Social Networks 101 P ROF. J ASON H ARTLINE AND P ROF. N ICOLE I MMORLICA

Lecture Fourteen: Mixed Nash Equilibria and fixed points.

( -1, 1 ) HeadsTails Heads Matching pennies ( 1, -1 )( -1, 1 ) ( 1, -1 ) Mr. Row Mrs. Column Q. Is there a Nash equilibrium?

( -1, 1 ) HeadsTails Heads Matching pennies ( 1, -1 )( -1, 1 ) ( 1, -1 ) (heads, heads) (heads, tails) (tails, heads) (tails, tails)

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Mixed strategies Each player picks a strategy randomly. Prob. p Prob. (1-p) Prob. q Prob. (1-q) HeadsTailsHeadsTails

Matching pennies equilibrium Ifplays Heads with prob. q, then if plays Heads, he gets [-q + (1 – q)] plays Tails, he gets [q + -(1 – q)]

Matching pennies equilibrium Mr. Row prefers Heads if (1 – 2q) > (2q – 1) or q ½, he prefers Tails.

Matching pennies equilibrium Mr. Row’s plan of play: q < ½: Play Heads q > ½: Play Tails q = ½: Indifferent ( -1, 1 ) HeadsTails Heads ( 1, -1 )( -1, 1 ) ( 1, -1 ) q (1-q)

Matching pennies equilibrium Similarly, Mrs. Columns’s plan of play: p < ½: Play Tails p > ½: Play Heads p = ½: Indifferent ( -1, 1 ) HeadsTails Heads ( 1, -1 )( -1, 1 ) ( 1, -1 ) p (1-p)

Matching pennies equilibrium Players flip coins to pick their strategies: (p, q) = (½, ½)

Germany-Argentina, WorldCup ‘06

GermanyArgentina Kicker GoalieGoal? Kicker GoalieGoal? Right Left Right Left Right GOAAL Left Right Left Right Left Right Left GOAAL NO!!! GOAAL NO!!!

Penalty kicks ( 0.58, 0.42 ) LeftRight Left ( 0.93, 0.07 )( 0.70, 0.30 ) ( 0.95, 0.05 ) Goalie Kicker

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. ( 0.58, 0.42 ) LeftRight Left ( 0.93, 0.07 )( 0.70, 0.30 ) ( 0.95, 0.05 ) Goalie Kicker

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: a)an input to this procedure such that the players simply stay put, b)the limit of this procedure if it converges (cf. PageRank), c)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]). f(x) line x=y Fixed point 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.

Existence of fixed points Given function f, draw g(x) = f(x) – x. Note g(x) maps [0,1] to [-1,1]. g(x) = f(x) - x Fixed point g(0) = f(0) – 0 ≥ 0 g(1) = f(1) – 1 ≤ 0

Existence of fixed points Divide x-axis interval into segments. Label endpoint + if g > 0, - if g < 0. g(x) = f(x) - x Function must cross in this segment.

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.

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Finding mixed NE ( 4, 2 ) LeftRight Down Up ( 3, 3 )( 5, 1 ) ( 0, 6 ) p(1-p) q (1-q)

( 4, 2 ) LeftRight Down Up ( 3, 3 )( 5, 1 ) ( 0, 6 ) p(1-p) q (1-q) will only set 0 < p < 1 if 2q + 3(1-q) = 6q + (1-q)

( 4, 2 ) LeftRight Down Up ( 3, 3 )( 5, 1 ) ( 0, 6 ) p(1-p) q (1-q) will only set 0 < q < 1 if 4p = 3p + 5(1-p) 2q + 3(1-q) = 6q + (1-q)

( 4, 2 ) LeftRight Down Up ( 3, 3 )( 5, 1 ) ( 0, 6 ) p(1-p) q (1-q) 4p = 3p + 5(1-p) 2q + 3(1-q) = 6q + (1-q) (p=5/6,q=1/3) is a mixed NE

Next time Selfish routing and price of anarchy.