Game Theory Notes
2
Table of Contents
1.
2.
3.
Background Stuff
Overview
Extensive Form
3
5
6
4.
5.
6.
Strategies
Preferences
Normal Form
10
12
13
7.
8.
9.
Efficiency
Beliefs
Mixed Strategies
16
17
18
10.
11.
12.
Expected Utility
Dominance
Best Response
19
20
22
13.
14.
15.
Nash Equilibrium
Mixed-Strategy Nash Equilibrium
Existence of Nash Equilibrium
23
27
30
16.
17.
18.
Cournot Model
Hotelling Model
Bertrand Model
32
34
35
19.
20.
21.
Weaknesses of Nash Equilibrium
Subgames
Subgame-Perfect Nash Equilibrium
40
41
42
22.
23.
24.
Stackelberg Model
Limit Capacity
Advertising
43
44
47
25.
26.
27.
Price Guarantees
Strictly Competitive Games
Equivalent Nash Equilibria
48
50
51
28.
29.
30.
Security Strategies
Parlour Games
One-Card Poker
52
54
55
3
Background Material for Introductory Game Theory
Calculus
As with most standard microeconomic theory, game theory typically assumes that an
agent behaves optimally (from its own point of view). In formal modelling, this
frequently involves taking a derivative of some objective function (the thing the agent
cares about) with respect to some choice variable(s) (the thing(s) the agent directly
affects) and setting that derivative equal to zero (in hopes of characterizing the maximum
of the objective function). The bottom line is that you want to have a solid grounding in
introductory calculus for this course.
Economics
You need suprisingly little background in economics for this course. It helps to be
familiar with utility and the way economists think about efficiency.
Set Theory
Set theory is used in game theory largely for compact notation.
A set is an unordered collection of things. The things that make up the set are the
elements of the set, sometimes called the members of the set. A set can consist of a
collection of anything, even other sets. For the purpose of this course, sets will normally
consist of strategies.
Notation:
A = {1, 5, 7} means the set A has elements 1, 5, and 7. Curly brackets are normally used
to enclose the elements of a set and the elements are separated by commas. (Round
brackets are used for vectors for which the order of the elements matters.)
The order of elements doesn't matter:
If B = {5, 1, 7} then A = B and we can also write B = {1, 5, 7}.
Repeated elements shouldn't be there:
Writing A = {1, 5, 5, 7} is bad. Not really that it's wrong--its most likely effect is to
confuse the reader.
Sets can be members of sets:
If C = {2, 3} and D = {A, C} then D = {{1, 5, 7}, {2, 3}}.
The 'nesting' of elements within sets within sets matters:
If E = {1, 5, 7, 2, 3} then D ≠ E. {{1}} ≠ {1}.
The empty set:
There is something known as the 'empty set', denoted φ (the Greek letter phi, sometimes
written as ϕ ) which has no members. It's also sometimes called the 'null set'.
4
Additional notation:
∈ means 'is a member of'. e.g. 1 ∈ A, A ∈ D
∉ means 'is not a member of. e.g. 1 ∉ C
∪ is an operator denoting the union of two sets. e.g. E=A ∪ C, {1, 5} ∪ {5, 7}=A
∩ is an operator denoting the intersection of two sets. e.g. C ∩ {2, 4}={2}, A ∩ C= φ
⊆ means 'is a subset of'. e.g. A ⊆ E, A ⊆ A
⊂ means 'is a proper subset of'. e.g. A ⊂ E
× is an operator denoting the cross product of two or more sets. The cross product of n
sets is the set of all n-element vectors where each successive element of a vector is a
member of each successive set. e.g. A × C = {(1, 2), (1, 3), (5, 2), (5, 3), (7, 2), (7, 3)}
and A × C × C = A × C 2 ={(1,2,2),(1,2,3),(1,3,2),(1,3,3),(5,2,2),(5,2,3),(5,3,2),(5,3,3),
(7,2,2),(7,2,3),(7,3,2),(7,3,3)}
Logical notation:
~ means "not".
∋ means "such that".
also means "such that".
∃ means "there exists at least one". This is normally read as just "there exists".
∀ means "for every".
Examples of the use of this notation:
A ⊆ B ⇔ ∀ x ∋ x ∈ A, x ∈ B defines the subset relation. "A is a subset of B if and only if
for every element of A, that element is also a member of B."
A ⊂ B ⇔ ∀ x ∋ x ∈ A, x ∈ B, ∃ y ∈ B ∋ y ∉ A defines the proper subset relation. "A is a
proper subset of B if and only if for every element of A, that element is also a member of
B and there exists an element of B not contained in A.
{
}
If E = {1, 5, 7, 2, 3}, then defining F = x x ∈ E , x > 4 we have F={5, 7}. "F is the set
of numbers such that they are members of E and greater than 4."
{
}
Cross product can be defined by A × B = ( x , y ) x ∈ A, y ∈ B .
Other set notation:
: the set of real numbers.
: the set of integers.
Continuous intervals: [ a, b ] = { x ∈
x ≥ a, x ≤ b} , ( a, b ] = { x ∈
x > a, x ≤ b} , etc.
5
Overview of Game Theory
A game is a formal representation of strategic interaction between rational players.
Strategic interaction requires players to reason about the other players—to anticipate,
accommodate, react to, and/or otherwise deal with the actions of the other players.
Generally, players of a game care about how other individual players act. As such a
game requires more than one player (or else there is no interaction between players) but
not too many (because normally then the actions of one individual player have an
insignificant effect on any other individual player). Examples of possible games include
most sports, most parlour games (i.e. board games, card games, etc.), aspects of
international relations, market interaction (by a small enough number of participants), the
design of law, public policy, and contracts, many social relationships, and evolution and
learning.
The formal representation of a game requires abstracting away everything not necessary
for the analysis of how the players might (or perhaps should) play. The essential parts of
a game for formal analysis are:
1) a set of players
2) a set of rules that include:
- actions available for the players to take
- rules about when players can or must take those actions or choose between actions
- the information players know when they take those actions or choose between them
- rules about how the actions players take lead to outcomes
3) preferences players have about the outcomes
Assumptions that are normally made in game theory, and that we will use in this course
include:
- The players are rational. This is taken to mean self-interested and intelligent. The
intelligence assumption is often abandoned when using game theory to model evolution
or learning.
- The players have perfect recall. They remember everything they have ever known. In
particular, as a game progresses they remember what they have done and what they have
seen others do.
- The players know the game. All the formal aspects of the game are known to the
players. e.g. There are no actions possible to take that a player might be able to identify
or dream up but are hidden. The consequences of actions are understood. The
preferences of all players are known by each player. Note that this does not require that
there be no randomness to the rules of the game or even the preferences, but if there is
randomness, the players know the probabilities involved.
6
The Extensive Form of a Game
This is one way of graphically representing a game. An extensive form representation
includes all of the information necessary to formally represent the game. It is probably
best to explain via examples. You can probably figure out most of what's going on
intuitively.
The terminology you should get from this is:
- decision node
- initial node
- terminal node
- successor node
- predecessor node
- immediate successor
- immediate predecessor
- information set (every decision node is in some information set)
- payoff
Prisoners' Dilemma
Two people (player 1 and player 2) are arrested and charged with two crimes, a serious
one and a minor one. The police separate them and tell each of them the following thing:
"You can choose between ratting out your accomplice or keeping quiet. If you both keep
quiet, we have enough evidence to convict you both of the minor crime and you both
serve one year's prison time. If one of you rats the other out but the other keeps quiet, the
quiet one will take the whole punishment and do 5 years time, and to show our gratitude,
we will let the ratter go without charge. If you both rat each other out, you will both be
convicted of the major crime, but the punishment will be shared so that you each do 3
years time." The only thing player i cares about is the amount of time he serves. (Less is
better.)
1
Q
R
2
Q
R
Q
R
( ) ( )( ) ( )
-1
-5
0
-3
-1
0
-5
-3
7
The Ultimatum Game
Two players have to split a pie. Player 1 proposes a division of the pie: a fraction, x, that
player 1 gets to keep and, consequently, a fraction that player 2 gets to keep (the
remainder). This proposal is then presented to player 2, who has two possible actions: to
accept the proposal (in which case player 1 gets x of the pie and player 2 gets 1-x of the
pie) or reject it resulting in neither player receiving any of the pie. The only thing player
i cares about is how much of the pie he gets.
1
x
2
A
( )
x
1-x
R
( )
0
0
8
A Sequential Entry Game
There is a market with no firms servicing it, but two potential entrants. First Firm 1
decides whether or not to enter the market. Firm 2 observes this choice and then decides
whether to enter the market. Firm 1 observes that choice. Then both firms
simultaneously decide whether to build large or small plants for production. The payoffs
(or profits) to Firm i are:
4 - if firm i alone enters and builds a big plant
3 - if firm i alone enters and builds a small plant
2 - if firm i has a large plant and firm j has a small plant 2
1 - if both have small plants 1
0 - if firm i doesn't enter 0
-1 - if both have large plants
-2 - if firm i has a small plant and firm j has a large plant
1
E
N
2
2
EE
EN
NE
1
1
SEN
NN
2
()
()()()()
()()()()
BEE
BEN
SEE
BNE
0
SNE
0
2
BEE
SEE
BEE
SEE
-1
2
-2
1
-1
-2
2
1
4
3
0
0
0
0
4
3
9
Some rules about the extensive form:
-The various actions as represented by the arrows coming from a decision node must be
things the relevant player can actually distinguish between. e.g. A batter isn't choosing
between 'be prepared' and 'don't be prepared' for a particular type of pitch if he doesn't
know what sort of pitch is coming. In paper-scissors-rock, you can't choose 'whatever
wins' you can only choose between 'paper', 'scissors', and 'rock'.
