Chapter 30 Game Applications.

Chapter 30 Game Applications

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.

Nash Equilibrium 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 aA2.

Best Responses B can choose between actions aB1 and aB2.
There are 4 possible action pairs; (aA1, aB1), (aA1, aB2), (aA2, aB1), (aA2, 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 aB1 are UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4. Similarly, suppose that UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7.

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7.

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If B chooses action aB1 then A’s best response is ??

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If B chooses action aB1 then A’s best response is action aA1 (because 6 > 4).

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If B chooses action aB2 then A’s best response is ??

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If B chooses action aB2 then A’s best response is action aA2 (because 5 > 3).

Best Responses If B chooses aB1 then A chooses aA1.
A’s best-response “curve” is therefore aA2 + A’s best response aA1 + aB1 aB2 B’s action

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7.

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If A chooses action aA1 then B’s best response is ??

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If A chooses action aA1 then B’s best response is action aB2 (because 5 > 4).

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If A chooses action aA2 then B’s best response is ??.

Best Responses UA(aA1, aB1) = 6 and UB(aA1, aB1) = 4 UA(aA1, aB2) = 3 and UB(aA1, aB2) = 5 UA(aA2, aB1) = 4 and UB(aA2, aB1) = 3 UA(aA2, aB2) = 5 and UB(aA2, aB2) = 7. If A chooses action aA2 then B’s best response is action aB2 (because 7 > 3).

Best Responses If A chooses aA1 then B chooses aB2.
B’s best-response “curve” is therefore aA2 A’s action aA1 aB1 aB2 B’s best response

Best Responses If A chooses aA1 then B chooses aB2.
B’s best-response “curve” is therefore Notice that aB2 is a strictly dominant action for B. aA2 A’s action aA1 aB1 aB2 B’s best response

Best Responses & Nash Equilibria
How can the players’ best-response curves be used to locate the game’s Nash equilibria? A’s response A’s choice B A aA1 aA2 aB2 aB1 + aA2 aA1 aB1 aB2 B’s choice B’s response

How can the players’ best-response curves be used to locate the game’s Nash equilibria? Put one curve on top of the other. A’s response A’s choice B A aA1 aA2 aB2 aB1 + aA2 aA1 aB1 aB2 B’s choice B’s response

How can the players’ best-response curves be used to locate the game’s Nash equilibria? Put one curve on top of the other. A’s response aA2 + Is there a Nash equilibrium? aA1 + aB1 aB2 B’s response

How can the players’ best-response curves be used to locate the game’s Nash equilibria? Put one curve on top of the other. A’s response aA2 + Is there a Nash equilibrium? Yes, (aA2, aB2). Why? aA1 + aB1 aB2 B’s response

How can the players’ best-response curves be used to locate the game’s Nash equilibria? Put one curve on top of the other. A’s response aA2 + Is there a Nash equilibrium? Yes, (aA2, aB2). Why? aA2 is a best response to aB2. aB2 is a best response to aA2. aA1 + aB1 aB2 B’s response

Player B Here is the strategic form of the game. aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 aA2 is the only best response to aB2. aB2 is the only best response to aA2.

Player B Here is the strategic form of the game. aB1 aB2 Is there a 2nd Nash eqm.? aA1 6,4 3,5 Player A aA2 4,3 5,7 aA2 is the only best response to aB2. aB2 is the only best response to aA2.

Player B Here is the strategic form of the game. aB1 aB2 Is there a 2nd Nash eqm.? No, because aB2 is a strictly dominant action for Player B. aA1 6,4 3,5 Player A aA2 4,3 5,7 aA2 is the only best response to aB2. aB2 is the only best response to aA2.

Player B aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 Now allow both players to randomize (i.e., mix) over their actions.

Player B A1 is the prob. A chooses action aA1. B1 is the prob. B chooses action aB1. aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 Now allow both players to randomize (i.e., mix) over their actions.

Player B 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? aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7

Player B 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? aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 EVA(aA1) = 6B1 + 3(1 - B1) = 3 + 3B1.

Player B 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? aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 EVA(aA1) = 6B1 + 3(1 - B1) = 3 + 3B1. EVA(aA2) = 4B1 + 5(1 - B1) = 5 - B1.

