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Cautious Behavior. Nash Equilibrium
81
• I “ t1, . . . , mu
• J “ t1, . . . , nu
• u1pi, jq “ aij, and we call matrix
is a set of strategies of the rst player,
is a set of strategies of the second player,
m,n
||aij||
the payo matrix of the game
i,j“1
. So, the rst
players selects a row, and the second player a column of the matrix.
Also, we assume the rst player acts to maximize his gain, and the second one to minimize
the payo of the rst player.
7.2. Cautious Behavior. Nash Equilibrium
Assume, each of the players suppose the opponent can somehow predict his actions and will use
the best strategy against him/her. That is, given the rst player acts
with
jpiq P J
The value
The best strategy the rst player can use is
such that the rst player gets the least possible payo:
0
α
0
α
is the rst player's
i
“ a
i
guaranteed payo for strategyi.
i,jpiq
“ min
jPJ
aij, i P I.
i0that provides him with maximal guaranteed
payo, which is equal to
α0“ max
iPI
min
jPJ
aij“ α
0
.
i
0
i P I
, the second replies
Strategy
i0P I
is called
a maximin
strategy of the rst player.
Similarly, if the second participant plays cautiously and suspects that his strategyjmay be
countered with the opponent's strategy
he would use his
minimax
strategy
ipjq
, which maximizes the payo,
0
β
j
“ a
ipjq,j
“ max
iPI
aij,
j0, which is associated with minimal guaranteed payo to
the opponent:
0
.
j
0
in a game: players keep in mind the worst
The observations above describe
β0“ min
jPJ
max
iPI
cautious behavior
aij“ β
possible scenario and choose their safest strategies to play. By following this policy, the rst
player is guaranteed to gain a payo of value at least
and the second is guaranteed to make the payo not greater than
value
.
Lemma 7.1
(On minimax and maximin).Let
α0, which is
f : X ˆ Y Ñ R
the lower value
β0, which is the game's
of the game,
upper
be a bounded function that
achieves
min
fpx, yq “ f px, ypxqq,
yPY
max
fpx, yq “ f pxpyq, yq,
xPX
for each
x P X
and
y P Y
. Then
max
xPX
min
fpx, yq ď min
yPY
yPY
max
xPX
fpx, yq.
82
Game Theory
Proof.
The following relations hold.
max
min
xPX
fpx, yq “ f px˚, ypx˚qq “ min
yPY
fpx˚, yq ď min
yPY
yPY
max
xPX
fpx, yq.
Therefore, the lower value of an arbitrary game does not exceed its upper value. But what
if they are equal? In this case, the game possesses an interesting quality: there is a balance of
interests situation determined by the pair of cautious strategies
pi0, j0q
. Indeed, the rst player
gets exactly the value the opponent cannot but let him get. Moreover, in this case, outcome
pi0, j0q
i0, it is unbenecial for the opponent to use a strategy other than
an outcome
strategy) Nash equilibrium
Example 7.1.
is stable in some sense: if one of the players, say, player one, uses his cautious strategy
j0. Due to these qualities,
pi0, j0q
of a game, the lower and upper values of which are equal, is called
and is considered as
a (pure strategy) solution
of the game.
a (pure
Consider the following game: two players simultaneously show 1, 2, or 3 ngers.
If the total number of the ngers is odd, the second player pays this number to the rst. If the
number is even, it is payed by the rst player to the second. The payo matrix is as follows:
Player 2
1 2 3
1 -2 3 -4
2 3 -4 5
3 -4 5 -6
Player 1
Calculate the upper and the lower values of the game:
α0“ max
i“1,2,3
β0“ min
j“1,2,3
0
α
“ maxt´4, ´4, ´6u “ ´4,
i
0
β
“ mint3, 5, 5u “ 3.
j
That is,
Example 7.2.
α0ă β0, and the game has no pure strategy solution.
Two conicting sides are involved in military actions. One of the sides attack
with three types of planes, the other side defends with three types of missiles. Eectiveness
(hit probability) of each type of missiles against each type of planes is known and given in the
table below.
