Static Games of Complete Information
Mixed Strategies
62
Let’s Mix It
As pointed out before, neither of the solution concepts has to exist in
pure strategies
Example: Rock-Paper-sCissors
R P C
R 0, 0 −1, 1 1, −1
P 1, −1 0, 0 −1, 1
C −1, 1 1, −1 0, 0
There are no strictly dominant pure strategies
No strategy is strictly dominated (IESDS removes nothing)
Each strategy is a best response to some strategy of the opponent
(rationalizability removes nothing)
No pure Nash equilibria: No pure strategy profile allows each player
to play a best response to the strategy of the other player
How to solve this?
Let the players randomize their choice of pure strategies ....
63
Probability Distributions
Definition 19
Let A be a finite set. A probability distribution over A is a function
σ : A → [0, 1] such that
�
a∈A σ(a) = 1.
We denote by Δ(A) the set of all probability distributions over A.
We denote by supp(σ) the support of σ, that is the set of all a ∈ A
satisfying σ(a) > 0.
Example 20
Consider A = {a, b, c} and a function σ : A → [0, 1] such that
σ(a) = 1
4 , σ(b) = 3
4 , and σ(c) = 0. Then σ ∈ Δ(A) and
supp(σ) = {a, b}.
64
Mixed Strategies
Let us fix a strategic-form game G = (N, (Si)i∈N , (ui)i∈N).
From now on, assume that all Si are finite!
Definition 21
A mixed strategy of player i is a probability distribution σ ∈ Δ(Si) over
Si. We denote by Σi = Δ(Si) the set of all mixed strategies of player i.
We define Σ := Σ1 × · · · × Σn, the set of all mixed strategy profiles.
Recall that by Σ−i we denote the set Σ1 × · · · Σi−1 × Σi+1 × · · · × Σn
Elements of Σ−i are denoted by σ−i = (σ1, . . . , σi−1, σi+1, . . . , σn).
We identify each si ∈ Si with a mixed strategy σ that assigns
probability one to si (and zero to other pure strategies).
For example, in rock-paper-scissors, the pure strategy R corresponds
to σi which satisfies σi(X) =



1 X = R
0 otherwise
65
Mixed Strategies
Sometimes we assume Si = {1, . . . , mi}, here mi ∈ {1, 2, . . .}, for
all i ∈ N.
Then every mixed strategy σi is a vector
σi = (σi(1), . . . , σi(mi))� ∈ [0, 1]mi so that
σi(1) + · · · + σi(mi) = 1
66
Mixed Strategy Profiles
Let σ = (σ1, . . . , σn) be a mixed strategy profile.
Intuitively, we assume that each player i randomly chooses his pure
strategy according to σi and independently of his opponents.
Thus for s = (s1, . . . , sn) ∈ S = S1 × · · · × Sn we have that
σ(s) :=
n�
i=1
σi(si)
is the probability that the players choose the pure strategy profile s
according to the mixed strategy profile σ, and
σ−i(s−i) :=
n�
k�i
σk (sk )
is the probability that the opponents of player i choose s−i ∈ S−i when
they play according to the mixed strategy profile σ−i ∈ Σ−i.
(We abuse notation a bit here: σ denotes two things, a vector of mixed
strategies as well as a probability distribution on S (the same for σ−i)
67
Mixed Strategies – Example
R P C
R 0, 0 −1, 1 1, −1
P 1, −1 0, 0 −1, 1
C −1, 1 1, −1 0, 0
An example of a mixed strategy σ1: σ1(R) = 1
2 , σ1(P) = 1
3 , σ1(C) = 1
6 .
Sometimes we write σ1 as (1
2 (R), 1
3 (P), 1
6 (C)), or only (1
2 , 1
3 , 1
6 ) if the
order of pure strategies is fixed.
Consider a mixed strategy profile (σ1, σ2) where
σ1 = (1
2 (R), 1
3 (P), 1
6 (C)) and σ2 = (1
3 (R), 2
3 (P), 0(C)).
Then the probability σ(R, P) that the pure strategy profile (R, P) will
be chosen by players playing the mixed profile (σ1, σ2) is
σ1(R) · σ2(P) =
1
2
·
2
3
=
1
3
68
Expected Payoff
... but now what is the suitable notion of payoff?