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) = 3 + 3B1. EVA(aA2) = 5 - B B B1 as B1 ?? > = <

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) = 3 + 3B1. EVA(aA2) = 5 - B B B1 as B1 ½. > = <

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) = 3 + 3B1. EVA(aA2) = 5 - B B B1 as B1 ½. A’s best response is: aA1 if B1 > ½ aA2 if B1 < ½ aA1 or aA2 if B1 = ½ > = <

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) = 3 + 3B1. EVA(aA2) = 5 - B B B1 as B1 ½. A’s best response is: aA1 (i.e. A1 = 1) if B1 > ½ aA2 (i.e. A1 = 0) if B1 < ½ aA1 or aA2 (i.e. 0  A1  1) if B1 = ½ > = <

A’s best response is: aA1 (i.e. A1 = 1) if B1 > ½ aA2 (i.e. A1 = 0) if B1 < ½ aA1 or aA2 (i.e. 0  A1  1) if B1 = ½ A’s best response A1 1 B1 1

A’s best response is: aA1 (i.e. A1 = 1) if B1 > ½ aA2 (i.e. A1 = 0) if B1 < ½ aA1 or aA2 (i.e. 0  A1  1) if B1 = ½ A’s best response A1 1 This is A’s best response curve when players are allowed to mix over their actions. B1 1

Player B 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? aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7

Player B 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? aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 EVB(aB1) = 4A1 + 3(1 - A1) = 3 + A1.

Player B 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? aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7 EVB(aB1) = 4A1 + 3(1 - A1) = 3 + A1. EVB(aB2) = 5A1 + 7(1 - A1) = 7 - 2A1.

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) = 3 + A1. EVB(aB2) = 7 - 2A A A1 as A1 ?? > = < > = <

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) = 3 + A1. EVB(aB2) = 7 - 2A A1 < 7 - 2A1 for all 0  A1  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? EVB(aB1) = 3 + A1. EVB(aB2) = 7 - 2A A1 < 7 - 2A1 for all 0  A1  1. B’s best response is: aB2 always (i.e. B1 = 0 always).

B’s best response is aB2 always (i.e. B1 = 0 always). A1 1 This is B’s best response curve when players are allowed to mix over their actions. B1 1 B’s best response

A’s best response A1 A1 1 1 B1 B1 1 1 B’s best response

Is there a Nash equilibrium? B A A’s best response A1 A1 1 1 B1 B1 1 1 B’s best response

Is there a Nash equilibrium? B A A1 1 B1 A’s best response A1 1 B1 B’s best response

Is there a Nash equilibrium? A’s best response A1 1 B1 B’s best response

Is there a Nash equilibrium? Yes. Just one. (A1, B1) = (0,0); i.e. A chooses aA2 only & B chooses aB2 only. A’s best response A1 1 B1 B’s best response

Player B Let’s change the game. aB1 aB2 aA1 6,4 3,5 Player A aA2 4,3 5,7

Player B Here is a new 2×2 game. aB1 aB2 aA1 6,4 3,5 3,1 Player A aA2 4,3 5,7

Player B Here is a new 2×2 game. Again let A1 be the prob. that A chooses aA1 and let B1 be the prob. that B chooses aB1. What are the NE of this game? aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 Notice that Player B no longer has a strictly dominant action.

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVA(aA1) = ?? EVA(aA2) = ??

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVA(aA1) = 6B1 + 3(1 - B1) = 3 + 3B1. EVA(aA2) = ??

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVA(aA1) = 6B1 + 3(1 - B1) = 3 + 3B1. EVA(aA2) = 4B1 + 5(1 - B1) = 5 - B1.

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVA(aA1) = 6B1 + 3(1 - B1) = 3 + 3B1. EVA(aA2) = 4B1 + 5(1 - B1) = 5 - B B B1 as B1 ½. > = < > = <

EVA(aA1) = 6B1 + 3(1 - B1) = 3 + 3B1. EVA(aA2) = 4B1 + 5(1 - B1) = 5 - B B B1 as B1 ½. > = < A’s best response A1 1 B1 1

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVB(aB1) = ?? EVB(aB2) = ??

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVB(aB1) = 4A1 + 3(1 - A1) = 3 + A1. EVB(aB2) = ??

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVB(aB1) = 4A1 + 3(1 - A1) = 4 + A1. EVB(aB2) = A1 + 7(1 - A1) = 7 - 6A1.