Plane
1 2 3
1 0.5 0.6 0.8
2 0.9 0.7 0.8
3 0.7 0.5 0.6
Missile
Calculate the upper and the lower values of the game:
In this game,
α0“ max
β0“ min
α0“ β0“ 0.7
α
i“1,2,3
j“1,2,3
, and pair
0
“ maxt0.6, 0.7, 0.5u “ α
i
0
β
“ mint0.9, 0.7, 0.8u “ β
j
p2, 2q
is Nash equilibrium. That means, the most rational
0
“ 0.7,
2
0
“ 0.7.
2
behavior in this conict is, for the second player, to use planes of the second type to attack,
and for the rst player, to use missiles of the second type to defend.
Mixed Strategies
83
Denition 7.1.
that is,
a
i0j
0
Theorem 7.1.
Given
A P M
a
ij
0
ď a
mˆn
i0j
0
pRq
, a row-column pair
ď a
, i “ 1, . . . , m, j “ 1, . . . , n,
i0j
pi0, j0q
is, simultaneously, a row minimum and a column maximum.
Given a game with payo matrix
A P M
mˆn
values of the game are equal if and only ifAhas a saddle point.
Proof.
Let
α0“ β0, and
i0,
j0be cautious strategies. Prove the existence of a saddle point. By
denition,
that is,
and
a
Now, let
β0ě a
ď a
ij
0
i0j
ď a
i0j
0
pi0, j0q
α0“ max
i“1,...,m
β0“ min
j“1,...,n
ě α0. From
0
a
for each
i0j
α0“ β0, it follows that
ď max
ij
0
i“1,...,m
i “ 1, m,j “ 1, m
be a saddle point. Prove the equality of the upper and the lower values of
min
j“1,...,n
max
i“1,...,m
a
ij
0
aij“ min
aij“ max
“ a
i0j
j“1,...,n
i“1,...,m
“ min
0
j“1,...,n
. That is,
a
i0j
a
ij
ď a
pi0, j0q
0
ď a
ě a
i0j
the game. By denition, we have
is a
saddle pointofA
pRq
. The lower and the upper
,
i0j
0
,
i0j
0
ď a
i0j
,
is a saddle point.
, if
min
j“1,...,n
max
i“1,...,m
aijď max
i“1,...,m
min
j“1,...,n
a
ij
max
i“1,...,m
ď a
i0j
0
0
aijě max
i“1,...,m
ď min
j“1,...,n
j“1,...,n
min
a
i0j
ď max
i“1,...,m
aij.
min
j“1,...,n
aij.
That is,
β0“ min
j“1,...,n
max
i“1,...,m
aij“ max
i“1,...,m
min
j“1,...,n
aij“ α0.
7.3. Mixed Strategies
As we saw above, a payo matrix does not necessarily possess a saddle point, and, therefore,
there are games that does not have a pure strategy equilibrium (in fact, most of the games are
of this kind). To cope with such games a
Denition 7.2.Amixed strategy
is a probability distribution on the set of a player's pure
strategies.
In our case, since each player has a nite number of pure strategies, mixed strategies can
be expressed as vectors:
p “ pp1, p2, . . . , pmq P Pm“ tp :
q “ pq1, q2, . . . , qnq P Qn“ tq :
mixed strategy
ÿ
i“1
n
ÿ
j“1
concept is involved.
m
pi“ 1, piě 0u,
qj“ 1, qjě 0u,
84
Game Theory
where
pi,
qjare probabilities (or frequencies) of using pure strategiesiandjby the rst and
the second players, respectively.
