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Mechanism Design without Money. Lecture 2. Let’s start another way…. Everyone choose a number between zero and a hundred, and write it on the piece of paper you have, with your name. Then pass all the numbers to me. I will compute the average (hopefully correctly)

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let s start another way
Let’s start another way…
  • Everyone choose a number between zero and a hundred, and write it on the piece of paper you have, with your name.
  • Then pass all the numbers to me.
  • I will compute the average (hopefully correctly)
  • The winner is the one who is closest to two thirds of the average.
here is another game
Here is another game
  • No dominant strategy for any player…
so how can we predict something
So how can we predict something?
  • Imagine that the game is played many times.
  • Imagine that at some point, the profile (R2, C2) is being played
  • Then no player has incentive to move
nash equilibrium
Nash equilibrium
  • The strategy profile is called a Nash equilibrium.
  • Named after John Nash who proved existence in ‘51 (Nobel in ‘94). Original concept due to Von Neumann
  • Formally: A strategy profile s = (s1, …sn) is a Nash equilibrium, if for every i and i we haveUi(s) ≥ Ui(s-i, i)
  • So given that everyone else sticks, no player wants to move
examples of nash equilibrium
Examples of Nash Equilibrium
  • Suppose each player i has a dominant strategy di. Then (d1… dn) is a Nash equilibrium
multiple nash equilibria
Multiple Nash equilibria
  • The battle of sexes
    • Multiple Nash equilibria (see next slides)
proof
Proof
  • Claim: (Football, Football) is Nash equilibrium
  • Proof:
    • Consider AUalice(Football, Football) = 1 > Ualice (Play, Football) = 0
    • Consider BUbob(Football, Football) = 2> Ubob (Football,play) = 0
proof1
Proof
  • Claim: (Play,Play) is Nash equilibrium
  • Proof:
    • Consider A Ualice(Play, Play) = 2 > Ualice (Football, play) = 0
    • Consider BUbob(Play, Play) = 1 > Ubob(Play, Football) = 0
multiple nash equilibria1
Multiple Nash equilibria
  • Different equilibria are good for different players.
  • Can some equilibria be better than others for everyone?
what would be a good strategy
What would be a good strategy?
  • Well, you never want to put more than 66, right?
  • But if everyone never puts more than 66, you never want to put more than 44, right?
  • So everyone should play zero.
  • There is a difference between rationality and common knowledge of rationality
what do you usually get
What do you usually get?
  • Politiken played this with 19,196 for 5000 krones
no pure nash equilibrium
No (pure) Nash equilibrium
  • But players can randomize…
mixed nash equilibrium
Mixed Nash equilibrium
  • Reminder – Si is the set of possible strategies of player i.
  • Let pi : Si[0,1] be a probability distribution on strategies.
  • Functions p1…pnare a (mixed) Nash equilibrium if for every player i and strategy i
mixed nash properties 1
Mixed Nash – properties (1)
  • Claim: Every pure Nash is a mixed Nash. Proof: Let s1…snbe a pure Nash. Set pi(si)=1, pi(i) = 0 if i  si
mixed nash properties 2
Mixed Nash – properties (2)
  • Claim: Let p1…pn be a Mixed Nash equilibrium. Then for every probability distribution qi on Si
  • Proof is an exercise. Follows from linearity of expectation
mixed nash properties 3
Mixed Nash – properties (3)
  • Claim: Let p1…pn be a Mixed Nash equilibrium. Then if pi(i) > 0, or equivalently i is in the support of pi then
  • Proof is an exercise. Again follows from linearity of expectation.
back to rock paper scissors rps
Back to Rock Paper Scissors (RPS)
  • Let p1(R)=p1(P)=p1(S)= 1/3 p2(R)=p2(P)=p2(S)= 1/3
  • Then p1,p2 is a mixed Nash equilibrium for RPS, with utility 0 for both players.
  • Note that property 2 and 3 hold:
    • If player 1 switches to a distribution q1 she still gets expected utility 0