Player B A1 is the prob. that A chooses aA1. B1 is the prob. that B chooses aB1. aB1 aB2 aA1 6,4 3,1 Player A aA2 4,3 5,7 EVB(aB1) = 4A1 + 3(1 - A1) = 3 + A1. EVB(aB2) = A1 + 7(1 - A1) = 7 - 6A A A1 as A > = < 4 7 /

EVB(aB1) = 4A1 + 3(1 - A1) = 3 + A1. EVB(aB2) = A1 + 7(1 - A1) = 7 - 6A A A1 as A > = < 4 7 / A1 1 4 7 / B1 1 B’s best response

A’s best response A1 A1 1 1 4 7 / B1 B1 1 1 B’s best response

Is there a Nash equilibrium? B A A’s best response A1 A1 1 1 4 7 / B1 B1 1 1 B’s best response

Is there a Nash equilibrium? B A A1 1 B1 A’s best response A1 1 4 7 / B1 1 B’s best response

Is there a Nash equilibrium? A’s best response A1 1 4 7 / B1 1 B’s best response

Is there a Nash equilibrium? Yes. 3 of them. A’s best response A1 1 4 7 / B1 1 B’s best response

Is there a Nash equilibrium? Yes. 3 of them. (A1, B1) = (0,0) A’s best response A1 1 4 7 / B1 1 B’s best response

Is there a Nash equilibrium? Yes. 3 of them. Is there a Nash equilibrium? (A1, B1) = (0,0) (A1, B1) = (1,1) A’s best response A1 1 4 7 / B1 1 B’s best response

Is there a Nash equilibrium? Is there a Nash equilibrium? Yes. 3 of them. (A1, B1) = (0,0) (A1, B1) = (1,1) (A1, B1) = ( , ) A’s best response 4 7 / A1 1 4 7 / B1 1 B’s best response

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.

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. B MW B 8,4 1,2 Sissy MW 2,1 4,8

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. What are the players’ best-response functions? B MW B 8,4 1,2 Sissy MW 2,1 4,8

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. What are the players’ best-response functions? B MW B 8,4 1,2 Sissy MW 2,1 4,8 EVS(B) = 8JB + (1 - JB) = 1 + 7JB.

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. What are the players’ best-response functions? B MW B 8,4 1,2 Sissy MW 2,1 4,8 EVS(B) = 8JB + (1 - JB) = 1 + 7JB. EVS(MW) = 2JB + 4(1 - JB) = 4 - 2JB.

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. What are the players’ best-response functions? B MW B 8,4 1,2 Sissy MW 2,1 4,8 EVS(B) = 8JB + (1 - JB) = 1 + 7JB. EVS(MW) = 2JB + 4(1 - JB) = 4 - 2JB. 1 + 7JB JB as JB 1 3 / > = < > = <

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. B MW B 8,4 1,2 SB Sissy 1 MW 2,1 4,8 EVS(B) = 8JB + (1 - JB) = 1 + 7JB. EVS(MW) = 2JB + 4(1 - JB) = 4 - 2JB. 1 + 7JB JB as JB 1 3 / > = < > = < 1 3 / 1 JB

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. B MW B 8,4 1,2 SB Sissy Sissy 1 MW 2,1 4,8 EVS(B) = 8JB + (1 - JB) = 1 + 7JB. EVS(MW) = 2JB + 4(1 - JB) = 4 - 2JB. 1 + 7JB JB as JB 1 3 / > = < > = < 1 3 / 1 JB

SB Sissy SB Jock 1 1 1 3 / 1 3 / 1 JB 1 JB

The game’s NE are ?? SB Sissy SB Jock 1 1 1 3 / 1 3 / 1 JB 1 JB

The game’s NE are ?? SB Sissy SB JB 1 3 / Jock 1 1 3 / 1 JB

The game’s NE are ?? Sissy SB 1 1 3 / 1 3 / 1 JB Jock

The game’s NE are: (JB, SB) = (0, 0); i.e., (MW, MW) Sissy SB 1 1 3 / 1 3 / 1 JB Jock

The game’s NE are: (JB, SB) = (0, 0); i.e., (MW, MW) (JB, SB) = (1, 1); i.e., (B, B) Sissy SB 1 1 3 / 1 3 / 1 JB Jock

The game’s NE are: (JB, SB) = (0, 0); i.e., (MW, MW) (JB, SB) = (1, 1); i.e., (B, B) (JB, SB) = ( , ); i.e., both watch the ballet with prob. 1/9, both watch the mud wrestling with prob. 4/9, and with prob. 4/9 they watch different events. 1 3 / 1 3 / Sissy SB 1 1 3 / 1 3 / 1 JB Jock

