Silent.


Confess

SilentTom

TimSilent

Confess

For Tim, Confess strictly dominates Silent.For Tom, Confess strictly dominates Silent.

-10,0-2,-2

-5,-50,-10


Confess

SilentTom

TimSilent

Confess

For Tim, Confess strictly dominates Silent.For Tom, Confess strictly dominates Silent.The only NE is (Confess, Confess).

-10,0-2,-2

-5,-50,-10


Confess

SilentTom

TimSilent

Confess

For Tim, Confess strictly dominates Silent.For Tom, Confess strictly dominates Silent.The only NE is (Confess, Confess).

-10,0-2,-2

-5,-50,-10

But (Silence, Silence)is better for both Timand Tom.


Confess

SilentTom

TimSilent

Confess

Possible means include future punishments or enforceablecontracts.

-10,0-2,-2

-5,-50,-10

What is needed is ameans of rationallyassuring commitmentby both players tothe most beneficialcoordinated actions.

Coordination Games; Assurance Games

A simultaneous play game with two “coordinated” NE, one of which gives strictly greater payoffs to each player than does the other.

The question is: How can each player give the other an “assurance” that will cause the better NE to be the outcome of the game?


A common example is the “arms race” problem.

India and Pakistan can both increase their stockpiles of nuclear weapons. This is very costly.

Having nuclear superiority over the other gives a higher payoff, but the worst payoff to the other.

Not increasing the stockpile is best for both.


Stockpile

Don’tPakistan

IndiaDon’t

Stockpile

1,45,5

3,34,1


Stockpile

Don’tPakistan

IndiaDon’t

Stockpile

1,45,5

3,34,1

The game’s NE are (Don’t, Don’t) and (Stockpile, Stockpile).Which is the “likely” NE?


Stockpile

Don’tPakistan

IndiaDon’t

Stockpile

1,45,5

3,34,1

The game’s NE are (Don’t, Don’t) and (Stockpile, Stockpile).Which is the “likely” NE? What if India moved first? Whataction would it choose? Wouldn’t Don’t be best?

Coordination Games; Chicken

A simultaneous play game with two “coordinated” NE in which each player chooses the action that is not the action chosen by the other player.


Two drivers race their cars at each other. A driver who swerves is a “wimp”. A driver who does not swerve is “macho.”

If both do not swerve there is a crash and a very low payoff to both.

If both swerve then there is no crash and a moderate payoff to both.

If one swerves and the other does not then the swerver gets a low payoff and the non-swerver gets a high payoff.


No Swerve

SwerveDumber

DumbSwerve

NoSwerve

-2,41,1

-5,-54,-2



No Swerve

SwerveDumber

DumbSwerve

NoSwerve

-2,41,1

-5,-54,-2

The game’s pure strategy NE are (Swerve, No Swerve) and(No Swerve, Swerve). There is also a mixed strategy NE inwhich each chooses Swerve with probability ½.


No Swerve

SwerveDumber

DumbSwerve

NoSwerve

-2,41,1

-5,-54,-2

The game’s pure strategy NE are (Swerve, No Swerve) and(No Swerve, Swerve). There is also a mixed strategy NE inwhich each chooses Swerve with probability ½.

Can Dumb assurehimself of a payoff of4? Only by convincingDumber that Dumbreally will choose NoSwerve. What will beconvincing?

Games of Competition Simultaneous play games in which any

increase in the payoff to one player is exactly the decrease in the payoff to the other player.

These games are thus often called “constant (payoff) sum” games.

Games of Competition

D

L2

1U

R

2,-20,0

1,-1x,-x

An example is the game below. What NE can such agame possess?


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up ??


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left ??


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.Therefore, if x < 0 the NEis ??


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.Therefore, if x < 0 the NEis (Up, Left)


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.Therefore, if x < 0 the NEis (Up, Left) and if 0 < x < 1the NE is ??


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.Therefore, if x < 0 the NEis (Up, Left) and if 0 < x < 1the NE is (Down, Left).


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.Therefore, if x < 0 the NEis (Up, Left) and if 0 < x < 1the NE is (Down, Left).If x > 1 then ??