Definition 22
The expected payoff of player i under a mixed strategy profile σ ∈ Σ is
ui(σ) :=
�
s∈S
σ(s)ui(s)

=
�
s∈S
n�
k=1
σk (sk )ui(s)


I.e., it is the "weighted average" of what player i wins under each pure
strategy profile s, weighted by the probability of that profile.
Assumption: Every rational player strives to maximize his own
expected payoff.
(This assumption is not always completely convincing ...)
69
Expected Payoff – Example
Matching Pennies:
H T
H 1, −1 −1, 1
T −1, 1 1, −1
Each player secretly turns a penny to heads or tails, and then they reveal
their choices simultaneously. If the pennies match, player 1 (row) wins, if they
do not match, player 2 (column) wins.
Consider σ1 = (1
3 (H), 2
3 (T)) and σ2 = (1
4 (H), 3
4 (T))
u1(σ1, σ2) =
�
(X,Y)∈{H,T}2
σ1(X)σ2(Y)u1(X, Y)
=
1
3
1
4
1 +
1
3
3
4
(−1) +
2
3
1
4
(−1) +
2
3
3
4
1 =
1
6
u2(σ1, σ2) =
�
(X,Y)∈{H,T}2
σ1(X)σ2(Y)u2(X, Y)
=
1
3
1
4
(−1) +
1
3
3
4
1 +
2
3
1
4
1 +
2
3
3
4
(−1) = −
1
6 70
"Decomposition" of Expected Payoff
Consider the matching pennies example from the previous slide:
H T
H 1, −1 −1, 1
T −1, 1 1, −1
together with some mixed
strategies σ1 and σ2.
We prove the following important property of the expected payoff:
u1(σ1, σ2) =
�
X∈{H,T}
σ1(X)u1(X, σ2)
An intuition behind this equality is following:
� u1(σ1, σ2) is the expected payoff of player 1 in the following experiment:
Both players simultaneously and independently choose their pure
strategies X, Y according to σ1, σ2, resp., and then player 1 collects his
payoff u1(X, Y).
�
�
X∈{H,T} σ1(X)u1(X, σ2) is the expected payoff of player 1 in the
following: Player 1 chooses his pure strategy X and then uses it against
the mixed strategy σ2 of player 2. Then player 2 chooses Y according to
σ2 independently of X, and player 1 collects the payoff u1(X, Y).
As Y does not depend on X in neither experiment, we obtain the above
equality of expected payoffs. 71
"Decomposition" of Expected Payoff
Consider the matching pennies example from the previous slide:
H T
H 1, −1 −1, 1
T −1, 1 1, −1
together with some mixed
strategies σ1 and σ2.
A formal proof is straightforward:
u1(σ1, σ2) =
�
(X,Y)∈{H,T}2
σ1(X)σ2(Y)u1(X, Y)
=
�
X∈{H,T}
�
Y∈{H,T}
σ1(X)σ2(Y)u1(X, Y)
=
�
X∈{H,T}
σ1(X)
�
Y∈{H,T}
σ2(Y)u1(X, Y)
=
�
X∈{H,T}
σ1(X)u1(X, σ2)
(In the last equality we used the fact that X is identified with a mixed strategy
assigning one to X.)
72
"Decomposition" of Expected Payoff
Consider the matching pennies example from the previous slide:
H T
H 1, −1 −1, 1
T −1, 1 1, −1
together with some mixed
strategies σ1 and σ2.
Similarly,
u1(σ1, σ2) =
�
(X,Y)∈{H,T}2
σ1(X)σ2(Y)u1(X, Y)
=
�
X∈{H,T}
�
Y∈{H,T}
σ1(X)σ2(Y)u1(X, Y)
=
�
Y∈{H,T}
�
X∈{H,T}
σ1(X)σ2(Y)u1(X, Y)
=
�
Y∈{H,T}
σ2(Y)
�
X∈{H,T}
σ1(X)u1(X, Y)
=
�
Y∈{H,T}
σ2(Y)u1(σ1, Y)
73
Expected Payoff – "Decomposition" in General
Lemma 23
For every mixed strategy profile σ ∈ Σ and every k ∈ N we have
ui(σ) =
�
sk ∈Sk
σk (sk ) · ui(sk , σ−k ) =
�
s−k ∈S−k
σ−k (s−k ) · ui(σk , s−k )
Lemma 23 immediately implies that
� each ui(σ) is affine in each σk (sk ),
� Also, ui(σ) = ui(σ1, . . . , σn) is linear in each σk .
