Mechanism Design without
Money
Lecture 2
1
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
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
• No dominant strategy for any player…
So how can we predict something?
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
• 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
• 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 have
Ui(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
Multiple Nash equilibria
• The battle of sexes
– Multiple Nash equilibria (see next slides)
7
Proof
• Claim: (Football, Football) is Nash equilibrium
• Proof:
– Consider A
Ualice(Football, Football) = 1 > Ualice (Play, Football)
=0
– Consider B
Ubob(Football, Football) = 2 > Ubob (Football,play) =
0
8
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
9
Multiple Nash equilibria
• Different equilibria are good for different players.
• Can some equilibria be better than others for
everyone?
10
Yes - coordination games
Left
Right
11
Left
Right
3/3
0/0
0/0
1/1
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…
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 (𝑈𝑖 (𝑆)) ≥ 𝐸𝑆~𝑃 (𝑆 −𝑖 ; 𝜎𝑖 )
15
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
16
𝐸𝑆~𝑃 (𝑈𝑖 (𝑆)) ≥ 𝐸𝑆~𝑃 (𝑆 −𝑖 ; 𝜎𝑖 )
Mixed Nash – properties (2)
• Claim: Let p1…pn be a Mixed Nash equilibrium.
Then for every probability distribution qi on Si
𝐸𝑆~𝑃 (𝑈𝑖 (𝑆)) ≥ 𝐸𝑆~𝑃,𝜎𝑖~ 𝑞𝑖 (𝑆 −𝑖 ; 𝜎𝑖 )
• Proof is an exercise. Follows from linearity of
expectation
17
𝐸𝑆~𝑃 (𝑈𝑖 (𝑆)) ≥ 𝐸𝑆~𝑃 (𝑆 −𝑖 ; 𝜎𝑖 )
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
𝐸𝑆~𝑃 (𝑈𝑖 (𝑆)) = 𝐸𝑆~𝑃 (𝑆 −𝑖 ; 𝜎𝑖 )
• Proof is an exercise. Again follows from
linearity of expectation.
18
𝐸𝑆~𝑃 (𝑈𝑖 (𝑆)) ≥ 𝐸𝑆~𝑃 (𝑆 −𝑖 ; 𝜎𝑖 )
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.
19
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
20
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)
21
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
22
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 > 0
F(b) = f(b) – b < 0
• So there must be x such that F(x) = 0, or f(x)=x
23
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
24
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…
25
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!
26
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?
27
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
28
Questions about mixed Nash?
29
Back to the Prisoner’s Dilemma
• Both players confessing is a Nash equilibrium
– But it sux…
• Both players remaining silent is great
30
– But it’s not an equilibrium
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 as
Welfare (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)
31
When is this notion meaningful?
• In the first game, PoA = 3. In the second, 100
Left
Right
Left
Right
Left
3/3
0/0
Left
100/100
0/0
Right
0/0
1/1
Right
0/0
1/1
• But it’s really the same game…
• PoA makes sense only when there is a real
bound, based on the structure of the game
32
Back to the game we played…
20
A
X
B
0
n
Y
• Time to count the votes… •
•
•
•
33
n
20
Choose 1 for AXB
Choose 2 for AXYB
Choose 3 for AYXB
Choose 4 for AYB
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!
34
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
35
Proof continued
• Suppose player 21 chose AXYB
– U21(Nash) = -40
– U21(Nash-21,AXB) = U21 (Nash-21,AYB) = -41
– U21(Nash-21,AXYB) = -42
36
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?
37
Back to the simpler game
20
X
n
A
B
n
Y
20
• Choose 5 for AXB
• Choose 6 for AYB
• Let’s count the votes
38
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
39
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
40
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
41
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”
42
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
43
Routing game example
0
X
x
A
B
0
Y
1
• 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
44
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
𝑥𝑗 𝑖 𝑐𝑜𝑠𝑡(𝑝𝑖 )
𝑖
45
Two ways to compute social welfare
𝑗
46
𝑥𝑗 𝑖 𝑐𝑜𝑠𝑡(𝑝𝑖 ) =
𝑖
𝑐𝑜𝑠𝑡(𝑓𝑒 )
𝑒
Nash flow
• Theorem – in a Nash flow, for every demand j,
all paths from sj to tj have the same cost
47
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
48
• Theorem – A flow f is optimal in G, if and only
if it is Nash with respect to the marginal cost
49
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
50
Questions?
• Feedback
• Office hours
51
Extra Slides
52
Chicken
53
Road example
1 hour
A
N minutes
N minutes
B
1 hour
• 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
54
Nash in road example
1 hour
A
N minutes
N minutes
B
1 hour
• In the Nash equilibrium, 25 people would take
each route, for a travel time of 85 minutes
55
Braess’ paradox
1 hour
A
N minutes
N minutes
Free
B
1 hour
• Now suppose someone adds an extra road which takes
no time at all. Travel time goes to 100 minutes
56
Multiple Nash equilibria
• 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 as
Welfare (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
1 hour
A
N minutes
N minutes
B
1 hour
• 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
1 hour
A
N minutes
N minutes
B
1 hour
• In the Nash equilibrium, 25 people would take
each route, for a travel time of 85 minutes
Braess’ paradox
1 hour
A
N minutes
N minutes
Free
B
1 hour
• Now suppose someone adds an extra road which takes
no time at all. Travel time goes to 100 minutes
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
69
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
70
A game
20
A
n
B
0
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
71
X
20
Y
•
•
•
•
Choose 1 for AXB
Choose 2 for AXYB
Choose 3 for AYXB
Choose 4 for AYB
A simpler game
20
X
n
A
B
n
Y
20
• You need to get from A to B • Choose 5 for AXB
• Travelling on AX or YB takes • Choose 6 for AYB
20 minutes
• Travelling on AY or XB takes
n minutes, where there are
n travellers
72
Reminder
• Nash equilibrium:
A strategy profile s = (s1, …,sn) is a Nash
equilibrium, if for every i and i we have
Ui(s) ≥ Ui(s-i, i)
73