More Tutorials at www.DumbLittleDoctor.com
5.9.
Symmetric Games
Symmetric games are the games that are being considered as fair. The roles of both
players are completely symmetric. They have the same strategies available. If player 1
wins by using strategy A against player 2, who is using B, then player 1 will loose by
playing strategy B against player 2’s strategy A. Mathematically, this can be expressed as
follows:
5.9.1. Definition.
A two player zero-sum game G with matrix A is called symmetric, if A is skew
symmetric, i.e. if A = AT
An example of a symmetric game is, of course, “Rock, Paper, Scissor”.
5.9.2. Proposition.
Every symmetric game G has value 0.
Proof. We compute:
{ {
}
}
m 1
m 1
v ( G ) = max min pT Aq : q m 1 : p m 1
{ {( p Aq ) : q } : p }
= max {min {( q A p ) : q } : p }
= max {min {( q Ap ) : q } : p }
= min {max {( q Ap ) : q } : p }
= max min
T
T
T
T
T
m 1
m 1
m 1
m 1
T
T
T
T
m 1
m 1
= v ( G )
Hence v(G) = 0.
5.9.3. Proposition.
Let G be a symmetric game. Then ( pe , qe ) is a point of equilibrium if and only if
Ape 0
Aqe 0
91
Copyright strictly belongs to UC
More Tutorials at www.DumbLittleDoctor.com
Proof. Assume first that the two conditions hold. Then Ape 0 implies that
T
peT A = AT pe
T
= ( Ape )
T
= ( Ape )
(
)
0T
and hence, since all coordinates of qe are positive, we conclude that pe Aqe 0 . Also,
since all coordinates of pe are positive and since Aqe 0 , it follows that pe Aqe 0 , i.e.
pe Aqe = 0
If q m 1 is given, then all coordinates of q are positive. Since all coordinates of
peT A are also positive, we find that
pe Aq 0 = pe Aqe
Similarly, if p m 1 is given, then
pAqe 0 = pe Aqe
and therefore ( pe , qe ) is a point of equilibrium.
Conversely, if ( pe , qe ) is a point of equilibrium, then
peT Aqe = max pT Aqe : p m 1 = 0
If at least one coordinate of Aqe were strictly positive then there would be a (unit) vector
p m 1 so that pT Aqe > 0 , contradicting the fact that max pT Aqe : p m 1 = 0 .
Hence
Aqe 0
Similarly, we show that
Ape 0
{
}
{
}
Using the last theorem, it is now easy to find all saddle points of “rock, paper, and
scissor”. We have to find all solutions of
0 1 1 r 0 1 0 1 s 0 1 1 0 t 0 r + s+t =1
r, s,t 0
This set of equations leads to
92
Copyright strictly belongs to UC
More Tutorials at www.DumbLittleDoctor.com
ts
rt
sr
r + s+t =1
r, s,t 0
The only solution to this set of equations is given by r = s = t =
1
, and therefore the only
3
1 1 1
point of equilibrium obtained from pe = qe = , , .
3 3 3
93
Copyright strictly belongs to UC