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Introduction to Game Theory

3a. More on Normal-Form Games Dana Nau

University of Maryland

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More Solution Concepts   Last time, we talked about several solution concepts

  Pareto optimality   Nash equilibrium   Maximin and Minimax   Dominance   Rationalizability

  We’ll continue with several more   Trembling-hand perfect equilibrium   ε-Nash equilibrium   Rationalizability   Evolutionarily stable strategies

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Trembling-Hand Perfect Equilibrium   A solution concept that’s stricter than Nash equilibrium

  “Trembling hand”: Requires that the equilibrium be robust against slight errors or “trembles” by the agents

  I.e., small perturbations of their strategies

  Recall: A fully mixed strategy assigns every action a non-0 probability

  Let S = (s1, …, sn) be a mixed strategy profile for a game G

  S is a (trembling hand) perfect equilibrium if there is a sequence of fully mixed-strategy profiles S0, S1, …, that has the following properties:

  lim k→∞ Sk = S

  for each Sk = (s1k, …, si

k, …, snk), every strategy si

k is a best response to the strategies S−i

k

  The details are complicated, and I won’t discuss them

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ε-Nash Equilibrium   Another solution concept

  Reflects the idea that agents might not change strategies if the gain would be very small

  Let ε > 0. A strategy profile S = (s1, . . . , sn ) is an ε-Nash equilibrium if, for every agent i and for all strategies siʹ′ ≠ si,

ui (si , S−i ) ≥ ui (siʹ′, S−i ) − ε   ε-Nash equilibria always exist

  Every Nash equilibrium is surrounded by a region of ε-Nash equilibria for any ε > 0

  This concept can be computationally useful   Algorithms to identify ε-Nash equilibria need consider only a finite set of

mixed-strategy profiles (not the whole continuous space)   Because of finite precision, computers generally find only ε-Nash

equilibria, where ε is roughly the machine precision

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Problems with ε-Nash Equilibrium   For every Nash equilibrium, there are ε-Nash equilibria that approximate it, but the

converse isn’t true   There are ε-Nash equilibria that aren’t close to any Nash equilibrium

  Example: the game at right has just one Nash equilibrium: (D, R)   We can use strategy elimination to get it:

•  D dominates U for agent 1

•  On removing U, R dominates L for agent 2   (D, R) is also an ε-Nash equilibrium

  But there’s another ε-Nash equilibrium: (U, L)   In this equilibrium, neither agent’s payoff

is within ε of the agent’s payoff in a Nash equilibrium

  Problem:   In the ε-Nash equilibrium (U, L), agent 1 can’t gain more than ε by deviating

  But if agent 1 deviates, agent 2 can gain more than ε by best-responding to agent 1’s deviation

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Problems with ε-Nash Equilibrium   Some ε-Nash equilibria are very unlikely to arise

  Agent 1 might not care about a gain of ε/2, but might reason as follows: •  Agent 2 may expect agent 1 to to play D since it dominates U •  So agent 2 is likely to play R

•  If agent 2 plays R, agent 1 does much better by playing D rather than U

  In general, ε-approximation is much messier in games than in optimization problems

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Rationalizability   A strategy is rationalizable if a perfectly rational agent could justifiably

play it against perfectly rational opponents   The formal definition is complicated

  Informally, a strategy for agent i is rationalizable if it’s a best response to some beliefs that agent i could have about the strategies that the other agents will take   But agent i’s beliefs must take into account i’s knowledge of the

rationality of the others. This incorporates •  the other agents’ knowledge of i’s rationality, •  their knowledge of i’s knowledge of their rationality, •  and so on ad infinitum

  A rationalizable strategy profile is a strategy profile that consists only of rationalizable strategies

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Example Matching Pennies   Agent 1’s pure strategy Heads is rationalizable

  Let’s look at the chain of beliefs

  For agent 1, Heads is a best response to agent 2’s pure strategy Heads, …   … and believing that 2 would also play Heads is consistent with 2’s

rationality, for the following reasons   2 could believe that 1 would play Tails, to which 2’s best response is

Heads; …   … and it would be rational for 2 to believe that 1 would play Tails, for

the following reasons: •  2 could believe that 1 believed that 2 would play Tails, to which

Tails is a best response; …

–1, 1 1,–1

1,–1 –1, 1

Heads Tails

Heads

Tails

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Strategies that aren’t rationalizable Prisoner’s Dilemma   Strategy C isn’t rationalizable for agent 1