-Each node has only one predecessor (except the initial node which has none). i.e. There
is only one path to reach any given node (or outcome) from the initial node.
-Within an information set each node must belong to the same player (otherwise it would
be something like a player being uncertain who they are) and each node must have the
same actions leading from it (otherwise an intelligent player could deduce which node
they were actually at).
-You can (as the textbook does) draw the game as if it progresses sideways. Sometimes
it is handy to draw it progressing in multiple directions. It's the arrows that matter for
sorting out the sequence of decisions the players take.
-By convention, the payoffs should be listed in the same order as the players' order of
first moves. (Or the implicit order of the moves when simultaneous moves are
represented.)
It is a good idea not to label two nodes for the same player with the same actions. Above,
I have distinguished between, say, the entry decision for firm 2 when it knows firm 1 has
entered and the similar decision when firm 1 has not entered using subscripts. A
commonly used alternative method is to label nodes which share identical action
possibilities. The point of this is that, in the later analysis, you don't want to get
references to the actions from the two nodes confused just because they can be described
briefly with the same words.
10
Strategies
A strategy is a complete contingency plan for a player to play the game. It contains
information specifying the action to be taken by a player at every information set where
that player takes action. This includes decision nodes that the strategy itself may prevent
from being reached.
Note that a strategy does not have to be intelligent or justifiable.
A strategy does not specify the rationale for a particular course of action. The statement
"My strategy was to sacrifice my bishop and hope he wouldn't notice his queen was in
danger." is using the word strategy with a different meaning than the technical game
theoretic one.
We will denote a strategy of player i as si.
A strategy set for player i is the set of all possible strategies available to player i. This is
denoted Si.
A strategy profile is a vector of strategies, one for each player. It's denoted s. So with n
players we have s = (s1, s2,...,sn).
The set of all possible strategy profiles is denoted S. Since strategies are just contingent
plans of action (which may or may not be good plans), one player having some particular
strategy never rules out another player having any particular strategy. This means that
with n players S = S1 × S 2 × …× S n .
For the analysis we employ later, it is handy to define the notation s-i as being a strategy
profile involving every player except player i. S-i is the set of all such incomplete
strategy profiles.
11
Prisoners' Dilemma
We can have s1=R or s1=Q so S1={R,Q}. Similarly for player two, s2=R or s2=Q so
S2={R,Q}. This gives us four possible strategy profiles: s=(R,R), s=(R,Q), s=(Q,R), or
s=(Q,Q) and S={(R,R),(R,Q),(Q,R),(Q,Q)}.
Ultimatum Game
Player 1's strategy can be any real number between 0 and 1 inclusive. Thus S1=[0,1].
Player 2's strategy has to specify a contingent plan of action, so it has to specify whether
to accept or reject for every possible action of player 1. For example, one of player 2's
strategies is 'accept if and only if x>.25'. Another is 'accept if and only if x<.75'. Another
is 'accept if and only if x is a rational number'. Another is 'never accept'. Any function
that maps [0,1] into {accept, reject} will constitute a strategy of player 2's. Thus
S 2 = f f :[0,1] → { A, R} (the set of all such functions).
{
}
Sequential Entry Game
S1={(E,BEE,BEN), (E,BEE,SEN), (E,SEE,BEN), (E,SEE,SEN), (N,BEE,BEN), (N,BEE,SEN),
(N,SEE,BEN), (N,SEE,SEN)}
S2={(EE,EN,BEE,BNE), (EE,EN,BEE,SNE), (EE,EN,SEE,BNE), (EE,EN,SEE,SNE),
(EE,NN,BEE,BNE), (EE,NN,BEE,SNE), (EE,NN,SEE,BNE), (EE,NN,SEE,SNE), (NE,EN,BEE,BNE),
(NE,EN,BEE,SNE), (NE,EN,SEE,BNE), (NE,EN,SEE,SNE), (NE,NN,BEE,BNE), (NE,NN,BEE,SNE),
(NE,NN,SEE,BNE), (NE,NN,SEE,SNE)}
12
Preferences
Preferences are defined over outcomes, but since each strategy profile uniquely
determines an outcome, we can (and it's usually convenient to) instead define preferences
over strategy profiles. Player i's preferences are payoffs given by a function ui mapping
the strategy space S into the real numbers. Formally, ui : S → (where is the set of
real numbers).
Everything that the players care about must be represented in the payoffs. This heads off
such lines of reasoning as 'The prisoners might be friends in which case betraying their
friends is even worse than saving 2 years prison time so they would never do that." If
such things matter to the prisoners, this should be reflected in the payoffs along with the
effect of prison time.
Prisoners' Dilemma
u1(R,R) = -3, u1(R,Q) = 0, u1(Q,R) = -5, u1(Q,Q) = -1
u2(R,R) = -3, u2(R,Q) = -5, u2(Q,R) = 0, u2(Q,Q) = -1
Ultimatum Game
Defining the preferences explicitly over strategy profiles:
u1 s1 , f f ( s1 ) = A = s1
(
u (s , f
u (s , f
u (s , f
1
1
2
1
2
1
)
f (s ) = R) = 0
f ( s ) = A) = 1 − s
f (s ) = R) = 0
1
1
1
1
or alternatively:
If f(s1) = A then u1(s1,f) = s1 and u2(s1,f) = 1-s1
If f(s1) = R then u1(s1,f) = 0 and u2(s1,f) = 0.
13
Normal Form
A game represented in normal form gives the set of players, their strategy sets, and their
payoff functions. If a game has two players and a finite number of strategies, the normal
form is normally done as a payoff matrix. By convention, the first number listed in a
payoff matrix's cell is the payoff for the player on the left side; the second one is for the
player on top.
The normal form doesn't give direct information about the timing of moves or
information sets. As such, if these things matter, it may be that the normal form is not the
best way to look at a game. However, for games in which the players each have only one
information set at which they choose (simultaneous move games), the normal form isn't
missing any formal aspect of the game.
Prisoners' Dilemma
Player 2
Q
R
Q
-1 , -1
-5 , 0
R
0 , -5
-3 , -3
Player 1
Ultimatum Game
S1=[0,1]
S 2 = f f :[0,1] → R
If f(s1) = A then u1(s1,f) = s1 and u2(s1,f) = 1-s1
If f(s1) = R then u1(s1,f) = 0 and u2(s1,f) = 0.
{
}
Matching Pennies
Two players simultaneously choose either heads or tails. If they choose the same thing,
player 1 wins and player 2 loses. If they choose different things, player 2 wins and player
1 loses. The only thing they care about is winning (good) and losing (bad).
Player 2
H
T
H
1,0
0,1
T
0,1
1,0
Player 1
14
Battle of the Sexes
A couple are going to see a movie and there are two choices: an action movie (A) and a
romance movie (R). If they see the same movie, the boy prefers to see the action movie
and the girl prefers to see the romance movie. For both of them the worst outcomes are
that they don't see the same movie.
Girl
A
R
A
2,1
0,0
R
0,0
1,2
Boy
Chicken
Two maniacs are driving down the same road on motorcycles in Bangkok in the middle
of the street so that they can go faster than the cars. They are headed towards each other
and there is not enough room for them to pass each other in the middle. They have to
each choose whether to continue driving fast (F) in the middle or slowing down (S) and
pulling into the cars' area to avoid a potential collision. They both have preferences such
that: The best thing that could happen is that I go straight and the other guy pulls over.
The next best thing is that we both pull over. The next best thing is that I pull over but
that other jerk gets to keep going fast. The worst thing that could happen is that neither
of us pull over and we collide.
Maniac 2
F
S
F
0,0
5,1
S
1,5
2,2
Maniac 1
Meeting in New York
Two people have arranged to meet in New York City. Neither can remember where they
planned to meet, but both can remember that it is either Grand Central Station or the
Empire State Building. The only thing either of them care about is that they meet; they
don't care where.
Player 2
GCS
ESB
GCS
1,1
0,0
ESB
0,0
1,1
Player 1
15
Pareto Coordination
This is a variant on meeting in NY where both people prefer one particular meeting place
over the other. The worst outcome for both is that they don't meet, in which case neither
of them care where they are.
Player 2
A
B
A
1,1
0,0
B
0,0
2,2
Player 1
16
Efficiency
In economics efficiency normally refers to 'Pareto efficiency' which is sometimes also
referred to as 'Pareto optimality'.
A situation is Pareto optimal if and only if it is impossible to make someone better off
without making anyone else worse off.
Prisoners Dilemma
(Q,R), (R,Q), & (Q,Q) are efficient.