Expected (or mean) payo of the rst player is
m
n
ÿ
ÿ
Epp, qq “
i“1
The cautious policy prescribes the players to use such mixed strategies
aijpiqj“ pAqτ.
j“1
p˚,
q˚that
mean guaranteed gain of the rst player,
αppq “ min
and
q˚minimizes mean guaranteed gain of the rst player,
βpqq “ max
Epp, qq Ñ max
qPQ
n
Epp, qq Ñ min
pPP
m
pPP
qPQ
,
m
.
n
Similarly to the pure strategies case, we introduce the lower value of the game:
α “ αpp˚q “ max
pPP
m
min
qPQ
Epp, qq,
n
and the upper value of the game:
p˚maximizes
β “ βpq˚q “ min
which, according to Lemma 7.1, are connected by the relation:
Theorem 7.2
pp˚, q˚q
Proof.
such that:
1.
Epp, q˚q ď Epp˚, q˚q ď Epp˚, qq
2.
Epp˚, q˚q “ α “ β “ ν
Without loss of generality we may assume that all the elements of payo matrixAare
(Von Neumann).For each matrix game there exists a pair of mixed strategies
for each
, whereνis
the value
positive. Otherwise, we transformAinto non-negative matrix
aijă 0u ` 1
Epp, qq
to each of its elements. Easy to see that expected payos
by constant valueCfor each mixed strategiespandq. That means, optimal strategies
max
qPQ
pPP
n
p P Pm,
of the game.
Epp, qq,
m
q P Qn;
α ď β
.
A1by adding
E1pp, qq
C “ maxt´aij:
for
A1dier from
are the same for both of the matrices, though the lower and the upper values associated with
matrix
A1are greater byC.
Consider the problem of the rst player, who maximizes his mean guaranteed payo:
where
αppq “ min
qPQ
αppq Ñ max
Epp, qq ď Epp, qjq “
n
pPP
,
m
m
ÿ
aijpi, j “ 1, . . . , n.
i“1
(7.1)
Here
qj“ p0, . . . , 0, 1j, 0, . . . , 0q
Introduce variables
is a mixed strategy corresponding to some pure strategyj.
p
ui“
i
ě 0, i “ 1, . . . , m.
αppq
Mixed Strategies
85
From (7.1), it follows that
m
ÿ
aijuiď 1, j “ 1, . . . , n.
i“1
Also, by summing values
uiover alli, we get:
m
ÿ
ui“
i“1
m
ř
i“1
αppq
p
i
1
“
.
αppq
Altogether that gives us a linear program associated with the rst player:
m
ÿ
fpuq “
m
ÿ
uiÑ min
i“1
tuu
,
aijuiě 1, j “ 1, . . . , n,
i“1
uiě 0, i “ 1, . . . , m.
Similarly, for the second player,
n
ÿ
ϕpvq “
n
ÿ
vjÑ max
j“1
tvu
,
aijvjď 1, i “ 1, . . . , m,
j“1
vjě 0, j “ 1, . . . , n,
(7.2)
(7.3)
(7.4)
(7.5)
(7.6)
(7.7)
where
vj“ qj{βpqq,q P Qn. Easy to see that problems (7.2)(7.4) and (7.5)(7.7) are dual
to one another, both have feasible solutions and are not unbounded. That implies, there are
feasible vectors
u˚and
v˚, which are optimal for corresponding problems,
˜
m
ÿ
˚
v
¨
j
i“1
˜
n
ÿ
˚
u
¨
i
j“1
˚
aiju
´ 1¸“ 0, j “ 1, . . . , n,
i
˚
aijv
´ 1¸“ 0, i “ 1, . . . , m,
j
fpu˚q “ ϕpv˚q
, and
due to complementary slackness. Note that by summing equalities of each group overjandi,
respectively, and dividing the results by
Epp˚, q˚q “
where
p˚,
q˚are mixed strategies such that
p
fpu˚q ¨ ϕpv˚q
˚
“
i
fpu˚q
, we get
1
fpu˚q
˚
u
i
, q
˚
j
“
“
1
ϕpv˚q
˚
v
j
ϕpv˚q
,
.