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. B MW B 8,4 1,2 Sissy MW 2,1 4,8 For Sissy the expected value of the NE (JB, SB) = ( , ) is 8× × × × = < 4 and 8. 1 9 / 2 4 10 3 1 3 / 1 3 /

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. B MW B 8,4 1,2 Sissy MW 2,1 4,8 For Sissy the expected value of the NE (JB, SB) = ( , ) is 8× × × × = < 4 and 8. 1 9 / 2 4 10 3 1 3 / 1 3 / For Jock the expected value of the NE (JB, SB) = ( , ) is 4× × × × = ; 4 < < 8. 1 9 / 2 4 14 3

Jock SB is the prob. that Sissy chooses ballet. JB is the prob. that Jock chooses ballet. B MW B 8,4 1,2 So, is the mixed strategy NE a focal point for the game? Sissy MW 2,1 4,8 For Sissy the expected value of the NE (JB, SB) = ( , ) is 8× × × × = < 4 and 8. 1 9 / 2 4 10 3 1 3 / 1 3 / For Jock the expected value of the NE (JB, SB) = ( , ) is 4× × × × = ; 4 < < 8. 1 9 / 2 4 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).

Tom Silent Confess Silent -2,-2 -10,0 Tim Confess 0,-10 -5,-5 For Tim, Confess strictly dominates Silent.

Tom Silent Confess Silent -2,-2 -10,0 Tim Confess 0,-10 -5,-5 For Tim, Confess strictly dominates Silent. For Tom, Confess strictly dominates Silent.

Tom Silent Confess Silent -2,-2 -10,0 Tim Confess 0,-10 -5,-5 For Tim, Confess strictly dominates Silent. For Tom, Confess strictly dominates Silent. The only NE is (Confess, Confess).

Tom Silent Confess But (Silence, Silence) is better for both Tim and Tom. Silent -2,-2 -10,0 Tim Confess 0,-10 -5,-5 For Tim, Confess strictly dominates Silent. For Tom, Confess strictly dominates Silent. The only NE is (Confess, Confess).

Tom What is needed is a means of rationally assuring commitment by both players to the most beneficial coordinated actions. Silent Confess Silent -2,-2 -10,0 Tim Confess 0,-10 -5,-5 Possible means include future punishments or enforceable contracts.

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.

Pakistan Don’t Stockpile Don’t 5,5 1,4 India Stockpile 4,1 3,3

Pakistan Don’t Stockpile Don’t 5,5 1,4 India Stockpile 4,1 3,3 The game’s NE are ??

Pakistan Don’t Stockpile Don’t 5,5 1,4 India Stockpile 4,1 3,3 The game’s NE are (Don’t, Don’t) and (Stockpile, Stockpile). Which is the “likely” NE?

Pakistan Don’t Stockpile Don’t 5,5 1,4 India Stockpile 4,1 3,3 The game’s NE are (Don’t, Don’t) and (Stockpile, Stockpile). Which is the “likely” NE? What if India moved first? What action 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.

Dumber No Swerve Swerve Swerve 1,1 -2,4 Dumb No Swerve 4,-2 -5,-5 The game’s NE are ??

Dumber No Swerve Swerve Swerve 1,1 -2,4 Dumb No Swerve 4,-2 -5,-5 The game’s pure strategy NE are (Swerve, No Swerve) and (No Swerve, Swerve). There is also a mixed strategy NE in which each chooses Swerve with probability ½.

Dumber No Swerve Can Dumb assure himself of a payoff of 4? Only by convincing Dumber that Dumb really will choose No Swerve. What will be convincing? Swerve Swerve 1,1 -2,4 Dumb No Swerve 4,-2 -5,-5 The game’s pure strategy NE are (Swerve, No Swerve) and (No Swerve, Swerve). There is also a mixed strategy NE in which each chooses Swerve with probability ½.