(R,R) is not.
Ultimatum Game
Any strategy profile in which player 2 accepts player 1's proposal is efficient.
Any strategy profile in which player 2 rejects player 1's proposal is not efficient.
Matching Pennies
All outcomes are efficient.
Battle of the Sexes
When they go to the same movie, the outcome is efficient.
When they go to different movies it is not.
Chicken
The outcome in which they collide is not efficient. All the other ones are.
Meeting in NY
If and only if they meet is it an efficient outcome.
Pareto Coordination
It is an efficient outcome if and only if they meet in the preferred spot.
17
Beliefs
A belief of player i in the context of game theory is a probability distribution over the
strategies of the other players S-i. A belief of player i is symbolized by μ −i (the lowercase Greek letter mu). A belief of player i's assigns to each strategy profile for the other
players a probability that they will employ that strategy. So a belief is a function that
maps the set of other players' strategy profiles into [0,1] (formally we have
μ −i : S −i → [0,1] ). Since it is a probability distribution it has the property that if you add
up the probabilities for all possible strategy profiles of the other players, you must get 1.
(Formally, we have that ∑s ∈S μ −i ( s −i ) = 1 .)
−i
−i
A belief does not have to be sensible, intelligent, well-founded, or justifiable. It just has
to be a probability distribution over the other players' strategies.
It may reflect uncertainty on the part of the player holding the belief. This uncertainty
could be grounded in ignorance or a belief that other players' behaviour is actually
random.
18
Mixed Strategies
So far we have only dealt with what are called pure strategies, where players plans of
action dictate a certain course of action.
Mixed strategies expand the idea of plans of action to include randomization over actions.
We denote a mixed strategy for player i as σ i . A mixed strategy is a probability
distribution over pure strategies. So, similar to beliefs, it is a function mapping a set of
possible strategies (in this case Si) into [0,1] such that the sum of probabilities associated
with all possible strategies is 1. (Formally σ i : S i → [0,1] ∋ ∑s ∈S σ i ( si ) = 1 .)
i
i
19
Expected Utility
The payoffs we've looked at so far reflect the players' preferences about certain
outcomes. We need to expand our analysis to include uncertainty about outcomes. To do
this we invoke the Expected Utility Theorem devised by John von Neumann and Oskar
Morgenstern.
What it says is that if
-L, M, & N are any three outcomes (or states of the world)
-someone has preferences such that L>M implies that the lottery aL+(1-a)N > the lottery
aM+(1-a)N (This is referred to as the Axiom of Independence.)
then there is a utility function representing their preferences such that:
u(aL+(1-a)N) = au(L) + (1-a)u(N)
So if our payoffs are utilities and we assume the independence axiom holds, we can say
that players seeking to maximize their utility will also seek to maximize their expected
utility. So for example, a mixed strategy that provides a player with utility of 1 half of
the time and 2 the other half of the time has an expected utility of 1.5 and that this
expected utility is what players care about.
For notation then, we redefine ui to be expected utility.
For our prisoners' dilemma:
I'll use (R,Q) to order the values the beliefs and mixed strategies can take.
if σ 1 = ( 13 , 23 ) , then u1 (σ 1 , R ) = 13 ( −3) + 23 ( −5) = −313 .
if σ 1 = ( 13 , 23 ) and μ −1 = ( 25 , 53 ) , then
u1 (σ 1 , μ −1 ) = 13 [25 ( −3) + 53 (0)] + 23 [25 ( −5) + 53 ( −1)] = −1532 .
Any two expected utility functions u and v such that u = a + bv where b > 0 represent the
same preferences. This is sometimes handy as a time-saving device.
It works the other way around too. If two expected utility functions u and v represent the
same preferences, then ∃b > 0, a ∈ u = a + bv .
20
Dominance
A pure strategy is dominated if there is a different strategy (pure or mixed) that does
better no matter what the other players do.
Formally, si is dominated if ∃σ i ∈ ΔSi ∋ ui (σ i , s−i ) > ui ( si , s−i ) ∀s− i ∈ S− i .
In the prisoners' dilemma, Q is dominated by R.
In the game below:
Player 2
P
l
a
y
e
r
A
X
3,*
Y
2,*
Z
1,*
B
3,*
4,*
2,*
C
4,*
3,*
2,*
1
A is dominated. This is rather straightforward to see since it is dominated by the pure
strategy C. (Note it isn't dominated by the pure strategy B since for X they give the same
payoff.)
Consider the game below:
Player 2
P
l
a
y
e
r
A
X
1,*
Y
2,*
Z
3,*
B
0,*
3,*
3,*
C
4,*
1,*
4,*
1
Is A dominated?
It's not dominated by either of the other pure strategies. Is it dominated by a mixed
strategy?
21
First off, note that if it is, it'll be dominated by a mixed strategy assigning positive
probability only to strategies that aren't A (as well as ones that do assign positive
probability to A). So we only need to check mixed strategies not using A. (This is a
principle that extends to all games.)
Secondly, we're only interested in comparing A and mixed strategies over B and C when
combined with pure strategies for player 2.
So, let p represent a mixed strategy over B and C assigning probability p to B and 1-p to
C. For p to dominate A we need:
u1(A,X)<u1(p,X)
1< (0)(p) + 4(1-p)
p<3/4
u1(A,Y)<u1(p,Y)
2< 3p + 1(1-p)
p>1/2
u1(A,Z)<u1(p,Z)
3<3p+4(1-p)
p<1
So any p that satisfies all three of those conditions gives us a strategy that dominates A.
i.e. any p where 1/2<p<3/4 will do. So A is dominated. If it weren't possible to have a p
satisfying all three conditions, A would not be dominated.
For example in the game below:
Player 2
P
l
a
y
e
r
A
X
1,*
Y
4,*
Z
3,*
B
0,*
5,*
3,*
C
4,*
0,*
4,*
1
u1(A,X)<u1(p,X)
1< (0)(p) + 4(1-p)
p<3/4
u1(A,Y)<u1(p,Y)
4< 5p + 0(1-p)
p>4/5
Since no p will satisfy both of these conditions, we conclude that A is not dominated.
22
Best Response
A player's best response to strategies for other players is the set of strategies that
maximize his own payoff given the other players' strategies.
{
}
Formally, BRi ( s− i ) = si ∈ Si ui ( si , s− i ) ≥ ui ( si′, s− i ) ∀si′ ∈ Si .
Strictly speaking, the best response is a set of strategies and so should be written in curly
brackets even when there is a unique best response. But that gets things unnecessarily
cluttered and so often the brackets will get dropped. (In a fit of excessive ambition for
tedium, the text claims at the bottom of p50 it'll write best responses in curly brackets.
By p72 they've given up.) For a lot of games, BRs have one element and it's common to
refer to the 'best response function' and drop the curly brackets.
23
Nash Equilibrium
Here we look at our first "solution concept": the Nash equilibrium. (A solution concept is
a way of predicting what people will do in a game.)
A Nash equilibrium is a strategy profile such that no player can change his strategy
(holding all other players' strategies constant) to increase his payoff.
Formally, s* is a Nash equilibrium if and only if
∀ i , ∀ si ′ ∈Si , ui ( si* , s−*i ) ≥ ui ( si ′ , s−*i ) .
A strategy profile is a Nash equilibrium if and only if every player's strategy is a best
response to the other players' strategies.
For now, we will only consider pure strategies.
With a two player game with strategy sets (i.e. one that can be represented in normal
form as a payoff matrix) finding a Nash equilibrium is straightforward. Put a mark of
some sort by the payoff of a player when the strategy profile has that player playing a
best response. Any strategy profile with two marks will be a Nash equilibrium.
Player 2
A
Player 1
B
A
3,3
5,0
* .
B
0,5
*
1,1
* *
BR1(A)={B} so a star under the 5 in the bottom left.
BR1(B)={B} so a star under the 1 on the left.
BR2(A)={B} so a star under the 5 in the top right.
BR2(B)={B} so a star under the 1 on the right.
Only the strategy profile (B,B) has two stars, so it is the only Nash equilibrium.
Consider the following version of the Nash Dating Game. There are five Princeton math
students at a bar, all of whom are boys. They are the players of this game. There are six
girls together at the bar, one is blonde and the other five are brunettes. Each of the math
students simultaneously chooses one girl to ask out on a date, which is the only move
they make in this game. If more than one boy asks the same girl on a date, she refuses to
go on a date with any of them. If a girl is asked out by only one boy, she accepts and
they have a date.
Each of the math students only cares about whether or not he has a date and whether or
not it is with the blonde or a brunette. They all have the same preferences and believe
that the best outcome is a date with the blonde, the second best outcome is a date with a
brunette, and the worst outcome is no date.
24
What is the set of pure-strategy Nash equilibria for this game?
First off we can rule out any strategy profile where there is more than one math student
asking out the same girl. In such a situation, any of those students would be better off
asking out one of the unasked girls (and there must be at least two such girls) and is
therefore not playing a best response. Thus in any Nash equilibrium, we must have each
math student asking out a different girl.
Consider now a strategy profile in which each math student is asking out a different
brunette. (This is the outcome predicted in the movie A Beautiful Mind.) Any of them
would be better off asking out the blond instead. Therefore none of them are playing a
best response and it is not a Nash equilibrium.