  It isn’t a best response to any of agent 2’s strategies

The 3x3 game we used earlier   M is not a rationalizable strategy for agent 1

  It is a best response to one of agent 2’s strategies, namely R

  But there’s no belief that agent 2 could have about agent 1’s strategy for which R would be a best response

5, 0 1, 1

3, 3 0, 5

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Comments   The formal definition of rationalizability is complicated because of the

infinite regress   But we can say some intuitive things about rationalizable strategies

  Nash equilibrium strategies are always rationalizable   So the set of rationalizable strategies (and strategy profiles) is always

nonempty   In two-player games, rationalizable strategies are simply those that survive

the iterated elimination of strictly dominated strategies   In n-agent games, this isn’t so

  Rather, rationalizable strategies are those that survive iterative removal of strategies that are never a best response to any strategy profile by the other agents

  Example: the p-beauty contest

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The p-Beauty Contest   At the start of my first class, I asked you to do the following:

  Choose a number in the range from 0 to 100   Write it on a piece of paper, along with your name   In a few minutes, I’ll ask you to pass your papers to the front of the

room   After class, I’ll compute the average of all of the numbers   The winner(s) will be whoever chose a number that’s closest to 2/3 of

the average   I’ll announce the results in a subsequent class

  This game is famous among economists and game theorists   It’s called the p-beauty contest   I used p = 2/3

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The p-Beauty Contest   Recall that in n-player games,

  Rationalizable strategies are those that survive iterative removal of strategies that are never a best response to any strategy profile by the other agents

  In the p-beauty contest, consider the strategy profile in which everyone else chooses 100   Every number in the interval [0,100) is a best response   Thus every number in the interval [0,100) is rationalizable

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Nash Equilibrium for the p-Beauty Contest   Iteratively eliminate dominated strategies

  All numbers ≤ 100 => 2/3(average) < 67 => any strategy that includes numbers ≥ 67 isn’t a best response to any

strategy profile, so eliminate it   The remaining strategies only include numbers < 67

=> for every rationalizable strategy profile, 2/3(average) < 45

=> any strategy that includes numbers ≥ 45 isn’t a best response to any strategy profile, so eliminate it

  Rationalizable strategies only include numbers < 45 => for every rationalizable strategy profile, 2/3(average) < 30

. . .   The only strategy profile that survives elimination of dominated strategies:

  Everybody chooses 0

  Therefore this is the unique Nash equilibrium

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p-Beauty Contest Results   (2/3)(average) = 21   winner = Giovanni

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Another Example of p-Beauty Contest Results

  Average = 32.93   2/3 of the average = 21.95   Winner: anonymous xx

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We aren’t rational   Most of you didn’t play Nash equilibrium strategies

  We aren’t game-theoretically rational agents   Huge literature on behavioral economics going back to about 1979

  Many cases where humans (or aggregations of humans) tend to make different decisions than the game-theoretically optimal ones

  Daniel Kahneman received the 2002 Nobel Prize in Economics for his work on that topic

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Choosing “Irrational” Strategies

  Why choose a non-equilibrium strategy?

  Limitations in reasoning ability •  Didn’t calculate the Nash equilibrium correctly •  Don’t know how to calculate it •  Don’t even know the concept

  Hidden payoffs •  Other things may be more important than winning

›  Want to be helpful ›  Want to see what happens ›  Want to create mischief

  Agent modeling (next slide)

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Agent Modeling

  A Nash equilibrium strategy is best for you if the other agents also use their Nash equilibrium strategies

  In many cases, the other agents won’t use Nash equilibrium strategies   If you can forecast their actions accurately, you may be

able to do much better than the Nash equilibrium strategy

  I’ll say more about this in Session 9   Incomplete-information games

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Evolutionarily Stable Strategies   An evolutionarily stable strategy (ESS) is a mixed strategy that’s “resistant to

invasion” by new strategies   This concept comes from evolutionary biology

  Consider how various species’ relative “fitness” causes their proportions of the population to grow or shrink

  For us, an organism’s fitness = its expected payoff from interacting with a random member of the population

  An organism’s strategy = anything that might affect its fitness

•  size, aggressiveness, sensory abilities, intelligence, …

  Suppose a small population of “invaders” playing a different strategy is added to a population

  The original strategy is an ESS if it gets a higher payoff against the mixture of the new and old strategies than the invaders do