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 L 2 1 U R 2,-2 0,0 1,-1 x,-x
An example is the game below. What NE can such a game possess? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up ?? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left ?? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. Therefore, if x < 0 the NE is ?? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. Therefore, if x < 0 the NE is (Up, Left) D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. Therefore, if x < 0 the NE is (Up, Left) and if 0 < x < 1 the NE is ?? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. Therefore, if x < 0 the NE is (Up, Left) and if 0 < x < 1 the NE is (Down, Left). D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. Therefore, if x < 0 the NE is (Up, Left) and if 0 < x < 1 the NE is (Down, Left). If x > 1 then ?? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

An example is the game below. What NE can such a game possess? If x < 0 then Up dominates Down. If x < 1 then Left dominates Right. Therefore, if x < 0 the NE is (Up, Left) and if 0 < x < 1 the NE is (Down, Left). If x > 1 then there is no NE in pure strategies. Is there a mixed-strategy NE? D L 2 1 U R 2,-2 0,0 1,-1 x,-x

The probability that 2 chooses Left is L. The probability that 1 chooses Up is U. x > 1. D L 2 1 U R 2,-2 0,0 1,-1 x,-x

EV1(U) = 2(1 - L). EV1(D) = xL L. The probability that 2 chooses Left is L. The probability that 1 chooses Up is U. x > 1. D L 2 1 U R 2,-2 0,0 1,-1 x,-x

Games of Competition > = < D L 2 1 U R 2,-2 0,0 1,-1 x,-x
EV1(U) = 2(1 - L). EV1(D) = xL L. The probability that 2 chooses Left is L. The probability that 1 chooses Up is U. x > 1. 2 - 2l (x - 1)L as L /(1 + x). > = < D L 2 1 U R 2,-2 0,0 1,-1 x,-x > = <

EV1(U) = 2(1 - L). EV1(D) = xL L. The probability that 2 chooses Left is L. The probability that 1 chooses Up is U. x > 1. 2 - 2l (x - 1)L as L /(1 + x). > = < D L 2 1 U R 2,-2 0,0 1,-1 x,-x > = < EV2(L) = - x(1 - U). EV2(R) = - 2U - (1 - U).

EV1(U) = 2(1 - L). EV1(D) = xL L. The probability that 2 chooses Left is L. The probability that 1 chooses Up is U. x > 1. 2 - 2l (x - 1)L as L /(1 + x). > = < D L 2 1 U R 2,-2 0,0 1,-1 x,-x > = < EV2(L) = - x(1 - U). EV2(R) = - 2U - (1 - U). - x + xU U as (x – 1)/(1 + x) U. > = < > = <

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). D L 2 1 U R 2,-2 0,0 1,-1 x,-x

Games of Competition 1: 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 L 1 2 1 1 L 1 U

Games of Competition 1: 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 L 1 2 1 1 L 1/(1+x) 1 U (x-1)/(1+x)

Games of Competition 1: 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 L 1 2 1 1 (x-1)/(1+x) L 1/(1+x) (x-1)/(1+x) 1 U

Games of Competition 1: 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 1 2 1 1 (x-1)/(1+x) L 1/(1+x) 1 L

Games of Competition 1: 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). L 1/(1+x) U 1 U 1 2 1 (x-1)/(1+x) 1 L

Games of Competition 1: 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 1 When x > 1 there is only a mixed-strategy NE in which 1 plays Up with probability (x – 1)/(x + 1) and 2 plays Left with probability 1/(1 + x). (x-1)/(1+x) L 1/(1+x) 1

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.

Bear 2 Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 Are there NE in pure strategies?

Bear 2 Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 Are there NE in pure strategies? Yes (Hawk, Dove) and (Dove, Hawk). Notice that purely peaceful coexistence is not a NE.

Bear 2 Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 Is there a NE in mixed strategies?

Bear 2 1H is the prob. that 1 chooses Hawk. 2H is the prob. that 2 chooses Hawk. What are the players’ best-response functions? Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 Is there a NE in mixed strategies?

Bear 2 1H is the prob. that 1 chooses Hawk. 2H is the prob. that 2 chooses Hawk. What are the players’ best-response functions? Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 EV1(H) = -52H + 8(1 - 2H) = 2H. EV1(D) = 4 - 42H.