    • For every pure strategy  in the support of p1, the expected utility for player 1 playing  is 0.
existence of mixed nash equilibrium
Existence of mixed Nash equilibrium
  • Theorem (John Nash, 1951): In an N player game, if the strategy state is finite* a mixed equilibrium always exists
    • Nobel prize in 1994
    • The proof gives something a bit stronger
  • The proof is based on Brower’s fixed point theorem
brower s fixed point theorem
Brower’s fixed point theorem
  • Brower Fixed point theorem: Let f be a continuous function from a compact set B Rn to itself. Then there exists xB with f(x)=x
  • Examples:
    • B = [0,1], f(x) = x2
    • B=[0,1], f(x) = 1-x
    • B=[0,1]2, f(x,y) = (x2, y2)
brower s fixed point theorem1
Brower’s fixed point theorem
  • All the conditions are necessary
    • Consider B = R, and f(x) = x+1
    • Consider B = (0,1] and f(x) = x/2
    • Consider B = the circle defined by x2+ y2=1, and f(x) is a rotation
brower s fixed point theorem2
Brower’s fixed point theorem
  • Proof in 1D
  • Let B = [a,b]
  • If f(a) = a or f(b) = b we are done.
  • Else define F(x) = f(x) – x
  • Note:F(a) = f(a) – a > 0F(b) = f(b) – b < 0
  • So there must be x such that F(x) = 0, or f(x)=x
intuition for proving nash given brower
Intuition for proving Nash given Brower
  • Define an n dimensional function, which takes as input n strategies, and outputs n new strategies
  • The Fi(s1…sn) is player i’s best response to s1…s-i…sn
  • If F has a fixed point, it’s a Nash equilibrium
ok so mixed nash always exists
OK, so mixed Nash always exists
  • But can we find it?
    • Note it’s not NP complete or anything – it always exists
  • For over 50 years, economists tried to come up with natural dynamics which would lead to a Nash equilibrium
    • They always failed…
finding special types of nash
Finding special types of Nash
  • Are there two Nash equilibria?
  • Is there a Nash equlibrium where a player i gets utility more than k?
  • Is there an equilibrium where player i has support larger than k?
  • Is there an equilibrium where player i sometimes plays i?
  • These are all NP complete!
so what can we say about finding just one nash
So what can we say about finding just one Nash?
  • Daskalakis Papadimitriou and Goldberg showed that finding a Nash is PPAD complete even for two player games
  • Finding a Nash is just as hard as finding the fixed point
    • And the proof we gave was not constructive…
  • If there are two players, and we know the support of pi for each player, then finding the probabilities is just solving an LP
    • Who knows about LP’s?
but what s ppad
But what’s PPAD?
  • Suppose you have a directed graph on 2n vertices (dente them 0, 1, … 2n-1), with the following properties:
  • The in degree and the out degree of each vertex is at most 1
  • Vertex 0 has out degree 1, and in degree 0
  • You want to find a vertex x with in degree 1 and out degree 0
    • Such an x always exists
  • Finding it is PPAD complete
back to the prisoner s dilemma
Back to the Prisoner’s Dilemma
  • Both players confessing is a Nash equilibrium
    • But it sux…
  • Both players remaining silent is great
    • But it’s not an equilibrium
how much can a nash equilibrium suck
How much can a Nash equilibrium Suck?
  • Or in a more clean language:
    • Consider a game G. The social welfare of a profile s is defined asWelfare (s) = iUi(s)
    • Let O be the profile with maximal social welfare
    • Let N be the Nash equilibrium profile with minimal social welfare (worst Nash solution)
    • The Price of Anarchy of G is defined to beWelfare(O) / Welfare(N)
when is this notion meaningful
When is this notion meaningful?
  • In the first game, PoA = 3. In the second, 100
  • But it’s really the same game…
  • PoA makes sense only when there is a real bound, based on the structure of the game
back to the game we played
Back to the game we played…