Indeed, assuming w.l.o.g. that Sk = {1, . . . , mk },
ui(σ) =
�
sk ∈Sk
σk (sk ) · ui(sk , σ−k ) =
mk�
�=1
σk (�) · ui(�, σ−k )
is the scalar product of the vector σk = (σk (1), . . . , σk (mk )) with
the vector (ui(1, σ−k ), . . . , ui(mk , σ−k )).
74
Expected Payoff – Pure Strategy Bounds
Before proving Lemma 23, we prove the following simple corollary.
Corollary 24
For all i, k ∈ N and σ ∈ Σ we have that
� minsk ∈Sk
ui(sk , σ−k ) ≤ ui(σ) ≤ maxsk ∈Sk
ui(sk , σ−k )
� mins−k ∈S−k
ui(σk , s−k ) ≤ ui(σ) ≤ maxs−k ∈S−k
ui(σk , s−k )
Proof.
We prove ui(σ) ≤ maxsk ∈Sk
ui(sk , σ−k ) the rest is similar. Define
B := maxsk ∈Sk
ui(sk , σ−k ). Then
ui(σ) =
�
sk ∈Sk
σk (sk ) · ui(sk , σ−k )
=
�
sk ∈Sk
σk (sk ) · (B − (B − ui(sk , σ−k )))
≤
�
sk ∈Sk
σk (sk ) · B
= B
75
Proof of Lemma 23
ui(σ) =
�
s∈S
σ(s)ui(s) =
�
s∈S
n�
�=1
σ�(s�)ui(s)
=
�
s∈S
σk (sk )
n�
��k
σ�(s�)ui(s)
=
�
sk ∈Sk
�
s−k ∈S−k
σk (sk )
n�
��k
σ�(s�)ui(sk , s−k )
=
�
sk ∈Sk
�
s−k ∈S−k
σk (sk )σ−k (s−k )ui(sk , s−k )
76
Proof of Lemma 23 (cont.)
The first equality:
ui(σ) =
�
sk ∈Sk
�
s−k ∈S−k
σk (sk )σ−k (s−k )ui(sk , s−k )
=
�
sk ∈Sk
σk (sk )
�
s−k ∈S−k
σ−k (s−k )ui(sk , s−k )
=
�
s−k ∈S−k
σk (sk )ui(sk , σ−k )
77
Proof of Lemma 23 (cont.)
The second equality:
ui(σ) =
�
sk ∈Sk
�
s−k ∈S−k
σk (sk )σ−k (s−k )ui(sk , s−k )
=
�
s−k ∈S−k
�
sk ∈Sk
σk (sk )σ−k (s−k )ui(sk , s−k )
=
�
s−k ∈S−k
σ−k (s−k )
�
sk ∈Sk
σk (sk )ui(sk , s−k )
=
�
s−k ∈S−k
σ−k (s−k )ui(σk , s−k )
78
Solution Concepts
We revisit the following solution concepts in mixed strategies:
� strict dominant strategy equilibrium
� IESDS equilibrium
� rationalizable equilibria
� Nash equilibria
From now on, when I say a strategy I implicitly mean a
mixed strategy.
In order to deal with efficiency issues we assume that the size of the game G
is defined by |G| := |N| +
�
i∈N |Si| +
�
i∈N |ui| where |ui| =
�
s∈S |ui(s)| and
|ui(s)| is the length of a binary encoding of ui(s) (we assume that rational
numbers are encoded as quotients of two binary integers)
Note that, in particular, |G| > |S|.
79
Strict Dominance in Mixed Strategies
Definition 25
Let σi, σ�
i
∈ Σi be (mixed) strategies of player i. Then σ�
i
is
strictly dominated by σi (write σ�
i
≺ σi) if
ui(σi, σ−i) > ui(σ�
i , σ−i) for all σ−i ∈ Σ−i
Example 26
X Y
A 3 0
B 0 3
C 1 1
Is there a strictly dominated strategy?