This leaves us only with strategy profiles in which one math student asks out the blond
and the others each ask out a different brunette. Any such strategy profile is a Nash
equilibrium. Consider the student asking out the blond. He cannot improve his payoff
since it is already as high as possible. Consider any one of the students asking out a
brunette. He cannot improve his payoff since the only way to do so is to get a date with
the blond which cannot be done since someone else is asking her out.
Alien Roulette
Two aliens (1 and 2) simultaneously choose two real numbers (x1 and x2 respectively)
from the closed interval [0,1]. As a result, they get "survival points" given by:
p1 = 3x1 – 2x12 – x1x2
p2 = 6x1x2 – 2x2 – x22
Alien 1 gets fatally shot with a ray gun with probability p1 p+1p2 . If Alien 1 isn't fatally
shot, then Alien 2 is. Each alien wants to maximize the probability of not being shot.
What are the Nash equilibria?
At any Nash equilibrium, both players must be playing a best response. We will start by
determining the players' best responses.
For player 1, his payoff is maximized (ignoring for now any boundary issues) when he
chooses x1 such that:
∂ p1
= 0.
∂ x1
∂ p1
= 3 − 4 x1 − x2 = 0 ⇒ x1 = 34 − 14 x2
∂ x1
Note that for all possible values of x2 (from 0 to 1) this can be satisfied with x1 also
between 0 and 1 inclusive (so there are no boundary issues after all). So we have player
1's best response function:
BR1 = 43 − 41 x2
Now for player 2:
∂ p2
= 6 x1 − 2 − 2 x2 = 0 ⇒ x2 = 3 x1 − 1
∂ x2
25
This however can give values for x2 outside of the range. When it suggests a value
greater than one, player 2's best response is to set x2=1 and when it suggests a value less
than zero, player 2's best response is to set x2=0. (This can be confirmed by graphing p2
as a function of x2.) Thus we have player 2's best response function:
BR2 = 0
if x1 < 13
BR2 = 3x1 − 1 if
BR2 = 1
1
3
≤ x1 ≤
if x1 >
2
3
2
3
(Or more compactly but less easily interpreted: BR2 = max {0, min {3x1 − 1,1}} .)
A Nash equilibrium will be found for any pair (x1, x2) that satisfies these two BR's.
Because of the conditional nature of one of the BR's, it is probably quickest to use a
diagram to shorten the search for intersections. For player 1's BR, we have:
X1
1
3/4
BR1
1/4
X2
1
26
For player 2's BR, we have:
X2
1
BR2
1/3
2/3
X1
1
Combining these we get:
X2
1
BR1
BR2
1/4 1/3
2/3 3/4
1
X1
So we have a single intersection of the BR's (i.e. a single pure-strategy Nash equilibrium)
which we can find by combining
x1 = 43 − 41 x2 and x 2 = 3x1 − 1 .
This gives us
x1 = 47 and x 2 = 75 .
So our unique pure-strategy Nash equilibrium for this game is the strategy profile
( 47 , 75 ) .
27
Mixed-Strategy Nash Equilibria
Consider the following game:
Player 2
C
D
A
4,1
1,2
B
1,3
5, 1
Player 1
It is straightforward to show that this game has no pure-strategy Nash equilibrium. I'll
use pA to denote player 1's mixed strategy such that the stategy A is used with probability
pA and so the probability B is used is 1-pA. pC will denote the probability player 2 uses
strategy C. (Note that these p's can be zero or one; the pure strategies are also mixed
strategies.)
To find any mixed-strategy Nash equilibrium, first we should find the players' best
responses.
For player 1, u1(pA,pC) = pA[4pC+1(1-pC)] + (1-pA)[1pC+5(1-pC)] = [7pC-4]pA + [5-4pC]
The best response will be found by choosing pA to maximize u1. We have three
possibilities:
if pC < 47
BR1 = {0}
BR1 = [0,1]
BR1 = {1}
if pC =
4
7
if pC >
4
7
PA
BR1
1
4/7
1
PC
For player 2, u2(pC,pA) = pC[1pA+3(1-pA)] + (1-pC)[2pA+1(1-pA)] = [2-3pA]pC + [1+pA]
The best response will be found by choosing pC to maximize u2. Again, we have three
possibilities:
28
BR2 = {1}
if p A <
2
3
BR2 = [0,1]
BR2 = {0}
if p A =
2
3
if p A >
2
3
PC
BR2
1
PA
2/3
1
Combining these graphs gives us:
PC
1
BR2
4/7
BR1
2/3
1
PA
For a unique mixed-strategy Nash equilibrium at the strategy profile (pA, pC) = (2/3,4/7).
Note that at the mixed-strategy Nash equilibrium, player 1 is choosing probabilities so
that player 2 is indifferent between his two pure strategies and vice versa. Generally it
will be true that whenever a player is playing a mixed strategy, the expected utilities for
29
any of the pure strategies involved have to be equal. Otherwise it would not be a best
response to play the mixed strategy. (One of the pure strategies would be better.)
In a mixed-strategy Nash a player must be indifferent between the pure strategies played
with positive probability (otherwise it wouldn't be a best response to randomize over
them). The expected values of a player's pure strategies are determined by the
probabilities used by the other player. So (in a 2-player game) the probabilities chosen
by player 1 make player 2 indifferent between his pure strategies and vice versa. This
opens up a short-cut for determining a mixed-strategy Nash equilibrium.
In the game above:
u1(A,pC) = u1(B,pC)
4pC + 1(1-pC) = 1pC + 5(1-pC)
pC = 4/7
u2(pA,C) = u2(pA,D)
1pA + 3(1-pA) = 2pA + 1(1-pA)
pA = 2/3
30
Existence of a Nash Equilibrium in Finite Games
If a game has a finite number of players each with a finite number of strategies, it must
have a Nash equilibrium—either a pure-strategy one or a mixed-strategy one (or both). I
will give a proof of this for a 2-player 2-strategy game.
Consider the generic payoff matrix:
Player 2
C
D
A
w1 , w2
x1 , x2
B
y1 , y2
z1 , z2
Player 1
We want to show that it has at least one equilibrium.
If the game has any pure-strategy Nash equilibria, then we would be done, so we'll
assume it has none and try to prove that it has to have a mixed-strategy equilibrium.
Assume BWOC (by way of contradiction) w1=y1. Thus BR1(C)={A,B}. So to avoid a
pure strategy Nash equilibrium we need x2>w2 (otherwise (A,C) would be Nash) and
z2>y2 (otherwise (B,C) would be Nash). Thus BR2(A)={D} and BR2(B)={D}. Now we
have a pure-strategy Nash equilibrium either at (A,D) (if x1>=z1) or at (B,D) (if x1<=z1).
So we must have either w1>y1 or w1<y1.
Assume WLOG (without loss of generality) w1>y1.
We can't have C in BR2(A) (because then (A,C) would be Nash) so we have x2>w2 and
BR2(A) = {D}.
We can't have A in BR1(D) (because then (A,D) would be Nash) so we have z1>x1 and
BR1(D) = {B}.
We can't have D in BR2(B) (because then (B,D) would be Nash) so we have y2>z2 and
BR2(B) = {C}.
So we have w1>y1, x2>w2, z1>x1, and y2>z2. Graphically, our best responses are:
Player 2
A
Player 1
B
C
w1 , w2
* .
y1 , y2
*
D
x1 , x2
*
z1 , z2
* .
Define pA as the mixed strategy for player 1 where A is played with probability pA (and
so B is played with probability 1-pA). Define pC as the mixed strategy for player 2 where
C is played with probability pC (and so D is played with probability 1-pC).
31
Consider the mixed-strategy profile where the two players are indifferent between their
two pure strategies (given the mixed strategy of the other guy).
This is where
u1(A,pC) = u1(B,pC)
pCw1 + (1-pC)x1 = pCy1 + (1-pC)z1
pC = (z1-x1) / [ (z1-x1) + (w1-y1) ]
Note that since z1>x1 and w1>y1 we have that 0<pC<1 so that it is a sensible probability
(and is not equivalent to a pure strategy).
Similarly we have
u2(C,pA) = u2(D,pA)
pAw2 + (1-pA)y2 = pAx2 + (1-pA)z2
pA = (y2-z2) / [ (y2-z2) + (x2-w2) ]
Again we have a sensible mixed strategy since y2>z2 and x2>w2 ensure that 0<pA<1.
Since u1(A,pC) = u1(B,pC) we have BR1(pC) = [0,1] so player 1 is playing a best response.
Similarly, u2(C,pA) = u2(D,pA) so BR2(pA) = [0,1] so player 2 is also playing a best
response. Thus this strategy profile is a (mixed-strategy) Nash equilibrium.
Nash proved the existence of a mixed-strategy Nash equilibrium for any game with a
finite number of players each with a finite number of pure strategies (by employing a
fixed point theorem).
He also argued that the concept of dominance was useful in identifying a Nash
equilibrium. Any dominated strategy is never a best response and thus can be removed
from consideration as part of any Nash equilibrium.