Bear 2 1H is the prob. that 1 chooses Hawk. 2H is the prob. that 2 chooses Hawk. What are the players’ best-response functions? Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 EV1(H) = -52H + 8(1 - 2H) = 2H. EV1(D) = 4 - 42H. 8 - 132H 2H as 2H 4/9. > = < < = >

Bear 2 1H is the prob. that 1 chooses Hawk. 2H is the prob. that 2 chooses Hawk. Hawk Dove Hawk -5,-5 8,0 1H Bear 1 Bear 1 1 Dove 0,8 4,4 EV1(H) = -52H + 8(1 - 2H) = 2H. EV1(D) = 4 - 42H. 8 - 132H 2H as 2H 4/9. > = < < = > 4/9 1 2H

Bear 1 2H Bear 2 1 1 4/9 1 2H 4/9 1 1H

Bear 1 2H Bear 2 1 1 4/9 4/9 1 2H 4/9 1 1H

1 4/9 Bear 1 1H 2H 1 4/9 Bear 2

The game has a NE in mixed-strategies in which each bear plays Hawk with probability 4/9. 1H 1 4/9 1 2H

Bear 2 Hawk Dove Hawk -5,-5 8,0 Bear 1 Dove 0,8 4,4 For each bear, the expected value of the mixed-strategy NE is (-5)× × × = , a value between -5 and +4. Is this NE focal? 81 16/ 81 20/ 81 25/ 81 180/

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 Games 5,9 c Player 1 has two actions, a and b.
Player 2 has two actions, c and d, following a, and two actions e and f following b. Player 1 chooses his action before Player 2 chooses her action. Game Tree 2 a d 5,5 Direction of play 1 7,6 e b 2 f 5,4

Commitment Games 5,9 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games 5,9 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? Yes. c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games 5,9 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? Yes. Is a claim by Player 2 that she will commit to choosing action e if Player 1 chooses b credible to Player 1? c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games 5,9 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? Yes. Is a claim by Player 2 that she will commit to choosing action e if Player 1 chooses b credible to Player 1? Yes. c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games So Player 1 should choose action ?? 5,9 c 2 a d 5,5 1
7,6 e b 2 f 5,4

Commitment Games So Player 1 should choose action b. 5,9 c 2 a d 5,5 1
7,6 e b 2 f 5,4

Commitment Games 5,3 Change the game. c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games 5,3 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games 5,3 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? No. If Player 1 chooses action a then Player 2 does best by choosing action d. What should Player 1 do? c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games 5,3 Is a claim by Player 2 that she will commit to choosing action c if Player 1 chooses a credible to Player 1? No. If Player 1 chooses action a then Player 2 does best by choosing action d. What should Player 1 do? Still choose b. c 2 a d 5,5 1 7,6 e b 2 f 5,4

Commitment Games Change the game. 5,9 5,3 c 2 a d 15,5 5,5 1 7,3 e b 2
f 5,12

Commitment Games Can Player 1 get 15 points? 5,9 c 2 a d 15,5 1 7,3 e
b 2 f 5,12

Commitment Games 5,9 Can Player 1 get 15 points? If Player 1 chooses a then Player 2 will choose c and Player 1 will get only 5 points. c 2 a d 15,5 1 7,3 e b 2 f 5,12

Commitment Games 5,9 Can Player 1 get 15 points? If Player 1 chooses a then Player 2 will choose c and Player 1 will get only 5 points. If Player 1 chooses b then Player 2 will choose f and again Player 1 will get only 5 points. c 2 a d 15,5 1 7,3 e b 2 f 5,12

Commitment Games If Player 1 can change payoffs so that a commitment by Player 2 to choose d after a is credible then Player 1’s payoff rises from 5 to 15, a gain of 10. 5,9 c 2 a d 15,5 1 7,3 e b 2 f 5,12

Commitment Games If Player 1 can change payoffs so that a commitment by Player 2 to choose d after a is credible then Player 1’s payoff rises from 5 to 15, a gain of 10. If Player 1 gives 5 of these points to Player 2 then Player 2’s commitment is credible. Player 1 cannot get 15 points. 5,9 c 2 a d 10,10 1 7,3 e b 2 f 5,12

Commitment Games Credible NE of this type are called subgame perfect. What exactly is this game’s SPE? It insists that every action chosen is rational for the player who chooses it. 5,9 c 2 a d 10,10 1 7,3 e b 2 f 5,12

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 Period 2: B offers x2. A responds. (x1,1-x1)
Y Y x1 x3 N (0,0) B B N A B x2 A A N Y Period 1: A offers x1. B responds. Period 3: A offers x3. B responds. (x2,1-x2)

Strategic Bargaining How should B respond to x3? (x3,1-x3) 1 Y x3 N
(0,0) B A Period 3: A offers x3. B responds.

Strategic Bargaining How should B respond to x3? Accept if 1 – x3 ≥ 0; i.e., accept any x3 ≤ 1. (x3,1-x3) 1 Y x3 N (0,0) B A Period 3: A offers x3. B responds.