D

L2

1U

R

2,-20,0

1,-1x,-x


If x < 0 then Up dominatesDown.If x < 1 then Left dominatesRight.Therefore, if x < 0 the NEis (Up, Left) and if 0 < x < 1the NE is (Down, Left).If x > 1 then there is no NEin pure strategies. Is therea mixed-strategy NE?


D

L2

1U

R

2,-20,0

1,-1x,-x

The probability that 2 choosesLeft is L. The probability that1 chooses Up is U. x > 1.


D

L2

1U

R

2,-20,0

1,-1x,-x


EV1(U) = 2(1 - L).EV1(D) = xL + 1 - L.


D

L2

1U

R

2,-20,0

1,-1x,-x


EV1(U) = 2(1 - L).EV1(D) = xL + 1 - L.

>=<

2 - 2l 1 + (x - 1)L

as L 1/(1 + x).

>=<


D

L2

1U

R

2,-20,0

1,-1x,-x


EV1(U) = 2(1 - L).EV1(D) = xL + 1 - L.

>=<

2 - 2l 1 + (x - 1)L

as L 1/(1 + x).

>=<

EV2(L) = - x(1 - U).EV2(R) = - 2U - (1 - U).


D

L2

1U

R

2,-20,0

1,-1x,-x


EV1(U) = 2(1 - L).EV1(D) = xL + 1 - L.

>=<

2 - 2l 1 + (x - 1)L

as L 1/(1 + x).

>=<

EV2(L) = - x(1 - U).EV2(R) = - 2U - (1 - U).

>=<- x + xU - 1 - U

as (x – 1)/(1 + x) U.

>=<


D

L2

1U

R

2,-20,0

1,-1x,-x

1 chooses Up if L > 1/(1 + x) and Down if L < 1/(1 + x).2 chooses Left if U < (x – 1)/(1 + x) and Right if U > (x – 1)/(1 + x).

Games of Competition1: Up if L > 1/(1 + x); Down if L < 1/(1 + x).2: Left if U < (x – 1)/(1 + x); Right if U > (x – 1)/(1 + x).

U

U

L

L

1 1

10000

1 2


U

U

L

L

1 21 1

10000 1/(1+x) (x-1)/(1+x)

(x-1)/(1+x)


U

L

1 21

100 1/(1+x) U

L1

00

(x-1)/(1+x)


U

L

1 21

100 1/(1+x)

U

L

1

00

(x-1)/(1+x)


1 2

1

U

L

1

00L1/(1+x)

U1

00

(x-1)/(1+x)


1

U

L

1

00

(x-1)/(1+x)

1/(1+x)

When x > 1 there is onlya mixed-strategy NE inwhich 1 plays Up withprobability (x – 1)/(x + 1)and 2 plays Left withprobability 1/(1 + x).

Coexistence Games Simultaneous play games that can be

used to model how members of a species act towards each other.

An important example is the hawk-dove game.

Coexistence Games; The Hawk-Dove Game

“Hawk” means “be aggressive.” “Dove” means “don’t be aggressive.” Two bears come to a fishing spot.

Either bear can fight the other to try to drive it away to get more fish for itself but suffer battle injuries, or it can tolerate the presence of the other, share the fishing, and avoid injury.


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1

Are there NE in pure strategies?

-5,-5


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1

Are there NE in pure strategies?Yes (Hawk, Dove) and (Dove, Hawk).Notice that purely peaceful coexistence is not a NE.

-5,-5


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1

Is there a NE in mixed strategies?

-5,-5


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1

Is there a NE in mixed strategies?

1H is the prob. that

1 chooses Hawk.2

H is the prob. that2 chooses Hawk.What are the players’best-responsefunctions?

-5,-5


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1


1 chooses Hawk.2


EV1(H) = -52H + 8(1 - 2

H) = 8 - 132H.

EV1(D) = 4 - 42H.

-5,-5


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1


1 chooses Hawk.2


EV1(H) = -52H + 8(1 - 2

H) = 8 - 132H.

EV1(D) = 4 - 42H.

8 - 132H 4 - 42

H as 2H 4/9.>=<

<=>

-5,-5


8,0

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1


1 chooses Hawk.2

H is the prob. that2 chooses Hawk.

Bear 11H

2H1

1

00 4/9

-5,-5

EV1(H) = -52H + 8(1 - 2

H) = 8 - 132H.

EV1(D) = 4 - 42H.

