Mechanism Design without Money

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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)• The winner is the one who is closest to two

thirds of the average.

Here is another game

• No dominant strategy for any player…

Rows/ Columns C1 C2 C3

R1 0 / 7 2 / 4 7 / 0

R2 4 / 2 5 / 5 4 / 2

R3 7 / 0 2 / 4 0 / 7

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

Rows/ Columns C1 C2 C3

R1 0 / 7 2 / 4 7 / 0

R2 4 / 2 5 / 5 4 / 2

R3 7 / 0 2 / 4 0 / 7

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

• Suppose each player i has a dominant strategy di. Then (d1… dn) is a Nash equilibrium

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Multiple Nash equilibria

• The battle of sexes– Multiple Nash equilibria (see next slides)

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Proof

• Claim: (Football, Football) is Nash equilibrium• Proof:– Consider A

Ualice(Football, Football) = 1 > Ualice (Play, Football) = 0

– Consider BUbob(Football, Football) = 2 > Ubob (Football,play) = 0

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Proof

• Claim: (Play,Play) is Nash equilibrium• Proof:– Consider A

Ualice(Play, Play) = 2 > Ualice (Football, play) = 0– Consider B

Ubob(Play, Play) = 1 > Ubob (Play, Football) = 0

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• Different equilibria are good for different players.• Can some equilibria be better than others for everyone?

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Yes - coordination gamesRight Left0/0 3/3 Left1/1 0/0 Right

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?

• Politiken played this with 19,196 for 5000 krones

No (pure) Nash equilibrium

0,0 1-,1 -1,1

-1,1 0,0 1-,1

1-,1 -1,1 0,0

• But players can randomize…

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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…pn are a (mixed) Nash equilibrium if for every player i and strategy i

𝐸𝑆 𝑃(𝑈 ¿¿ 𝑖 (𝑆))≥𝐸𝑆 𝑃(𝑆−𝑖 ;𝜎 𝑖)¿

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Mixed Nash – properties (1)

• Claim: Every pure Nash is a mixed Nash. Proof: Let s1…sn be a pure Nash. Set pi(si)=1, pi(i) = 0 if i si


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• 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

𝐸𝑆 𝑃(𝑈 ¿¿ 𝑖 (𝑆))≥𝐸𝑆 𝑃 ,𝜎𝑖 𝑞 𝑖(𝑆− 𝑖 ;𝜎 𝑖)¿


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• 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.

𝐸𝑆 𝑃(𝑈 ¿¿ 𝑖 (𝑆))=𝐸𝑆 𝑃(𝑆− 𝑖 ;𝜎 𝑖)¿


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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.

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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

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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 xB 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)

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• 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

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• 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

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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

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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…

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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!

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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?

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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

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Questions about mixed Nash?

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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

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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 be

Welfare(O) / Welfare(N)

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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

Right Left0/0 100/100 Left1/1 0/0 Right

Right Left0/0 3/3 Left1/1 0/0 Right

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Back to the game we played…

• Time to count the votes…

BA

Y

X

0

20

20n

n

• Choose 1 for AXB• Choose 2 for AXYB• Choose 3 for AYXB• Choose 4 for AYB

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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!

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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

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Proof continued

• Suppose player 21 chose AXYB– U21(Nash) = -40

– U21(Nash-21,AXB) = U21 (Nash-21,AYB) = -41

– U21(Nash-21,AXYB) = -42

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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?

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Back to the simpler game

• Let’s count the votes

• Choose 5 for AXB• Choose 6 for AYB

BA

Y

X20

20n

n

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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

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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

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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

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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”

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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

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Routing game example

• 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

BA

Y

X0

10

x

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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 ∑𝑖

𝑥 𝑗 (𝑖 )𝑐𝑜𝑠𝑡 (𝑝𝑖)

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Two ways to compute social welfare

∑𝑗∑𝑖

𝑥 𝑗 (𝑖 )𝑐𝑜𝑠𝑡 (𝑝𝑖)=∑𝑒

𝑐𝑜𝑠𝑡 ( 𝑓 𝑒)

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Nash flow

• Theorem – in a Nash flow, for every demand j, all paths from sj to tj have the same cost

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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 sj

to tj have the same marginal cost

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

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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 ge such 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

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Questions?

• Feedback• Office hours

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Extra Slides

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Chicken

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Road example

A B

1 hour

1 hour

N minutes

N minutes

• 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

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Nash in road example

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

A B

1 hour

1 hour

N minutes

N minutes

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Braess’ paradox

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

A B

1 hour

1 hour

N minutes

N minutes

Free


• The battle of sexes– See multiple equilibria on the board– Note different equilibria are better for some players

No (Pure) Nash equilibrium

0,0 1-,1 -1,1

-1,1 0,0 1-,1

1-,1 -1,1 0,0

• We will get back to this – players can randomize

Back to the Prisoner’s Dilemma

• Both players confessing is a Nash equilibrium– But it sux…

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 be

Welfare(O) / Welfare(N)

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

• 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

Questions?

• Feedback• Office hours

Extra Slides

Chicken

Road example

A B

1 hour

1 hour

N minutes

N minutes

• 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

Nash in road example

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

A B

1 hour

1 hour

N minutes

N minutes

Braess’ paradox

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

A B

1 hour

1 hour

N minutes

N minutes

Free

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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

http://u.cs.biu.ac.il/~avinatan/

http://u.cs.biu.ac.il/~avinatan/

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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

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A game

• 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

• Choose 1 for AXB• Choose 2 for AXYB• Choose 3 for AYXB• Choose 4 for AYB

BA

Y

X

0

20

20n

n

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A simpler game

• 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

• Choose 5 for AXB• Choose 6 for AYB

BA

Y

X20

20n

n

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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)

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Mechanism Design without Money