Social Networks 101

1. Social Networks 101 Prof. Jason Hartline and Prof. Nicole Immorlica

2. Lecture Fourteen: Mixed Nash Equilibria and fixed points.

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

4. Matching pennies Heads Tails ( -1 , 1 ) ( 1 , -1 ) Heads (heads, heads) ( 1 , -1 ) ( -1 , 1 ) Tails (heads, tails) (tails, heads) (tails, tails)

5. Time for Math Corner

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

7. 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)]

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

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

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

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

12. Germany-Argentina, WorldCup ‘06

13. 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

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

15. 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

16. 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)

17. Rates of play Kickers: kick left 40% Goalies: defend left 42% of the time

18. 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?

19. 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.’’)

20. 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’)

21. 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)

22. 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.

23. 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

24. 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

25. 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

26. 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.

27. 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.

28. Time for Math Corner

29. Finding mixed NE p (1-p) Left Right ( 4 , 2 ) ( 0 , 6 ) q Up ( 3 , 3 ) ( 5 , 1 ) (1-q) Down

30. 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)

31. 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)

32. 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)

33. Next time Selfish routing and price of anarchy.