32
The Cournot Oligopoly Model
There are n firms all producing the same good. They simultaneously choose quantities
(any non-negative real number) to produce, their joint output enters a market with pricetaking consumers giving rise to a downward-sloping market demand curve and is sold at
a market-clearing price. This earns each firm profit—its revenue (price times its output)
minus its costs of production—which is its payoff.
We'll look first at a simple 2-firm version with linear demand and identical linear costs:
Notation:
P – market price
qi – firm i's output
Q – total output (Q = q1 + q2)
Ci – firm i's costs of production
Demand: P = a – bQ
Costs of production: Ci = cqi
Payoff: ui = Pqi – Ci = Pqi – cqi = (a-bQ)qi – cqi = (a-b(qi+qj))qi – cqi = (a-c-bqj)qi – bqi2
∂ ui
a− c 1
= a − c − bq j − 2bqi = 0 ⇒ qi =
− 2 qj
∂ qi
2b
⎧a − c 1
⎫
BRi = max⎨
− 2 q j ,0⎬
⎩ 2b
⎭
Graphically, we see we have a single pure-strategy Nash equilibrium where:
q1 = a2−bc − 21 q 2 and q 2 = a2−bc − 21 q1
This solves for the strategy profile
( a3−bc , a3−bc ) .
Now let's do this for n firms with the same costs and demand as above. Following the
⎧a− c 1
⎫
same steps gives us: BRi = max⎨
− 2 ∑ q j ,0⎬ . Again, we'll have a single purej≠i
⎩ 2b
⎭
a− c 1
− 2 ∑ q j ∀ i ∈ {1,2,..., n} . We can take
strategy Nash equilibrium where: BRi =
2b
j≠i
these n equations, rearrange them, and sum them as follows:
a − c − bq1 − bq 2 − … − bq n = bq1
a − c − bq1 − bq 2 − … − bq n = bq 2
+ a − c − bq1 − bq 2 − … − bq n = bq n
_______________________
na − nc − nbq1 − nbq 2 − …− nbq n = bQ
33
Now sub in Q = q1+q2+...+qn and solve for Q:
n(a − c) − nbQ = bQ
Q = nn+1 a b− c
Note that at n=1 we have the standard monopoly outcome and as n approaches infinity
total output approaches the competitive level.
We can also solve for qi = n1 ab− c .
Try working this out for:
- two firms with different constant marginal costs: c1 and c2.
- all firms have costs Ci = cqi + (1/2)dqi2.
34
The Hotelling Location Model
The usual story to motivate this model goes something like this: There are two ice-cream
vendors preparing to set up shop somewhere along a beach. Each of them must chose a
location and stick with it. They sell the same thing at the same prices and the only
potential difference between them from any customer's point of view is their locations.
Customers are distributed uniformly along the beach and will go to whichever ice-cream
vendor is closest. Of the customers located the same distance from the two vendors, half
will go to one and half to the other.
Formally, the two players each chose a real number from some closed interval, say [0,1].
A player's payoff is the fraction of the line where the points are closer to him than the
other player. So it is a simultaneous-move game with
S1 = S2 = [0,1]
ui = 1-.5(s1+s2)
if si>sj
ui = .5(s1+s2)
if si<sj
The best response for player i is:
{
}
BRi ( s j ) = max{x ∈ [0,1] ∋ x < s j } = φ
if sj < 0.5
BRi ( s j ) = { 0.5}
if sj = 0.5
{
}
BRi ( s j ) = max{x ∈ [0,1] ∋ x < s j } = φ
if sj > 0.5
The unique pure-strategy Nash equilibrium is (s1,s2) = (0.5,0.5).
This model is often invoked to explain not just spatial location but product differentiation
(Coke and Pepsi, Mr.Sub and Subway, Crest and Colgate, etc.), political positions in a
democracy (Republicans and Democrats), and probably other things that I can't think of
right now.
Simple variations which get roughly the same results include:
- having the location space be discrete. For example, there are n locations in a line. Now
there are no empty BRs. If n is even, you get the players in different locations at the two
in the middle. If n is odd, you get them both in the same spot: the middle.
- having the customers not be uniformly distributed.
- having the two players move sequentially with perfect information.
A variation which isn't easy to work through:
- having more than 2 players.
35
Bertrand Model
In this model, firms simultaneously and independently select a price. The firm with the
lowest price serves the whole market. If more than one firm offers the same lowest price,
the market is shared. The quantity demanded is some decreasing function of the (lowest)
price. The firms' payoffs are their profits which are the difference between revenue
(price times quantity sold) and costs of production.
Basic Case
There are two firms. Price can be any non-negative real number. The quantity demanded
is a linear function of the price. Costs of production are some multiple of the quantity
produced. If the market is shared (i.e. when they offer the same price) it is evenly split
between the two firms. In normal form, this gives us:
Players: firm 1 and firm 2
Strategy sets: Si = [0, ∞ ) i = 1, 2
Payoffs:
Pi < Pj ⇒ ui = ( Pi − c )(a − Pi )
Pi = Pj ⇒ ui = 12 ( Pi − c )( a − Pi )
Pi > Pj ⇒ ui = 0
Best Responses:
Pj > a 2−c ⇒ BRi = { a 2−c
c < Pj ≤
a −c
2
}
⇒ BRi = φ
Pj = c ⇒ BRi = [c, ∞ )
Pj < c ⇒ BRi = ( Pj , ∞ )
Which gives us a unique Nash equilibrium at ( P1 , P2 ) = ( c, c) .
36
Basic Case with extension to n>2 firms:
Players: firm 1, firm 2, ... , firm n.
Strategy sets: Si = [0, ∞ ) i = 1, 2,…, n
Payoffs:
If Pi = min {Pj } j =1 ui = 1x ( Pi − c)(a − Pi ) where x is the number of firms with the same
n
price as firm i. Otherwise ui = 0.
Best Responses:
Let Pj = min P− i . Then we have:
Pj >
a −c
2
c < Pj ≤
⇒ BRi = { a 2−c
a −c
2
}
⇒ BRi = φ
Pj = c ⇒ BRi = [c, ∞ )
Pj < c ⇒ BRi = ( Pj , ∞ )
Any strategy profile where at least two firms set their prices equal to c and no firm sets a
price lower than c constitutes a Nash equilibrium. There is no other Nash equilibrium.
Basic Case with Discrete Strategy Space:
Back to 2 firms but the prices they choose must be non-negative integers.
Players: firm 1 and firm 2
Strategy sets: Si = {Pi Pi ≥ 0, Pi ∈
}
Payoffs:
Pi < Pj ⇒ ui = ( Pi − c )(a − Pi )
Pi = Pj ⇒ ui = 12 ( Pi − c )( a − Pi )
Pi > Pj ⇒ ui = 0
Best Responses:
If Pj > PM 1 BRi = {PM 1 , PM 2 }
If c + 1 < Pj ≤ PM 1 BRi = {Pj − 1}
If Pj = c + 1 BRi = {c + 1}
If Pj = c BRi = [ c, ∞ ) ∩
If Pj < c BRi = ( Pj , ∞ ) ∩
Where PM1 and PM2 are the integer(s) closest to (a-c)/2 with PM1>PM2 when there is not a
unique such integer.
Now we have two Nash equilibria: (c,c) and (c+1,c+1).
37
Basic Case with Different Costs
Firm i has costs of production ci with c1<c2.
Players: firm 1 and firm 2
Strategy sets: Si = [0, ∞ ) i = 1, 2
Payoffs:
If P1 < P2
u1 = ( P1 − c1 )( a − P1 ) u2 = 0
If P1 = P2
u1 = 12 ( P1 − c1 )( a − P1 ) u2 = 12 ( P2 − c2 )( a − P2 )
If P1 > P2 u1 = 0 u2 = ( P2 − c2 )( a − P2 )
Best Responses:
If Pj > a −2ci BRi = a −2ci
If ci < Pj ≤
a − ci
2
If Pj = ci
BRi = [ci , ∞ )
If Pj < ci
BRi = ( Pj , ∞ )
BRi = ϕ
No Nash equilibrium.
A variant on this changes it so that the payoffs are:
If P1 ≤ P2 u1 = ( P1 − c1 )( a − P1 ) u2 = 0
If P1 > P2 u1 = 0 u2 = ( P2 − c2 )( a − P2 )
Which gives best responses which are different by:
If c1 < P2 ≤ a −2c1 BR1 = P2 (Firm 1 doesn't need to undercut firm 2 to steal the market.)
If P1 < c2 BR2 = [ P1 , ∞ ) (Firm 2 doesn't need to set a higher price to avoid selling at a
loss.)
Resulting in a Nash equilibrium at (c2,c2).
38
Basic Case with Different Costs and Discrete Strategy Spaces
Firm i has costs of production ci with c1<c2 and c2-c1>1.
The prices they choose must be non-negative integers.