X

20

n

  • Time to count the votes…

A

B

0

n

20

Y

  • Choose 1 for AXB
  • Choose 2 for AXYB
  • Choose 3 for AYXB
  • Choose 4 for AYB
optimal solution
Optimal solution
  • Suppose there are 26 players.
  • 10 go for AYXB, and 16 for AXYB
  • AYXB players take 10 minutes each
  • AXYB players take 40 minutes each
  • Total time spend on the road is 16*40+10*20 = 840 minutes
  • But is this a Nash equilibrium?
  • No – if a player moves from AXYB to AYXB they save 18 minutes!
nash equilibrium1
Nash equilibrium
  • Utility of a player is minus the time spent on the road
  • Claim: The following is a Nash equilibrium:
    • 20 players take route AYXB
    • 6 players take route AXYB
  • Proof: For every player i and deviation si we need to show that Ui(Nash)≥Ui(Nash-i,si)
    • Suppose player 1 chose AYXB
    • U1(Nash) = -40
    • U1(Nash-1,AXB) = U1(Nash-1,AXB) = U1(Nash-1,AXB) = -40
proof continued
Proof continued
  • Suppose player 21 chose AXYB
    • U21(Nash) = -40
    • U21(Nash-21,AXB) = U21 (Nash-21,AYB) = -41
    • U21(Nash-21,AXYB) = -42
equilibrium analysis
Equilibrium Analysis
  • The total time spent on the road is 26*40 = 1040 minutes
  • Worse than the optimal time of 858 minutes, but not much worse
  • How does that compare to us?
back to the simpler game
Back to the simpler game

X

20

n

  • Let’s count the votes

A

B

n

20

Y

  • Choose 5 for AXB
  • Choose 6 for AYB
analysis of the simpler game
Analysis of the simpler game
  • Same optimal solution as for the game with XY:
    • 13 players use AYB, and 13 use AXB
    • Same time on the road, 26*33 = 858 minutes
nash equilibria of the simple game
Nash equilibria of the simple game
  • Claim: 13 people use AXB and 13 use AYB is a Nash equilibrium of the simpler game
  • Proof: Suppose player 1 uses AXB
    • U1(Nash) = -33
    • U1(Nash-1 , AYB) = -34
  • Similarly, for a player who uses AYB
braess paradox
Braess paradox
  • By adding a road, we made the situation worse
  • The paradox exists in road networks
    • Making 42nd street one way
    • Simulations on road networks in various cities
can we bound price of anarchy on the road
Can we bound Price of Anarchy on the road?
  • Yes, but we won’t finish this lesson.
  • Let’s begin by formally defining a “road network” and showing that a pure Nash exists
  • Based on “potential functions”
routing games
Routing games
  • The problem has three ingredients:
    • A graph G
    • Demands: Each demand (commodity) is of the form: sj,tj meaning j want to move 1 unit from a vertex sj to a vertex tj
    • Each edge has a cost function: a monotone continuous function from traffic to the real numbers
routing game example
Routing game example

X

  • G is given
  • We want to route 1 unit from A to B, through AXB or through AYB
  • The costs are given. AX=AY=0, XB=x, YB=1

0

x

A

B

0

1

Y

how much do you pay
How much do you pay?
  • Suppose we have a flow on the graph
  • Each edge now has a cost – the function evaluated on the flow
  • Each path has a cost – the sum of costs of all edges in the paths
  • Each demand has a cost. If for every pj the demand passes xj on it, the cost is
nash flow
Nash flow
  • Theorem – in a Nash flow, for every demand j, all paths from sjto tj have the same cost
marginal cost
Marginal cost
  • The cost of an edge is fecost(fe)
  • The marginal cost of an edge is(fecost(fe) )’ = cost(fe) + fecost’(fe)
  • Marginal cost of a path is the sum of marginal costs of its edges
  • Theorem: In an optimal flow, all paths from sjto tj have the same marginal cost
slide49