8 - 132H 4 - 42

H as 2H 4/9.>=<

<=>


Bear 1 Bear 22H

1H1

1

00 4/9

1H

2H1

1

00 4/9


Bear 1 Bear 22H

1H1

1

00 4/9

1H

2H1

1

00 4/9

4/9


Bear 1 Bear 21H

2H1

1

00

4/9

1H

2H1

1

00 4/9


1H

2H1

1

00 4/9

The game has a NE in mixed-strategies in whicheach bear plays Hawk with probability 4/9.


8,0-5,-5

4,40,8

Hawk

Dove

Hawk DoveBear 2

Bear 1

For each bear, the expected value of the mixed-strategy NE is(-5)× + 8× + 4× = , a value between-5 and +4

8116/ 8120/ 8125/ 81180/

Commitment Games Sequential play games in which

One player chooses an action before the other player chooses an action.

The first player’s action is both irreversible and observable by the second player.

The first player knows that his action is seen by the second player.

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f

Game TreePlayer 1 has twoactions, a and b.Player 2 has twoactions, c and d,following a, and two actions e and ffollowing b.

Player 1 chooseshis action beforePlayer 2 choosesher action.

Direction of play

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f

Is a claim by Player 2that she will commit tochoosing action c ifPlayer 1 chooses acredible to Player 1?

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f

Is a claim by Player 2that she will commit tochoosing action c ifPlayer 1 chooses acredible to Player 1?Yes.

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f


Is a claim by Player 2that she will commit tochoosing action e ifPlayer 1 chooses bcredible to Player 1?

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f


Is a claim by Player 2that she will commit tochoosing action e ifPlayer 1 chooses bcredible to Player 1?Yes.

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f

So Player 1 shouldchoose action ??

Commitment Games5,9

5,5

7,6

5,4

1

2

2

a

be

c

d

f

So Player 1 shouldchoose action b.

Commitment Games5,3

5,5

7,6

5,4

1

2

2

a

be

c

d

f

Change the game.

Commitment Games5,3

5,5

7,6

5,4

1

2

2

a

be

c

d

f


Commitment Games5,3

5,5

7,6

5,4

1

2

2

a

be

c

d

f


No. If Player 1 choosesaction a then Player 2does best by choosingaction d.What should Player 1do?

Commitment Games5,3

5,5

7,6

5,4

1

2

2

a

be

c

d

f


No. If Player 1 choosesaction a then Player 2does best by choosingaction d.What should Player 1do? Still choose b.

Commitment Games5,3

5,5

7,3

5,12

1

2

2

a

be

c

d

f

Change the game.5,9

15,5

Commitment Games5,9

7,3

5,12

1

2

2

a

be

c

d

f

Can Player 1 get 15points?

15,5

Commitment Games5,9

7,3

5,12

1

2

2

a

be

c

d

f


If Player 1 chooses athen Player 2 willchoose c and Player 1will get only 5 points.

15,5

Commitment Games5,9

7,3

5,12

1

2

2

a

be

c

d

f


If Player 1 chooses athen Player 2 willchoose c and Player 1will get only 5 points.

If Player 1 chooses bthen Player 2 willchoose f and againPlayer 1 will get only5 points.

15,5

Commitment Games5,9

7,3

5,12

1

2

2

a

be

c

d

f

If Player 1 can changepayoffs so that acommitment by Player2 to choose d after ais credible then Player1’s payoff rises from 5to 15, a gain of 10.

15,5

Commitment Games5,9

10,10

7,3

5,12

1

2

2

a

be

c

d

f

If Player 1 can changepayoffs so that acommitment by Player2 to choose d after ais credible then Player1’s payoff rises from 5to 15, a gain of 10.If Player 1 gives 5 ofthese points to Player2 then Player 2’scommitment iscredible. Player 1 cannot get 15 points.

Commitment Games5,9

10,10

7,3

5,12

1

2

2

a

be

c

d

f

Credible NE of thistype are calledsubgame perfect.What exactly is thisgame’s SPE? Itinsists that everyaction chosen isrational for the playerwho chooses it.

Bargaining Games Two players bargain over the division of

a pie of size 1. What will be the outcome?

Two approaches: Nash’s axiomatic bargaining. Rubinstein’s strategic bargaining.