Players: firm 1 and firm 2
Strategy sets: Si = {Pi Pi ≥ 0, Pi ∈
}
Payoffs:
If P1 < P2
u1 = ( P1 − c1 )( a − P1 ) u2 = 0
If P1 = P2
u1 = 12 ( P1 − c1 )( a − P1 ) u2 = 12 ( P2 − c2 )( a − P2 )
If P1 > P2 u1 = 0 u2 = ( P2 − c2 )( a − P2 )
Best Responses:
If Pj > PM 1 BRi = {PM 1 , PM 2 }
If ci + 1 < Pj ≤ PM 1 BRi = {Pj − 1}
If Pj = ci + 1 BRi = {ci + 1}
If Pj = ci BRi = [ci , ∞ ) ∩
If Pj < ci BRi = ( Pj , ∞ ) ∩
Where PM1 and PM2 are the integer(s) closest to (a-ci)/2 with PM1>PM2 when there is not a
unique such integer.
The set of Nash equilibria is {( x, x + 1) c1 ≤ x ≤ c2 , x ∈ }
39
Basic Case with Fixed Costs
There's a fixed component to costs that isn't paid if nothing is produced. The fixed costs
are assumed to be small enough that a single firm (i.e. monopolist) operating in the
market could make positive profits.
Players: firm 1 and firm 2
Strategy sets: Si = [0, ∞ ) i = 1, 2
Payoffs:
Pi < Pj ⇒ ui = ( Pi − c )( a − Pi ) − F
Pi = Pj ⇒ ui = 12 ( Pi − c )( a − Pi ) − F
Pi > Pj ⇒ ui = 0
Best Responses:
Define c* as the smaller price giving a payoff of zero when there is only one firm selling
in the market.
So c* =
If Pj >
a +c
2
( a + c )2
−
a −c
2
If c* < Pj ≤
4
BRi =
a −c
2
− ac − F
a −c
2
{
BRi = max x x < Pj , x ∈
}=φ
If Pj ≤ c * BRi = ( Pj , ∞ )
No Nash equilibrium.
But if we restrict the strategy space here to integers, we have a pair of Nash equilibria
where one firm sets a price equal to the smallest integer that is greater than c* and the
other firm sets a price one higher than that. If c* is an integer we have an additional pair
where one firm sets a price equal to c* and the other sets a price one above that.
40
Weaknesses of the Nash Equilibrium
The Nash equilibrium is a solution concept--a way of determining the outcome(s) of a
game. Its virtues are that it is consistent with rational self-interested behaviour and at
least one exists for every finite game. It may seem like the idea is rather straightforward,
but there were a lot of very clever people trying to identify such a useful solution concept
at the time, and they couldn't pull it off. (Look up some of John von Neumann's various
accomplishments if you want to see just how clever some of these people were.)
The downside of the Nash equilibrium is that for some games it identifies multiple
equilibria and offers no further suggestions.
Consider Meeting in NY.
Consider the Pareto Coordination game.
Consider games with a relevant dynamic element.
41
Subgames
A subgame is any part of a game that starts at a node such that neither that node nor any
of its successors are in an information set containing nodes that are not successors of that
subgame-starting node.
i.e. If you were to cut that subgame-starting node and all its successors out of the
extensive form of the larger game, you wouldn't be cutting any of the dotted lines
denoting an information set.
Thus a subgame can itself be considered a game. (Although it might have only one
player moving in which case we technically shouldn't be calling it a game.)
Note that a game is a subgame of itself.
A proper subgame is any subgame which is not the full game.
42
Subgame-Perfect Nash Equilibria (SPNE)
Consider the following game: There are two players, me and my little brother. First I
must decide whether or not to threaten to punch my brother if he doesn't clean up my
room. If I don't, then the game is over. If I do, then my bother has to decide whether or
not to clean up my room. If he does clean it, then the game is over. If he decides not to,
then I have to decide whether or not to punch him. Regardless of how this is decided, the
game ends.
My favourite outcome is that my brother cleans my room. My least favourite outcome is
that I punch him. (Then mom grounds me.)
My brother's favourite outcomes involve neither getting punched nor cleaning my room.
His least favourite outcome is getting punched.
What are the Nash equilibria?
A strategy profile which is a Nash equilibrium for every subgame of a game is a
subgame-perfect Nash equilibrium.
What are the subgame-perfect Nash for this game?
Backward Induction
Usually quickest way to find the SPNE for a game is to write it out in extensive form and,
starting with the smallest subgames, iteratively replace them with equilibrium payoffs
until you reach the initial node. The path(s) through these equilibrium payoffs are the
SPNE(s).
The Centipede Game
Players take turns deciding whether to halt the game or continue. Once a player halts, the
game is over. There is, eventually, a final possible "move" in which one of the players
cannot continue. When a player continues and then in the next move the other player
chooses halt, the continuing player's payoff is one less than it would have been had he
instead chosen to halt while the halting player's payoff is two greater than it would have
been.
Multiple SPNEs
Employing backward induction in a game with multiple SPNE can become complicated.
The trick is to just be careful about keeping track of what you're doing.
43
Stackelberg Model
Vaticrat
Similar to the Cournot model but the two firms move one after the other with perfect
information.
Consider a simple version with linear demand
P = a – bQ
and linear variable costs with no fixed costs
Ci = cqi.
We'll find the SPNE using backward induction. Firm 2's payoff is:
u2 = Pq2 – cq2 = (P-c)q2 = (a-bQ-c)q2 = (a-b(q1+q2)-c)q2 = (a-c-bq1)q2 – bq22
Set the partial derivative of this with respect to q2 equal to zero:
a – c – bq1 – 2bq2 = 0
and rearrange to:
q2 = (a-c)/2b – q1/2
Keeping in mind that these quantities can't be negative, we have firm 2's best response
function:
BR2(q1) = max{ (a-c)/2b-q1/2 , 0 }
Now we assess firm 1's payoff using this knowledge of how firm 2 will react to what firm
1 does. Note that if q1 is large enough that we can't use BR2=(a-c)/2b-q1/2 then P<c and
firm 1 is better off not producing. i.e. In looking for where u1 is at a maximum, we don't
have to worry about firm 2's non-negativity constraint.
Thus when firm 1 assesses its payoff
u1 = (a-c-bq2)q1 – bq12
it can predict how q2 depends on q1 and instead use
u1 = (a-c-b((a-c)/2b-q1/2))q1-bq12 = (1/2)((a-c)q1-bq12)
Setting the partial derivative equal to zero:
(1/2)(a-c-2bq1) = 0
Which solves for:
q1 = (a-c)/2b
Plugging this back into firm 2's best response will tell us how much firm 2 actually
produces in the SPNE:
q2 = (a-c)/2b – ((a-c)/2b)/2 = (a-c)/4b
So in the SPNE firm 1 produces more output than firm 2 and since payoff is (P-c)/q it
also earns more profit as a result of moving first.
The SPNE is
(s1, s2) = ( (a-c)/2b, max{ (a-c)/2b-q1/2, 0 } ).
It is not
(s1, s2) = ( (a-c)/2b, (a-c)/4b ).
44
Limit Capacity
2 Players: Firm 1 and Firm 2
Firm 1 decides to build either a large plant, a small plant, or no plant.
Firm 2 observes what firm 1 did and also decides to build either a large plant, a small
plant, or no plant.
Firm 1 observes what firm 2 did.
Now the firms engage in a Cournot game: they simultaneously decide some level of
output, the joint output is sold at the market-clearing price, and payoffs are profits earned.
If a firm doesn't build a plant, it can't produce anything.
If a firm built a small plant, it can produce up to some level of output, qS.
If a firm built a large plant, it has no capacity constraint.
Suppose:
Demand is P = 60 - Q
The cost of a small plant is cS and the cost of a large plant is cL. (With cL>cS.)
There are no variable costs of production.
The capacity constraint with the small plant is 10 units of output.
Looking for the subgame-perfect Nash equilibrium, we have 9 proper subgames of 6
different types:
If both firms select large plants: q1 = q2 = 20, P = 20, and u1= u2 = 400-cL.
If firm i has a large plant and firm j has a small plant: qi = 25, qj = 10, P = 25, ui = 625-cL,
uj = 250-cS.
If firm i has a large plant and firm j has no plant: qi = 30, qj = 0, P = 30, ui = 900-cL, uj =
0.
If both firms select small plants: q1 = q2 = 10, P = 40, and u1= u2 = 400-cS.
If firm i has a small plant and firm j has no plant: qi = 10, qj = 0, P = 50, ui = 500-cS, uj =
0.
If both firms select no plant, q1 = q2 = 0, and u1= u2 = 0.
45
Removing the proper subgames and replacing them with the payoffs from equilibrium
play we have:
1
L
S
2
2
L
N
S
N
L’
2
S’
N’
L”
S”
N”
( )( )( )( )( )( )( )( )( )
400-cL
400-cL
625-cL 900-cL
250-cS
0
250-cS
625-cL
400-cS
400-cS
500-cS
0
0
900-cL
0
500-cS
0
0
The SPNE at this point depends on the plant costs.