86
Game Theory
Show that pair
Let
p P Pnand
Epp, q˚q “
Epp˚, qq “
pp˚, q˚q
satises the statements of the Theorem.
q P Qn. From (7.3) and (7.6) we have:
m
ÿ
i“1
n
ÿ
j“1
n
ÿ
p
aijq
i
j“1
m
ÿ
q
aijp
j
i“1
m
ÿ
˚
“
j
i“1
ÿ
˚
“
i
j“1
n
ÿ
p
i
j“1
n
m
ÿ
q
j
i“1
aijv
ϕpv˚q
aiju
fpu˚q
˚
j
m
ÿ
1
ď
ϕpv˚q
˚
i
ě
fpu˚q
i“1
n
ÿ
1
j“1
That gives us the rst statement. Prove the second one. From
Epp, q˚q ď Epp˚, q˚q ď Epp˚, qq, p P Pm, q P Qn,
follows
max
Epp, q˚q ď Epp˚, q˚q ď min
pPP
m
qPQ
Epp˚, qq,
n
or,
β ď βpq˚q ď Epp˚, q˚q ď αpp˚q ď α.
That means,
β ď α
. Lemma 7.1 gives us opposite inequality, so,
7.4. Methods to Solve Games
pi“
“ Epp˚, q˚q;
ϕpv˚q
1
1
qj“
“ Epp˚, q˚q.
fpu˚q
α “ β “ Epp˚, q˚q
.
In this section, we demonstrate several approaches to nd a mixed-strategy solution of a game.
But rst we introduce a concept that allows to reduce payo matrices by eliminating some
obviously unpromising strategies.
Denition 7.3.
if
aijě akjfor each
Letiandkbe two pure strategies of the rst player. We say,idominatesk,
j “ 1, . . . , n
, and the inequality is strict for somej.
Similarly, for the second player.
Denition 7.4.
k
, if
aijď aikfor each
Letjandkbe two pure strategies of the second player. We say,jdominates
i “ 1, . . . , m
, and the inequality is strict for somei.
Clearly, if a pure strategy is dominated, it can neither be optimal in terms of pure Nash
equilibrium, nor can occur in an optimal mixed strategy with non-zero probability. Thus, the
strategy can not be considered. Sometimes, sequential removal of dominated strategies can
signicantly shorten a game.
Example 7.3.
Solve the game with payo matrix
Consider two rows,iandk,
i ą k
. For eachj,
||aij||
m,n
i,j“1
:
aij“ i ´ j
.
aij´ akj“ pi ´ jq ´ pk ´ jq “ i ´ k ą 0.
That is, a strategy with greater number dominates a strategy with smaller number, and, there-
fore, strategymdominates all the other strategies of the rst player.
Similarly, for the second player. If
j ą l
, then
amj´ aml“ pm ´ jq ´ pm ´ lq “ l ´ j ă 0,
and strategyjdominatesl. Therefore, strategyndominates all the other, and the matrix is
reduced to a single element,
j0“ n,ν0“ amn“ m ´ n
amn“ m ´ n
.
, which is the saddle point. The solution is
i0“ m
,
Methods to Solve Games
87
Remark
7.1.The concept of strategy dominance can be generalized. Letibe a pure strategy
of the rst player such that
m
ÿ
aijď
k“1, k‰i
ηkakj, @j “ 1, . . . , n,
ř
for some non-negative
tηku
such that
ηk“ 1
k
.
That is,iis dominated not by a particular one other strategy, but by a convex combination
of all the remaining strategies. Easy to see, strategyidoes not occur in an optimal mixed
strategy with non-zero probability, and, therefore, can be omitted from consideration.
Similarly, for the second player. If there is a non-negative sequence
such that
ř
l
λl“ 1
and
aijě
n
ÿ
l“1, l‰j
λkail, @i “ 1, . . . , n,
tλlu,l P t1, 2, . . . , nuztju
then pure strategyjof the second player can not occur in an optimal mixed strategy with
non-zero probability.
Example 7.4.
Consider a game with payo matrix
ˆ
A “
4 2 3 ´1
´4 0 ´2 2
˙
.
