Social Networks 101

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# Social Networks 101 - PowerPoint PPT Presentation

Social Networks 101. Prof. Jason Hartline and Prof. Nicole Immorlica. Lecture Fourteen : Mixed Nash Equilibria and fixed points. Matching pennies. Q . Is there a Nash equilibrium?. Mrs. Column. Heads. Tails. ( -1 , 1 ). ( 1 , -1 ). Heads. ( 1 , -1 ). ( -1 , 1 ).

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### Social Networks 101

Prof. Jason Hartline and Prof. Nicole Immorlica

Lecture Fourteen:

Mixed Nash Equilibria

and fixed points.

Matching pennies

Q. Is there a Nash equilibrium?

Mrs. Column

Tails

( -1 , 1 )

( 1 , -1 )

( 1 , -1 )

( -1 , 1 )

Tails

Mr. Row

Matching pennies

Tails

( -1 , 1 )

( 1 , -1 )

( 1 , -1 )

( -1 , 1 )

Tails

(tails, tails)

Time for

Math Corner

Mixed strategies

Each player picks a strategy randomly.

Prob. (1-p)

Prob. (1-q)

Prob. p

Prob. q

Tails

Tails

Matching pennies equilibrium

If plays Heads with prob. q, then if

plays Heads, he gets [-q + (1 – q)]

plays Tails, he gets [q + -(1 – q)]

Matching pennies equilibrium

(1 – 2q) > (2q – 1)

or q < ½. Else if q > ½, he prefers Tails.

Matching pennies equilibrium

Mr. Row’s plan of play:

q > ½: Play Tails

q = ½: Indifferent

(1-q)

q

Tails

( -1 , 1 )

( 1 , -1 )

( 1 , -1 )

( -1 , 1 )

Tails

Matching pennies equilibrium

Similarly, Mrs. Columns’s plan of play:

p < ½: Play Tails

p = ½: Indifferent

Tails

( -1 , 1 )

( 1 , -1 )

p

( 1 , -1 )

( -1 , 1 )

(1-p)

Tails

Matching pennies equilibrium

Players flip coins to pick their strategies:

(p, q) = (½, ½)

Germany-Argentina, WorldCup ‘06

Germany

Argentina

Goalie

Goal?

Goalie

Goal?

Kicker

Kicker

Right

Right

GOAAL

Left

Left

GOAAL

Left

GOAAL

Right

Right

NO!!!

Right

Right

Left

GOAAL

Left

Left

GOAAL

Right

GOAAL

Left

Right

NO!!!

Right

Penalty kicks

Goalie

Left

Right

( 0.58 , 0.42 )

( 0.95 , 0.05 )

Left

Kicker

( 0.93 , 0.07 )

( 0.70 , 0.30 )

Right

Penalty Kicks

Experiment: In your envelope is a card indicating if you are a Kicker or a Goalie. Your payoff will be the average over all of your opponents x 2.

Goalie

Left

Right

( 0.58 , 0.42 )

( 0.95 , 0.05 )

Left

Kicker

( 0.93 , 0.07 )

( 0.70 , 0.30 )

Right

Mixed Nash equilibrium

Kickers: kick to the left 39% (61% to the right)

Goalies: defend left 42% of the time (defend right 58% of the time)

Rates of play

Kickers: kick left 40%

Goalies: defend left 42% of the time

Equilibria

Last time: Dominant strategy equilibria and pure Nash equilibria don’t always exist.

Today, so far: Mixed Nash equilibria exist for matching pennies and penalty kicks.

Do mixed Nash equilibria always exist?

John Nash, Ph.D. Thesis, 1950

Mixed NE always exist in finite games.

Nobel Prize, 1994

(von Neumann: ``That’s trivial, you know. That’s just a fixed point theorem.’’)

Intuition

Procedure (for finding mixed NE):

Initially, players have mixed strategies (p, q)

Repeatedly,

Mr. Row picks a best-response p’ to q

Mrs. Column picks a best-response q’ to p

Set p = p’, q = q’

f(p,q) = (p’, q’)

Intuition

A Nash equilibrium is:

• an input to this procedure such that the players simply stay put,
• the limit of this procedure if it converges (cf. PageRank),
• a fixed-point of the “best-response function” f(p,q)
Intuition

Every well-behaved function has a fixed point.

NE are fixed pts of the best-response function mapping each strategy profile to a new profile in which every player plays his best-response.

Existence of fixed points

Consider a continuous function mapping real numbers between 0 and 1 to themselves.

(i.e., f(x) is a number in [0,1]).

line x=y

1

As defined, best-response function is not exactly continuous, but we can fix this using the fact that we are looking at mixed strategies.

This comes from the fact that we have finitely many players, each of which has finitely many pure strategies.

f(x)

Fixed point

0

0

1

Existence of fixed points

Given function f, draw g(x) = f(x) – x.

Note g(x) maps [0,1] to [-1,1].

1

g(0) = f(0) – 0 ≥ 0

Fixed point

0

0

1

g(x) = f(x) - x

-1

g(1) = f(1) – 1 ≤ 0

Existence of fixed points

Divide x-axis interval into segments.

Label endpoint + if g > 0, - if g < 0.

1

Function must cross in this segment.

0

+

+

+

-

-

-

-

g(x) = f(x) - x

-1

Existence of fixed points

For any labeling of the interval in which the left-most point is “+” and the right-most point is “-”, there must be a bi-chromatic segment (i.e., a segment with a “+” on one side and a “-” on the other side).

+

+

+

-

+

-

-

Bi-chromatic segments.

Higher dimensions

Puzzle: show there is a tri-chromatic triangle

(i.e., a triangle with a red, blue, and green vertex).

Color top node blue, left node green, right node red.

Nodes on the edges get a color equal to one of the endpoints.

Nodes in the middle get an arbitrary color.

Enter grey region in a bi-chromatic edge.

Continue traversing bi-chromatic edges until you hit tri-chromatic triangle.

Number of bi-chromatic edges along a side is odd, so we can always find a new entry.

Time for

Math Corner

Finding mixed NE

p

(1-p)

Left

Right

( 4 , 2 )

( 0 , 6 )

q

Up

( 3 , 3 )

( 5 , 1 )

(1-q)

Down

p

(1-p)

Left

Right

( 4 , 2 )

( 0 , 6 )

q

Up

( 3 , 3 )

( 5 , 1 )

(1-q)

Down

will only set 0 < p < 1 if

2q + 3(1-q) = 6q + (1-q)

p

(1-p)

Left

Right

2q + 3(1-q) = 6q + (1-q)

( 4 , 2 )

( 0 , 6 )

q

Up

( 3 , 3 )

( 5 , 1 )

(1-q)

Down

will only set 0 < q < 1 if

4p = 3p + 5(1-p)

p

(1-p)

Left

Right

( 4 , 2 )

( 0 , 6 )

q

Up

( 3 , 3 )

( 5 , 1 )

(1-q)

Down

2q + 3(1-q) = 6q + (1-q)

(p=5/6 ,q=1/3)

is a mixed NE

4p = 3p + 5(1-p)

Next time

Selfish routing

and price of anarchy.