An interesting case shows up if:
Monopoly profits are higher with a small plant than a large one. (500-cS>900-cL)
If there is a large plant and a small plant, the firm with the small one earns negative
profits. (250-cS<0)
If both firms have small plants, they earn positive profits. (400-cS>0)
With two small plants, profits per firm are less than a monopolist with a large plant. (900cL>400-cS)
Taken together this gives us: 500 > cL - cS > 400 > cS > 250
46
For example, if we use cS = 300 and cL = 750 we get:
1
L
S
2
2
L
N
S
N
L’
2
S’
N’
L”
S”
N”
( )( )( )( )( )( )( )( )( )
-350
-350
-125
-50
150
0
-50
-125
100
100
200
0
0
150
0
200
0
0
Player 2's equilibrium behaviour involves N, S', and S''. This reduces firm 1's initial
decision to:
1
L
S
N
( )( )( )
150
0
100
100
0
200
So its equilibrium strategy involves L.
Note what is observed in a market like this: a monopolist that, at first glance, appears to
not be maximizing its profits (since it should be building a small plant).
It is changing its behaviour in order to maintain its monopoly position.
If we analyse a variant of this game where they simultaneously choose plant size, and
then knowing what choices were made simultaneously choose quantities, the unique
SPNE changes to involve two small plants. So in this scenario, there is a value to being
able to commit early or move first.
47
Advertising
Just a simple variant on the Cournot model.
Firm 1 picks a level of advertising (non-negative real number) which affects demand but
is costly.
Then, having observed the level of advertising, the firms engage in a Cournot interaction.
P=a-Q
u1 = (a - q1 - q2)q1 - a3
u2 = (a - q1 - q2)q2
'a' is the level of advertising.
Using backward induction:
q1 = q2 = a/3
Putting this into player 1's payoff:
u1 = a2/9 - a3
∂u1 2
= 9 a − 3a 2 = 0 ⇒ a = 0 or a =
∂a
∂ 2u1 2
= 9 − 6a < 0 ⇒ a > 271
∂a 2
2
27
⎫
⎪⎪
⎬⇒a =
⎪
⎪⎭
2
27
Note that the advertising has a general public good quality: it provides benefits for both
firms but firm 1 only takes into account its own benefit when deciding on the level to
choose. This suggests it'll be underprovided when taking into consideration both firms
(so that the outcome isn't efficient).
Using v for joint profit:
2
v = 29a − a 3
= 94 a − 3a 2 = 0 ⇒ a = 247
So both firms could be better off a contract having firm 2 pay firm 1 to choose a higher
level of advertising.
∂v
∂a
48
Dynamic Monopoly - Price Guarantees
This is pretty much straight out of the textbook.
Two periods
Durable good
High demand consumers and low demand consumers (To keep things simple, suppose
there are an equal number of each.)
The firm offers a price in period 1.
Consumers decide whether or not to buy.
The firm offers a price in period 2.
Consumers decide whether or not to buy.
The durability of the good means that, if purchased in period 1, the consumer still has it
in period 2 so doesn't need to buy a second one.
Period 1
Period 2
High
1200
500
Low
500
200
The idea is:
The value to a high demander of getting the good in the 1st period is 1200+500=1700.
The value to a low demander of getting the good in the 1st period is 500+200=700.
The value to a high demander of getting the good in the 2nd period is 500.
The value to a low demander of getting the good in the 2nd period is 200.
Consumers will pay up to (and including) the value they place on the good.
They decide whether or not to buy in such a way as to maximize the difference between
the value placed on owning the good and the price paid (which is the payoff).
The firm's payoff is revenue generated.
3 Possible Pricing Schemes:
Get them all to buy in the first period.
This requires P1<=700.
So P1 = 700 maximizes profit at 1400.
Get the high demand types to buy in the first period and the low demand types in the
second.
P2 will be 200. The surplus available to the high demanders has to be at least 500200=300 if they buy in the first period. Otherwise, they will prefer to buy in the second
period. So P1 = 1200+500-300 = 1400. Profit is 1400+200 = 1600.
49
Get them all to buy in the second period.
P1 is high and P2 = 500 for profit of 500.
So profit maximized at 1600 with P1 = 1400 and P2 = 200.
Note that if they could commit to P1 = P2 = 1700 profit would be higher.
Think of rebates and price guarantees as a way of accomplishing this (at the consumers'
expense).
50
Strictly Competitive Games
A game is strictly competitive if the players have opposite rankings over any two strategy
profiles. i.e.
u1 ( s′) > u1 ( s′′) ⇔ u2 ( s′) < u2 ( s′′)
Note that this implies that in a strictly competitive game
u1 ( s′) ≥ u1 ( s′′) ⇔ u2 ( s′) ≥ u2 ( s′′) and u1 ( s ′ ) = u1 ( s ′′ ) ⇔ u2 ( s ′ ) = u2 ( s ′′ ) ∀ s ′ , s ′′ ∈ S .
Matching pennies is strictly competitive.
Player 1
H
T
Player 2
H
T
0,1
1,0
1,0
0,1
We can line up the strategy profiles under player 1's preferences and have that serve as
the opposite ordering under player 2's preferences.
u1 (TH ) ≥ u1 ( HT ) ≥ u1 ( HH ) ≥ u1 (TT )
u2 ( TH ) ≤ u2 ( HT ) ≤ u2 ( HH ) ≤ u2 (TT )
or, if you prefer:
u1 (TH ) = u1 ( HT ) > u1 ( HH ) = u1 (TT )
u2 ( TH ) = u2 ( HT ) < u2 ( HH ) = u2 (TT )
Being able to do this to a game is equivalent to it being strictly competitive.
It doesn't require that the sum of the payoffs be constant (although this is sufficient).
e.g.
Player 2
H
T
H 4,5
0 , 20
Player 1
T 1 , 17 2 , 12
Is strictly competitive since:
u1 ( HH ) ≥ u1 (TT ) ≥ u1 (TH ) ≥ u1 ( HT )
u2 ( HH ) ≤ u2 (TT ) ≤ u2 ( TH ) ≤ u2 ( HT )
or, if you prefer, replace the weak inequalities with strict ones.
The prisoners' dilemma is not strictly competitive since u1 ( CC ) > u1 ( DD ) but it's not the
case that u2 ( CC ) < u2 ( DD ) .
51
Equivalence of Nash Equilibria in Strictly Competitive Games
In a strictly competitive game, there may be more than one Nash equilibrium. However,
any Nash equilibrium has the same payoffs to the players as in any other Nash
equilibrium. (This is the sense in which they are 'equivalent'.)
Proof:
Let s′ and s′′ be any two strategy profiles in a strictly competitive game.
Define s1′, s2′ , s1′′, s2′′ so that s′ = ( s1′, s2′ ) , s′′ = ( s1′′, s2′′ )
Since s′ is a Nash equilibrium, s2′ ∈ BR2 ( s1′ ) so we have:
u2 ( s1′, s2′′ ) ≤ u2 ( s′)
Since the game is strictly competitive this gives us:
u1 ( s1′, s2′′ ) ≥ u1 ( s′)
(2)
(1)
Since s′′ is a Nash equilibrium, s1′′∈ BR1 ( s2′′ ) so we have:
u1 ( s′′) ≥ u1 ( s1′, s2′′ )
Combining the last two inequalities gives us:
u1 ( s′′) ≥ u1 ( s′)
(4)
(3)
Since s′′ is a Nash equilibrium, s2′′ ∈ BR2 ( s1′′) so we have:
u2 ( s′′) ≥ u2 ( s1′′, s2′ )
Since the game is strictly competitive this gives us:
u1 ( s′′) ≤ u1 ( s1′′, s2′ )
(6)
(5)
Since s′ is a Nash equilibrium, s1′ ∈ BR1 ( s2′ ) so we have:
u1 ( s1′′, s2′ ) ≤ u1 ( s′)
Combining the last two inequalities gives us:
u1 ( s′′) ≤ u1 ( s′)
(8)
(7)
Combining this with the 4th inequality of the proof gives us:
u1 ( s′′) = u1 ( s′)
(9)
And since the game is strictly competitive we have as well that:
u2 ( s′′) = u2 ( s′)
(10)
All done.
52
Security Strategies
Define wi(si) as the worst payoff that player i can receive if he plays the strategy si. i.e.
wi ( si ) = min ui ( si , s j ) .
s j ∈S j
Define a security strategy as follows. si* is a security strategy if it maximizes wi(si). A
security strategy for player i is one that maximizes his worst possible payoff. Formally,
si* solves
max wi ( si ) or max min ui ( si , s j ) .
si ∈Si
si ∈Si
s j ∈S j
Preliminary Result: If a game is strictly competitive then any Nash equilibrium involves
the players using security strategies.
Proof:
Let s′ be a Nash equilibrium.
Since it's a Nash equilibrium, player 2 is playing a best response, so:
u2 ( s1′, s2′ ) ≥ u2 ( s1′, s2′′ ) ∀s2′′ ∈ S2
(1)
Since it's a strictly competitive game this gives us:
u1 ( s1′, s2′ ) ≤ u1 ( s1′, s2′′ ) ∀s2′′ ∈ S2
(2)
And thus, by definition of wi(si) we have:
u1 ( s1′, s2′ ) = w1 ( s1′ ) .