,
Easy to see, the matrix has no saddle points and no dominated rows/columns. At the same
time,
„
2
0
„
3
´2
ě
ě
„
1
2
„
4
5
´4
´4
4
4
`
`
„
1
2
„
1
5
´1
´1
2
2
“„
“„
´1
´
3
2
14
,
3
.
5
That implies, the second and the third strategies of the second player can be eliminated, and
the matrix can be reduced to
ˆ
´4 2
4 ´1
˙
.
7.4.1. Linear Programming Approach
The most general method to solve games follows from Von Neumann's theorem and involves
solving a pair of dual linear programs. We illustrate this approach in the examples below.
Example 7.5
(Search Game).The rst player searches for an item hidden by the second player
in one ofnpossible places. If the rst player fails to nd the item, he gains nothing, otherwise
the player gains
aią 0
(if the item was located at thei-th place).
The game has a diagonal payo matrix:
¨
˛
a10 ¨ ¨ ¨ 0
‹
‹
.
‹
.
.
‚
.
n
A “
˚
0 a2¨ ¨ ¨ 0
˚
˚
.
.
.
˝
.
.
.
0 0 ¨ ¨ ¨ a
.
.
.
88
Game Theory
The matrix has no saddle point, so, the game has no solution in pure strategies. To nd a
mixed solution, consider pair of linear programs associated with both of the players.
For the rst player, it is as follows
n
ÿ
fpuq “
uiÑ min
i“1
tuu
,
aiuiě 1, i “ 1, . . . , n,
uiě 0, i “ 1, . . . , n.
n
Obviously, the solution is:
1
˚
u
“
i
for eachi, and
a
i
fpu˚q “
ř
1
. Therefore, the value of the
a
k
k“1
game is
1
ν “
k“1
n
ř
,
1
a
k
and the optimal mixed strategy is to search in placeiwith frequency
1
“
n
ř
k“1
a
i
, i “ 1, . . . , n.
1
a
k
˚
p
i
For the second player, we have the following linear program:
n
ÿ
ϕpvq “
vjÑ max
j“1
tvu
,
ajvjď 1, j “ 1, . . . , n,
vjě 0, j “ 1, . . . , n,
the solution of which denes optimal mixed strategy:
1
a
“
n
ř
k“1
¨
j
, j “ 1, . . . , n.
1
a
k
˛
Example 7.6.
˚
q
j
Solve the game from example 7.1. Recall the payo matrix:
´2 3 ´4
A “
˝
3 ´4 5
‚
.
´4 5 ´6
First, we add 6 to each element ofAto obtain a nonnegative matrix
A1:
A1“
¨
4 9 2
˝
9 2 11
2 11 0
˛
‚
.
Methods to Solve Games
Consider the rst player's linear program:
89
fpuq “ u1` u2` u3Ñ min
u1,u2,u3ě0
,
4u1` 9u2` 2u3ě 1,
9u1` 2u2`11u3ě 1,
2u1`11u
2
Note that summing the rst, the third, and twice the second inequalities gives
24u3ě 4
, so,
fpuq ě 1{6
and the minimum is
ě 1.
1{6
if there are nonnegative values
24u1` 24u2`
u1,
u2,
u3,
which satisfy all the constraints as equalities. Solving the linear system, we obtain:
1
˚
u
“
1
, u
24
1
˚
“
2
, u
12
1
˚
“
3
, f pu˚q “
24
1
,
6
from which it follows
1
˚
p
“
1
, p
4
1
˚
“
, p
2
2
1
˚
“
.
3
4
Obviously, solution of the program of the second player gives us the same probabilities:
1
˚
q
“
1
, q
4
1
˚
“
2
, q
2
1
˚
“
,
3
4
since the payo matrix is symmetrical and all of the constraints in the second player's program
must be satised as equalities (due to complementary slackness).
The value of the game with modied matrix is
ν1“
original game we must subtract previously added 6 from
1
“ 6
fpu˚q
ν1:
ν “ ν1´ 6 “ 0
. To nd the value of the
.
7.4.2. Active Strategies Theorem and Graphical Approach
Denition 7.5.