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Evolutionary Stability   Let G be a symmetric 2-player game

  Recall that the matrix shows u(r,r') = payoff for r against r'

  A strategy r' invades a strategy r at level x if fraction x of the population uses r' and fraction (1–x) of the population uses r

  fitness(r) = expected payoff for r against a random member of the population = (1–x)a + xb

  Similarly, fitness(r') = (1–x)c + xd

  r is evolutionarily stable against r' if there is an ε > 0 such that for every x < ε, fitness(r) > fitness(r')

  i.e., (1–x)a + xb > (1–x)c + xd

  As x → 0, (1–x)a + xb → a and (1–x)c + xd → c   For sufficiently small x, the inequality holds if either a > c, or a = c and b > d

  Thus r is evolutionarily stable against r' iff one of the following holds:   a > c

  a = c and b > d

r'

r

r' r

a

c

b

d

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Evolutionary Stability   More generally

  We’ll use a mixed strategy s to represent a population that is composed of several different species

  We’ll talk about s’s evolutionary stability against all other mixed strategies   s is evolutionarily stable iff for every mixed strategy s' ≠ s,

one of the following holds:

•  u(s,s) > u(s',s) •  u(s,s) = u(s',s) and u(s,s') > u(s',s')

  s is weakly evolutionarily stable iff for every mixed strategy sʹ′ ≠ s, one of the following holds:

•  u(s,s) > u(s',s) •  u(s,s) = u(s',s) and u(s,s') > u(s',s')

  Includes cases where the original strategy and invading strategy have the same fitness, so the population with the invading strategy neither grows nor shrinks

r'

r

r' r

a

c

b

d

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Example

  The Hawk-Dove game   2 animals contend for a piece of food   The animals are chosen at random from the entire population

•  Each animal may be either a hawk (H) or a dove (D)   The prize is worth 6 to each   Fighting costs each 5

  When a hawk meets a dove, the hawk gets the prize without a fight: payoffs 6, 0

  When 2 doves meet, they split the prize without a fight: payoff 3, 3   When 2 hawks meet, they fight (–5 for each), each with a 50% chance

of getting the prize ((0.5)(6) = 3): payoffs –2,–2   It’s easy to show that this game has a unique Nash equilibrium (s, s),

where s = (3/5, 2/5)   i.e., 60% hawks, 40% doves

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Example   To confirm that s is also an ESS, show that, for all sʹ′ ≠ s,

u1(s, sʹ′) = u1(sʹ′, s) and u1(s, sʹ′) > u1(sʹ′, sʹ′)

  u1(s,sʹ′) = u1(sʹ′,s) is true of any mixed strategy equilibrium with full support

  To show u1(s,sʹ′) > u1(sʹ′,sʹ′), find the sʹ′ that minimizes f (sʹ′) = u1(s,sʹ′) − u1(sʹ′,sʹ′)

  s = “play H with probability 3/5, D with probability 2/5”

  sʹ′ = “play H with probability p, D with probability 1–p”

  u1(s,s') = (3/5)[–2p + 6(1–p)] + (2/5)[0p + 3(1–p)]

  u1(s',s') = p[–2p + 6(1–p)] + (1–p)[0p + 3(1–p)]

  so f (sʹ′) = u1(s,sʹ′) − u1(sʹ′,sʹ′) is quadratic in p

  Set d f(s')/d p = 0, solve for p => p = 3/5

•  So the unique minimum occurs when sʹ′ = s

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Evolutionary Stability and Nash Equilibria Theorem. Let G be a symmetric 2-player game, and s be a mixed strategy. If s

is an evolutionarily stable strategy, then (s, s) is a Nash equilibrium of G.

Proof. By definition, an ESS s must satisfy u(s,s) ≥ u(sʹ′,s), i.e., s is a best response to itself, so it must be a Nash equilibrium.

Theorem. Let G be a symmetric 2-player game, and s be a mixed strategy. If (s,s) is a strict Nash equilibrium of G, then s is an evolutionarily stable strategy.

Proof. If (s,s) is a strict Nash equilibrium, then u(s,s) > u(sʹ′,s).

  This satisfies the first of the two alternative criteria of an ESS

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Summary   We’ve discussed several more solution concepts

  trembling-hand perfect equilibria   epsilon-Nash equilibria   rationalizability

•  the p-Beauty Contest   evolutionarily stable strategies

•  Hawk-Dove game