Strategic Bargaining The players have 3 periods in which to decide

how to divide the pie; else both get nothing. Player A discounts next period’s payoffs by . Player B discounts next period’s payoffs by . The players alternate in making offers, with

Player A starting in period 1. If the player who receives an offer accepts it

then the game ends immediately. Else the game continues to the next period.

Strategic Bargaining

0

1

A

0

1

B

0

1

AB B

A

x1

(x1,1-x1)

Y

N x2

x3

(x3,1-x3)

(x2,1-x2)

(0,0)

Y

Y

N

N

Period 1:A offers x1.B responds.

Period 2:B offers x2.A responds.



0

1

AB

x3

Y

N


How should B respond to x3?(x3,1-x3)

(0,0)


0

1

AB

x3

Y

N


How should B respond to x3?Accept if 1 – x3 ≥ 0; i.e.,accept any x3 ≤ 1.

(x3,1-x3)

(0,0)


0

1

AB

x3(0,0)

Y

N



Knowing this, what should Aoffer?

(x3,1-x3)

x3=1


0

A

B

(1,0)

(0,0)

Y

N

Period 3:A offers x3 = 1.B accepts.


Knowing this, what should Aoffer? x3 = 1.

x3=1


0

1

A

0

1

B

0

AB

A

x1

(x1,1-x1)

Y

N x2

(x2,1-x2)

Y

N



BY

N


(1,0)

(0,0)

x3=1


0

1

B

0

AAx2

(x2,1-x2)

Y

N


BY

N


In Period 3 A getsa payoff of 1. Inperiod 2, whenreplying to B’soffer of x2, thepresent-value toA of N is thus ??

(1,0)

(0,0)

x3=1


0

1

B

0

AAx2

(x2,1-x2)

Y

N


BY

N


In Period 3 A getsa payoff of 1. Inperiod 2, whenreplying to B’soffer of x2, thepresent-value toA of N is thus .

(1,0)

(0,0)


0

1

B Ax2

(x2,1-x2)

Y

N



What is the mostB should offer toA?


0

1

B Ax2=

(,1- )

Y

N

Period 2:B offers x2 = .A accepts.


What is the mostB should offer toA? x2 = .

x3=1


0

1

A

0

1

B

0

AB

A

x1

(x1,1-x1)

Y

N

Y

N


BY

N



x2=

(,1- )

(1,0)

(0,0)


0

1

A

0

1

BB

A

x1

(x1,1-x1)

Y

N

Y

N



x2=

(,1- )

In period 2 A will accept. Thus B will get thepayoff 1 - in period 2. What is the present-value to B in period 1 ofN ?


0

1

A

0

1

BB

A

x1

(x1,1-x1)

Y

N

Y

N



x2=

(,1- )

In period 2 A will accept. Thus B will get thepayoff 1 - in period 2. What is the present-value to B in period 1 ofN ? (1 - ).


0

1

A

0

1

BB

A

x1

(x1,1-x1)

Y

N

Y

N



x2=

(,1- )


What is the most that Ashould offer to B inperiod 1?


0

1

A

0

1

BB

A

x1

(1-(1 - ), (1 - ))

Y

N

Y

N



x2=

(,1- )


What is the most that Ashould offer to B inperiod 1?1 – x1 = (1 - ); i.e.x1 = 1 - (1 - ).B will accept.

x3=1


0

1

A

0

1

B

0

AB

A

Y

N

Y

N

Period 1: A offersx1 = 1-(1 - ).B accepts.

BY

N



x2=

(,1- )

(1-(1 - ), (1 - ))

x1=1-(1-)

(1,0)

(0,0)

Strategic Bargaining Notice that the game ends immediately,

in period 1. Player A gets 1 - (1 – ) units of the pie.

Player B gets (1 – ) units.Which is the larger?

x1 = 1 - (1 – ) ≥ ½ ≤ 1/2(1 - )so Player A gets more than Player B if Player B is “too impatient” relative to Player A.

Strategic Bargaining Suppose the game is allowed to continue

forever (infinitely many periods). Then using the same reasoning shows that the subgame perfect equilibrium results in Players 1 and 2 respectively getting

and pie units.

Player 1’s share rises as and .Player 2’s share rises as and .

11

1)1(

lec 24 game applications

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action ab1

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