We call a pure strategy
active
if it is used in some optimal mixed strategy
with non-zero probability.
The connection between games and linear programs stated in Von Neumann's theorem
allows us to make an observation: since a solution of a game can be found from solution of
two dual linear programs with
which contain at most
mintm, nu
of active strategies for both of the players does not exceed
Theorem 7.3
(On active strategies).If one of the players uses an optimal mixed strategy, his
m ˆ n
and
n ˆ m
matrices, there are optimal mixed strategies,
non-zero elements. That is, for any
mintm, nu
m ˆ n
.
game, the number
opponent gets the value of the game by using any mixed strategy involving active strategies only.
Proof.
and
qj“ p0, . . . , 0, 1j, 0, . . . , 0q
Let
p˚and
J1Ď J
be the set of active strategies of the second player. Denote
ν “ Epp˚, q˚q “
ě ν
q˚be optimal mixed strategies for the rst and the second player, respectively,
ÿ
j‰k
˚
q
j
. Clearly,
ÿ
i“1
˚
` q
νk“ ν
k
νj“ Epp˚, qjq
νjě ν
m
n
ÿ
aijp
j“1
n
ÿ
j“1
for any
n
q
j
˚
ÿ
˚
“
j
j“1
˚
` q
pνk´ νq “ ν ` q
k
˚
i
q
j P J
. On the other hand,
m
ÿ
˚
q
aijp
j
i“1
n
ÿ
˚
“
i
j“1
˚
k
˚
q
j
ÿ
νj“
j‰k
pνk´ νq ě ν.
˚
q
j
νj` q
k
, where
˚
ν
k
90
Game Theory
That is,
˚
q
pνk´ νq “ 0
k
. If
k P J1, then
Consider a mixed strategyqsuch that
m
ÿ
Epp˚, qq “
i“1
ÿ
jPJ
aijp
1
˚
q
k
ą 0
and, therefore,
qją 0
˚
qj“
i
implies
ÿ
jPJ
qjνj“ ν
1
νk“ ν
.
j P J1. Easy to see that
ÿ
qj“ ν.
1
jPJ
ν “ Epp˚, qq
The theorem above is useful to solve games of small size.
2 ˆ 2
Consider a game with
games
2 ˆ 2
payo matrix,
A “
ˆ
a11a
a21a
12
22
˙
,
which has no saddle points. In this case, all the players' strategies are active. Indeed, assume
i
is the only active strategy of the rst player. Clearly, if the second player also has a single
active strategyj, the pair
player are active, then, from Theorem 7.3,
pi, jq
is a saddle point. If both from the strategies of the second
ai1“ ai2, and, therefore, one of the matrix columns
is elementwise greater or equal to another. This leads us to contradiction, since corresponding
pure strategy is dominated and cannot be active.
Let
p˚“ px, 1 ´ xq,0 ă x ă 1
, be an optimal mixed strategy for the rst player. From
Theorem 7.3,
:
ν “ Epp˚, q1q “ a11x ` a21p1 ´ xq “ pa11´ a21qx ` a21,
ν “ Epp˚, q2q “ a12x ` a22p1 ´ xq “ pa12´ a22qx ` a22.
Solving the system above, we get:
x “ p
1 ´ x “ p
˚
“
1
pa11´ a21q ´ pa12´ a22q
˚
“
2
pa11´ a21q ´ pa12´ a22q
a22´ a
a11´ a
a22a21´ a12a
ν “
21
12
21
,
,
.
pa11´ a21q ´ pa12´ a22q
Note that the denominator is not zero. Otherwise,
a11´ a21“ a12´ a22, and
a11,
a12are both
greater or both less then corresponding elements of the second row. That implies, one of the
rst player's strategies is dominated. Also,
x P p0, 1q
the argument is similar to that above.
Similarly, for the second player,
˚
q
“
1
pa11´ a21q ´ pa12´ a22q
a22´ a
12
, q
˚
“
2
pa11´ a21q ´ pa12´ a22q
a11´ a
21
.
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