(3)
Since it's a Nash equilibrium, player 1 is playing a best response, so:
u1 ( s1′, s2′ ) ≥ u1 ( s1′′, s2′ ) ∀s1′′∈ S1
(4)
By definition of wi(si) we have:
u1 ( s1′′, s2′ ) ≥ w1 ( s1′′) ∀s1′′∈ S1
(5)
Combining the last three lines we have:
w1 ( s1′ ) = u1 ( s1′, s2′ ) ≥ u1 ( s1′′, s2′ ) ≥ w1 ( s1′′) ∀s1′′∈ S1
(6)
Getting rid of the stuff in the middle there, this is:
w1 ( s1′ ) ≥ w1 ( s1′′) ∀s1′′∈ S1
(7)
So we have our result.
53
Final Result: If a two-player game is strictly competitive, then all SPNEs have the same
payoffs for all players. i.e. In a win-loss game, either player 1 can guarantee that he wins
or player 2 can. In a win-loss-tie game, either player 1 can guarantee that he wins or
player 2 can or both players can guarantee themselves a draw.
Proof:
Consider such a game.
Let s' be a SPNE. (If there is no SPNE, we're done.)
s' is a Nash equilibrium for the game. (Any SPNE is also a Nash equilibrium.)
s' is a security strategy for both players. (Our preliminary result in action.)
s' solves max min ui ( si , s j ) . (By definition of a security strategy.)
si ∈Si
s j ∈S j
Let s'' be a different SPNE. (If it doesn't exist, we're done.)
s'' also solves max min ui ( si , s j ) . (As above with s'.)
si ∈Si
s j ∈S j
ui(s') = ui(s'') (Since the strategies solve that same maximization problem.)
uj(s') = uj(s'') (Since it's a strictly competitive game.)
Therefore any two SPNEs give the two players the same payoff.
Alternative Proof:
Consider such a game.
Let s' be a SPNE. (If there is no SPNE, we're done.)
s' is a Nash equilibrium for the game. (Any SPNE is also a Nash equilibrium.)
Let s" be a different SPNE. (If it doesn't exist, we're done.)
s" is a Nash equilibrium for the game. (Any SPNE is also a Nash equilibrium.)
s' and s" have the same payoffs for the two players. (All Nash equilibria have the same
payoffs for the players in strictly competitive games.)
Therefore any two SPNEs give the two players the same payoff.
Here's a summary of results we've developed and the class of games for which we've
proven them. Note that the class is not necessary for the result. e.g. There are infinite
games with SPNEs. There are games in which the players receive the same payoff at all
SPNEs but which are not strictly competitive.
Any SPNE is also Nash: all games.
Existence of Nash: finite.
Existence of SPNE: finite.
Existence of pure-strategy SPNE: finite and perfect information.
Equivalent payoffs of Nash: strictly competitive.
Equivalent payoffs of SPNE: strictly competitive.
Existence and Equivalent payoffs of SPNE: finite, perfect information, strictly
competitive.
54
Parlour Games
There's a board with n rows and m columns of squares. Initially there is a playing piece
in the bottom-right square. Player one must move the playing piece either i) up one to the
square directly above or ii) left one to the square immediately to the left or iii) up and left
diagonally to the square that's bottom-right corner touches the initial square's top-left
corner. Then player two must move the playing piece similarly from the square that
player one moved it to. They alternate moving the playing piece until a player moves it
into the square in the top-left corner of the board. The player who moves it into that final
square loses.
How does the victor in the set of SPNEs depend on n and m?
Chomp
Two players take turns placing a playing piece an n by m board of squares. Once a
square has been moved into, all squares below and/or to the right are out of play. The
player forced to move into the top-left square loses.
55
One-Card Poker
There are three cards, an ace (A), king (K), and queen (Q). Two players ante one dollar
each into the pot. Each player receives one of the cards. They each see their own card,
but not their opponent's and not the one that has not been dealt. Player one either bets
(and places one additional dollar into the pot) or folds (in which case the game is over
and player two gets the pot). If player one bets, then player two either bets (and places
one additional dollar into the pot) or folds (in which case the game is over and player one
gets the pot). If they have both bet (now there is four dollars in the pot) they reveal their
cards to each other and the player with the higher card gets the pot. The ace is higher
than the king which in turn is higher than the queen. The payoff for each player is the
amount of money won.
Vaticrat
The extensive form of one-card poker looks like this:
Nature
1/6
AK
AQ
1/6
1
KQ
1/6
QA
1/6
1/6
QK
1/6
1
Ba
Fa
Ba
2
2
Bk
KA
Bq
Fk
Fa
1
Bk
Fk
Bk
Fk
Bq
Fq
Bq
Fq
2
Fq
Ba
Bq
Fa
Fq
Ba
Fa
Bk
Fk
( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )
2
1
-1
2
1
-1
-2
1
-1
2
1
-1
-2
1
-1
-2
1
-1
The payoffs listed are for player 1. (Player 2's payoff is the negative of player 1's.)
1's Str
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
B
B
B
B
B
B
B
B
B
B
B
B
2's Str
AK
AQ
KA
KQ
QA
QK
Total
BABKBQ
2
2
-2
2
-2
-2
0
BABKBQ
2
2
-2
2
-1
-1
2
BABKBQ
2
2
-1
-1
-2
-2
-2
BABKBQ
2
2
-1
-1
-1
-1
0
BABKBQ
-1
-1
-2
2
-2
-2
-6
BABKBQ
-1
-1
-2
2
-1
-1
-4
BABKBQ
-1
-1
-1
-1
-2
-2
-8
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
56
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
BABKBQ
BABKFQ
BABKFQ
BABKFQ
BABKFQ
BABKFQ
BABKFQ
BABKFQ
BABKFQ
BAFKBQ
BAFKBQ
BAFKBQ
BAFKBQ
BAFKBQ
BAFKBQ
BAFKBQ
BAFKBQ
BAFKFQ
BAFKFQ
BAFKFQ
BAFKFQ
BAFKFQ
BAFKFQ
BAFKFQ
BAFKFQ
FABKBQ
FABKBQ
FABKBQ
FABKBQ
FABKBQ
FABKBQ
FABKBQ
FABKBQ
FABKFQ
FABKFQ
FABKFQ
FABKFQ
FABKFQ
FABKFQ
FABKFQ
FABKFQ
FAFKBQ
FAFKBQ
FAFKBQ
FAFKBQ
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
-1
2
2
2
2
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
2
2
2
2
-1
-1
-1
-1
2
2
2
2
-1
-1
-1
-1
1
1
1
1
-1
1
1
1
1
-1
-1
-1
-1
2
2
2
2
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
2
2
2
2
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
2
2
2
2
-1
-2
-2
-1
-1
-2
-2
-1
-1
-2
-2
-1
-1
-2
-2
-1
-1
-2
-2
-1
-1
-2
-2
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
-1
1
1
-1
-1
1
1
-1
-1
2
2
-1
-1
2
2
-1
-1
1
1
-1
-1
1
1
-1
-1
2
2
-1
-1
2
2
-1
-1
1
1
-1
-1
1
1
-1
-1
2
2
-1
-1
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
-1
-2
-1
-2
-1
-2
-1
-2
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
-2
-1
1
-1
1
-1
-6
-2
0
-3
-1
-7
-5
-8
-6
2
1
0
-1
-3
-4
-5
-6
0
-1
-1
-2
-4
-5
-5
-6
6
5
1
0
0
-1
-5
-6
4
3
0
-1
-1
-2
-5
-6
8
4
3
-1
57
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
FAFKFQ
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
FAFKBQ
FAFKBQ
FAFKBQ
FAFKBQ
FAFKFQ
FAFKFQ
FAFKFQ
FAFKFQ
FAFKFQ
FAFKFQ
FAFKFQ
FAFKFQ
B
B
B
B
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
2
2
-1
-1
1
1
-1
-1
1
1
-1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
3
-1
-2
-6
6
2
2
-2
2
-2
-2
-6
58
Use the above table to fill in the normal form matrix:
PLAYER 2
BABKBQ
BABKFQ
BAFKBQ
BAFKFQ
FABKBQ
FABKFQ
FAFKBQ
F AF KF Q
BABKBQ
0
-2
2
0
6
4
8
6
BABKFQ
2
0
1
-1
5
3
4
2
BAFKBQ
-2
-3
0
-1
1
0
3
2
BAFKFQ
0
-1
-1
-2
0
-1
-1
-2
FABKBQ
-6
-7
-3
-4
0
-1
3
2
FABKFQ
-4
-5
-4
-5
-1
-2
-1
-2
FAFKBQ
-8
-8
-5
-5
-5
-5
-2
-2
F AF KF Q
-6
-6
-6
-6
-6
-6
-6
-6
B
B
B
B
B
P
L
A
Y
E
R
1
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
A strategy of player i, si', is dominated by another strategy of player i, si'' if and only if
ui(si'', sj)>ui(si',sj) for ever strategy of the other player, sj. A dominated strategy will never
be part of any Nash equilibrium since it is never a best response. Here, we can strike out
the bottom five pure strategies of player 1 on the grounds that they are dominated. Then
the right-most four strategies of player 2 and the other two that involve betting with the
queen. Now we can strike out the strategy that involves folding with the king for player
one. This leaves us with two strategies for each player, and no pure strategy Nash
equilibrium. The unique Nash equilibrium for this game is a mixed strategy.