Question: Is there a game with at least one strictly dominated
strategy but without strictly dominated pure strategies?
80
Strictly Dominant Strategy Equilibrium
Definition 27
σi ∈ Σi is strictly dominant if every other mixed strategy of player i is
strictly dominated by σi.
Definition 28
A strategy profile σ ∈ Σ is a strictly dominant strategy equilibrium if
σi ∈ Σi is strictly dominant for all i ∈ N.
Proposition 2
If the strictly dominant strategy equilibrium exists, it is unique, all its
strategies are pure, and rational players will play it.
Proof.
Let σ∗
= (σ∗
1
, . . . , σ∗
n) ∈ Σi be the strictly dominant strategy equilibrium.
By Corollary 24, for every i ∈ N and σ−i ∈ Σ−i, there must exist si ∈ Si
such that ui(σ∗
i
, σ−i) ≤ ui(si, σ−i).
But then σ∗
i
= si since σ∗
i
is strictly dominant.
�
81
Computing Strictly Dominant Strategy Equilibrium
How to decide whether there is a strictly dominant strategy
equilibrium s = (s1, . . . , sn) ∈ S ?
I.e. whether for a given si ∈ Si, all σi ∈ Σi � {si} and all σ−i ∈ Σ−i :
ui(si, σ−i) > ui(σi, σ−i)
There are some serious issues here:
� Obviously there are uncountably many possible σi and σ−i.
� ui(σi, σ−i) is nonlinear, and for more that two players even
ui(si, σ−i) is nonlinear in probabilities assigned to pure strategies.
First, we prove the following useful proposition using Lemma 23:
Lemma 29
σi strictly dominates σ�
i
iff for all pure strategy profiles s−i ∈ S−i:
ui(σi, s−i) > ui(σ�
i , s−i)
Proof: Simple application of the second equality from Lemma 23.
In other words, it suffices to check the strict dominance only with
respect to all pure profiles of opponents. 82
Computing Strictly Dominant Strategy Equilibrium
How to decide whether for a given si ∈ Si, all σi ∈ Σi � {si} and all
s−i ∈ S−i we have
ui(si, s−i) > ui(σi, s−i)
Lemma 30
ui(si, s−i) > ui(σi, s−i) for all σi ∈ Σi � {si} and all s−i ∈ S−i
iff
ui(si, s−i) > ui(s�
i
, s−i) for all s�
i
∈ Si � {si} and all s−i ∈ S−i.
Proof: Simple application of the first equality from Lemma 23.
Thus it suffices to check whether ui(si, s−i) > ui(s�
i
, s−i) for all s�
i
∈ Si
and all s−i ∈ S−i.
This can easily be done in time polynomial w.r.t. |G|.
83
IESDS in Mixed Strategies
Define a sequence D0
i
, D1
i
, D2
i
, . . . of strategy sets of player i.
(Denote by Gk
DS
the game obtained from G by restricting the pure strategy
sets to Dk
i
, i ∈ N.)
1. Initialize k = 0 and D0
i
= Si for each i ∈ N.
2. For all players i ∈ N: Let Dk+1
i
be the set of all pure strategies of
Dk
i
that are not strictly dominated in Gk
DS
by mixed strategies.
3. Let k := k + 1 and go to 2.
We say that si ∈ Si survives IESDS if si ∈ Dk
i
for all k = 0, 1, 2, . . .
Definition 31
A strategy profile s = (s1, . . . , sn) ∈ S is an IESDS equilibrium if each
si survives IESDS.
84
IESDS – Algorithm
Note that in step 2 it is not sufficient to consider pure strategies.
Consider the following zero sum game:
X Y
A 3 0
B 0 3
C 1 1
C is strictly dominated by (σ1(A), σ1(B), σ1(C)) = (1
2 , 1
2 , 0) but no
strategy is strictly dominated in pure strategies.
However, there are uncountably many mixed strategies that may
dominate a given pure strategy ...
Recall ui(σi, σ−i) is linear in σi. So to decide strict dominance, we use
linear programming ...
