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APPENDIX B
A PROCEDURE FOR GENERATING AN EQUILIBRIUM
POINT FOR 2-PERSON GAMES(That sometimes works!)
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Basic Idea• Player II speaking:
• "I'll design my strategy, call it y* such that the expected payoff to Player I will be constant regardless of what she will do!
• This will produce a stable situation for me!!!!!!
• Thus, to compute y* I have to solve the following equation:
xAy* = Constant
• Any such y* will certainly satisfy xAy* ≤ x*Ay*, for all x in S, regardless of what Player I is selecting for x* (her best strategy)."
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• Needless to say, Player I will attempt to do the same thing and construct a strategy x* such that x*By is independent of y.
• How then do we solve xAy* = Constant ???Let z = Ay*
Then xAy* = xz = (x1, x2, ..., xm) (z1, z2, ..., zm)t
= (x1z1 + x2z2 + ... xmzm)
Now if z1 = z2 = ... = zm , this becomes
xAy* = (x1 + x2 + ... xm) zm = zm = constant REGARDLESS OF WHAT x IS!!How do we use this? We can try putting all
components of Ay* equal to each other.
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Particular case: n = m = 2
A =
a b
c d
⎡
⎣ ⎢
⎤
⎦ ⎥
y* = (y*1, y*2 )= (y*1, 1 – y*1)
z =
a b
c d
⎡
⎣ ⎢
⎤
⎦ ⎥
y *1
1 − y *1
⎡
⎣ ⎢
⎤
⎦ ⎥
= (ay*1 + b(1 – y*1), cy*1 + d(1 – y*1))
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•(z1, z2) = (ay*1 + b(1 – y*1), cy*1 + d(1 – y*1)
•For z1 = z2 we obtain•ay*1 + b(1 – y*1) = cy*1 + d(1 – y*1)•So, provided a + d – (b + c) = 0
•NOTE: Beware! This y* formula is obtained from matrix A. The x* will use matrix B.
y *1 d b
a d (b c )
y *2 a c
a d (b c)since y*2 = 1 – y*1
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Example 1.8.2 (Continued)
A =
8 1
6 13
⎡
⎣ ⎢
⎤
⎦ ⎥
y *1
=
d − b
a + d − ( b + c )
=
13 − 1
8 + 13 − ( 6 + 1 )
=
6
7
y *2
= 1 − y *1
=
1
7
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checking z=Ay* is constant
8 1
6 13
⎡
⎣
⎢
⎤
⎦
⎥
6
7
1
7
⎡
⎣
⎢
⎤
⎦
⎥
= ( 7 , 7 )
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a bc d
Exercise• Derive the recipe for x* in the case n = m = 2.
• Answer: If B = ( )• x* = ((d–c)/(a+d–b–c), (a–b)/(a+d–b–c))
Example:• Find an equilibrium pair (X*, Y*) of mixed strategies for the 2-person non-zero sum game with payoff matrix
• See lecture for solution.
(8,−1) (3, 7) (0,3)(7,6) (−1,2) (2,8)(1,4) (6,1) (5,0)
⎛
⎝
⎜ ⎜
⎞
⎠ ⎟ ⎟
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Appendix A
• See lecture for discussion and examples.
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Solution concepts of NON-zero-sum games
• We have two concepts–Security level pairs, Equilibrium pairs
• The security level idea is not really very good here, because it assumes a player is simultaneously trying to maximize their own payoff, whilst minimizing their opponents payoff. These two objectives are sometimes diametrically opposed. It is no longer true that a player can get rich only by keeping their opponent poor.
• Also we know that the payoff for secuity level is
≤ that for equilibrium pair.
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• Equilibrium pairs - more acceptable concept, but difficulties with these too.
• E.g.
• Only one equilibrium pair, payoff (1,1). But clearly the payoff (5,5) better for both.
• No satisfactory simple notion of ‘optimal strategy’ and ‘value’ as there is with zero sum games.
(5,5) (0,10)
(10,0) (1,1)
Ê Ë Á ˆ
¯