Theorem – A flow f is optimal in G, if and only if it is Nash with respect to the marginal cost

using the theorem
Using the theorem
  • We know how to find optimal flows (greedy algorithm works)
  • Can we use this to get Nash flows?
  • For every e, we want a function gesuch that ge’ = ce
  • Then we can find the optimal flows according to the cost function ge
  • Using the theorem it’s a Nash flow for the cost ce
questions
Questions?
  • Feedback
  • Office hours
road example
Road example
  • 50 people want to get from A to B
  • There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it

1 hour

N minutes

A

B

1 hour

N minutes

nash in road example
Nash in road example
  • In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

1 hour

N minutes

A

B

1 hour

N minutes

braess paradox1
Braess’ paradox
  • Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes

1 hour

N minutes

A

B

Free

1 hour

N minutes

multiple nash equilibria2
Multiple Nash equilibria
  • The battle of sexes
    • See multiple equilibria on the board
    • Note different equilibria are better for some players
no pure nash equilibrium1
No (Pure) Nash equilibrium
  • We will get back to this – players can randomize
back to the prisoner s dilemma1
Back to the Prisoner’s Dilemma
  • Both players confessing is a Nash equilibrium
    • But it sux…
how much can a nash equilibrium suck1
How much can a Nash equilibrium Suck?
  • Or in a more clean language:
    • Consider a game G. The social welfare of a profile s is defined asWelfare (s) = iUi(s)
    • Let O be the profile with maximal social welfare
    • Let N be the Nash equilibrium profile with minimal social welfare (worst Nash solution)
    • The Price of Anarchy of G is defined to beWelfare(O) / Welfare(N)
price of anarchy
Price of Anarchy
  • Suppose a 100 people want to get from BIU to Jerusalem
  • The train takes two hours
  • Driving the car takes 1 hour+1 minute for every other driver
  • How many people will drive to Jerusalem?
the train game
The train game
  • Each player has two strategies – Car and Train.
  • The utility of each player is 0 for taking the train (regardless of the number of passengers)
  • The utility of taking the car is 60 – the number of car drivers.
  • The Nash equilibrium is that 60 drivers take the car and 40 take the train. Social welfare = 0
  • The optimal solution is that 30 people take the car, for a social welfare of 30*30 = 900
  • Think about the new entrance to Tel Aviv
  • Rigorous treatment of Price of Anarchy later
questions1
Questions?
  • Feedback
  • Office hours
road example1
Road example
  • 50 people want to get from A to B
  • There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it

1 hour

N minutes

A

B

1 hour

N minutes

nash in road example1
Nash in road example
  • In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

1 hour

N minutes

A

B

1 hour

N minutes

braess paradox2
Braess’ paradox
  • Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes

1 hour

N minutes

A

B

Free

1 hour

N minutes

feedback points
Feedback points
  • Lectures online: look in http://u.cs.biu.ac.il/~avinatan/
  • Slides are numbered
  • Algorithmic versus game theoretic focus – my initial plan was to give a few lectures of background, but will try to give juice today
  • Relevant book chapters – 17,18 (In Algorithmic Game Theory). Focus on 18.3
a scheduling problem
A scheduling problem
  • I can’t do Tuesday next week.
  • Two options:
    • Will ask someone to fill me in
    • Will move to a different time
  • I prefer the second, but it depends on you
  • Two stage vote – first we find a good time, and then you vote if you want a filler or not
a game
A game

X

20

n

  • You need to get from A to B
  • Travelling on AX or YB takes 20 minutes
  • Travelling on AY or XB takes n minutes, where there are n travellers
  • XY takes no time

A

B

0

n

20

Y

  • Choose 1 for AXB
  • Choose 2 for AXYB
  • Choose 3 for AYXB
  • Choose 4 for AYB
a simpler game
A simpler game

X

20

n

  • You need to get from A to B
  • Travelling on AX or YB takes 20 minutes
  • Travelling on AY or XB takes n minutes, where there are n travellers

A

B

n

20

Y

  • Choose 5 for AXB
  • Choose 6 for AYB
reminder
Reminder
  • Nash equilibrium: A strategy profile s = (s1, …,sn) is a Nash equilibrium, if for every i and i we haveUi(s) ≥ Ui(s-i, i)