85
Intermezzo: Linear Programming
Linear programming is a technique for optimization of a linear
objective function, subject to linear (non-strict) inequality constraints.
Formally, a linear program in so called canonical form looks like this:
maximize
m�
j=1
cjxj
subject to
m�
j=1
aijxj ≤ bi 1 ≤ i ≤ n
xj ≥ 0 1 ≤ j ≤ m
(objective function)
(constraints)
Here aij, bk and cj are real numbers and xj’s are real variables.
A feasible solution is an assignment of real numbers to the variables
xj, 1 ≤ j ≤ m, so that the constraints are satisfied.
An optimal solution is a feasible solution which maximizes
the objective function
�m
j=1 cjxj.
86
Intermezzo: Complexity of Linear Programming
We assume that coefficients aij, bk and cj are encoded in binary
(more precisely, as fractions of two integers encoded in binary).
Theorem 32 (Khachiyan, Doklady Akademii Nauk SSSR, 1979)
There is an algorithm which for any linear program computes an
optimal solution in polynomial time.
The algorithm uses so called ellipsoid method.
In practice, the Khachiyan’s is not used. Usually simplex algorithm
is used even though its theoretical complexity is exponential.
There is also a polynomial time algorithm (by Karmarkar) which has
better complexity upper bounds than the Khachiyan’s and sometimes
works even better than the simplex.
There exist several advanced linear programming solvers (usually
parts of larger optimization packages) implementing various
heuristics for solving large scale problems, sensitivity analysis, etc.
For more info see
http://en.wikipedia.org/wiki/Linear_programming#Solvers_
and_scripting_.28programming.29_languages
87
IESDS Algorithm – Strict Dominance Step
So how do we use linear programming to decide strict dominance in
step 2 of IESDS procedure?
I.e. whether for a given si there exists σi such that for all σ−i we have
ui(σi, σ−i) > ui(si, σ−i)
Recall that by Lemma 29 we have that σi is strictly dominates σ�
i
iff for
all pure strategy profiles s−i ∈ S−i:
ui(σi, s−i) > ui(σ�
i , s−i)
In other words, it suffices to check the strict dominance only with
respect to all pure profiles of opponents.
88
IESDS Algorithm – Strict Dominance Step
Recall that ui(σi, s−i) =
�
s�
i
∈Si
σi(s�
i
)ui(s�
i
, s−i).
So to decide whether si ∈ Si is strictly dominated by some mixed
strategy σi, it suffices to solve the following system:
�
s�
i
∈Si
xs�
i
· ui(s�
i , s−i) > ui(si, s−i) s−i ∈ S−i
xs�
i
≥ 0 s�
i ∈ Si
�
s�
i
∈Si
xs�
i
= 1
(Here each variable xs�
i
corresponds to the probability σi(s�
i
) assigned
by the strictly dominant strategy σi to s�
i
)
Unfortunately, this is a "strict linear program" ... How to deal with
the strict inequality?
89
IESDS Algorithm – Complexity
Introduce a new variable y to be maximized under the following
constraints:
�
s�
i
∈Si
xs�
i
· ui(s�
i , s−i) ≥ ui(si, s−i) + y s−i ∈ S−i
xs�
i
≥ 0 s�
i ∈ Si
�
s�
i
∈Si
xs�
i
= 1
y ≥ 0
Now si is strictly dominated iff a solution maximizing y satisfies y > 0
The size of the above program is polynomial in |G|.
So the step 2 of IESDS can be executed in polynomial time.
As every iteration of IESDS removes at least one pure strategy,
IESDS runs in time polynomial in |G|.
90
IESDS in Mixed Strategie – Example
X Y
A 3 0
B 0 3
C 1 1
Let us have a look at the first iteration of IESDS.
Observe that A, B are not strictly dominated by any mixed strategy.
Let us construct the linear program for deciding whether C is strictly
dominated: The program maximizes y under the following constraints:
3xA + 0xB + xC ≥ 1 + y
0xA + 3xB + xC ≥ 1 + y
xA , xB , xC ≥ 0
xA + xB + xC = 1
y ≥ 0
The maximum y = 1
2
is attained at xA = 1
2
and xB = 1
2
.
91