Strategic Bargaining How should B respond to x3? Accept if 1 – x3 ≥ 0; i.e., accept any x3 ≤ 1. Knowing this, what should A offer? (x3,1-x3) 1 Y x3 N (0,0) B A Period 3: A offers x3. B responds.

Strategic Bargaining How should B respond to x3? Accept if 1 – x3 ≥ 0; i.e., accept any x3 ≤ 1. Knowing this, what should A offer? x3 = 1. (1,0) Y x3=1 B N (0,0) A Period 3: A offers x3 = 1. B accepts.

Strategic Bargaining Period 2: B offers x2. A responds. (1,0)
(x1,1-x1) Y 1 1 x3=1 Y B x1 N B N (0,0) A B x2 A A N Y Period 1: A offers x1. B responds. Period 3: A offers x3 = 1. B accepts. (x2,1-x2)

Strategic Bargaining Period 2: B offers x2.
A responds. In Period 3 A gets a payoff of 1. In period 2, when replying to B’s offer of x2, the present-value to A of N is thus ?? (1,0) Y 1 x3=1 B N (0,0) B x2 A A N Y Period 3: A offers x3 = 1. B accepts. (x2,1-x2)

A responds. In Period 3 A gets a payoff of 1. In period 2, when replying to B’s offer of x2, the present-value to A of N is thus . (1,0) Y 1 x3=1 B N (0,0) B x2 A A N Y Period 3: A offers x3 = 1. B accepts. (x2,1-x2)

A responds. In Period 3 A gets a payoff of 1. In period 2, when replying to B’s offer of x2, the present-value to A of N is thus . What is the most B should offer to A? 1 B x2 A N Y (x2,1-x2)

Strategic Bargaining Period 2: B offers x2 = .
A accepts. In Period 3 A gets a payoff of 1. In period 2, when replying to B’s offer of x2, the present-value to A of N is thus . What is the most B should offer to A? x2 = . 1 x2= B A N Y (,1- )

Strategic Bargaining Period 2: B offers x2 = . A accepts. (1,0)
(x1,1-x1) Y 1 1 x3=1 Y B x1 N B N x2= (0,0) A B A A N Y Period 1: A offers x1. B responds. Period 3: A offers x3 = 1. B accepts. (,1- )

Strategic Bargaining In period 2 A will accept . Thus B will get the payoff 1 -  in period 2. What is the present- value to B in period 1 of N ? Period 2: B offers x2 = . A accepts. (x1,1-x1) 1 1 Y x1 B N x2= A B A N Y Period 1: A offers x1. B responds. (,1- )

Strategic Bargaining In period 2 A will accept . Thus B will get the payoff 1 -  in period 2. What is the present- value to B in period 1 of N ? (1 - ). Period 2: B offers x2 = . A accepts. (x1,1-x1) 1 1 Y x1 B N x2= A B A N Y Period 1: A offers x1. B responds. (,1- )

Strategic Bargaining In period 2 A will accept . Thus B will get the payoff 1 -  in period 2. What is the present- value to B in period 1 of N ? (1 - ). What is the most that A should offer to B in period 1? Period 2: B offers x2 = . A accepts. (x1,1-x1) 1 1 Y x1 B N x2= A B A N Y Period 1: A offers x1. B responds. (,1- )

Strategic Bargaining In period 2 A will accept . Thus B will get the payoff 1 -  in period 2. What is the present- value to B in period 1 of N ? (1 - ). What is the most that A should offer to B in period 1? 1 – x1 = (1 - ); i.e. x1 = 1 - (1 - ). B will accept. Period 2: B offers x2 = . A accepts. (1-(1 - ), (1 - )) 1 1 Y x1 B N x2= A B A N Y Period 1: A offers x1. B responds. (,1- )

Strategic Bargaining Period 2: B offers x2 = . A accepts.
(1-(1 - ), (1 - )) (1,0) Y 1 x1=1- (1-) 1 x3=1 Y B N B N x2= (0,0) A B A A N Y Period 1: A offers x1 = 1-(1 - ). B accepts. Period 3: A offers x3 = 1. B accepts. (,1- )

Strategic Bargaining Notice that the game ends immediately, in period 1.

Strategic Bargaining 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.

Strategic Bargaining Player 1’s share rises as  and  . Player 2’s share rises as  and  .

Chapter 30 Game Applications.

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