1

2

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

8

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

¯