A very little Game Theory

Math 20Linear Algebra and Multivariable

CalculusOctober 13, 2004

A Game of Chance

You and I each have a six-sided die

We roll and the loser pays the winner the difference in the numbers shown

If we play this a number of times, who’s going to win?

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The Payoff Matrix Lists one player’s

(call him/her R) possible outcomes versus another player’s (call him/her C) outcomes

Each aij represents the payoff from C to R if outcomes i for R and j for C occur (a zero-sum game).

C’s outcomes

1 2 3 4 5 6

R’s outcom

es

1 0 -1 -2 -3 -4 -5

2 1 0 -1 -2 -3 -4

3 2 1 0 -1 -2 -3

4 3 2 1 0 -1 -2

5 4 3 2 1 0 -1

6 5 4 3 2 1 0

Expected Value

Let the probabilities of R’s outcomes and C’s outcomes be given by probability vectors

€

p = p1 p2 L pn[ ]

€

q =

q1

q2

M

qn

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Expected Value

The probability of R having outcome i and C having outcome j is therefore piqj.

The expected value of R’s payoff is

E(p,q) = π ι αιϕθϕι, ϕ=1

ν

∑ = πΑθ

Expected Value of this Game

€

1

6

1

6

1

6

1

6

1

6

1

6

⎡ ⎣ ⎢

⎤ ⎦ ⎥⋅

0 −1 −2 −3 −4 −51 0 −1 −2 −3 −42 1 0 −1 −2 −33 2 1 0 −1 2

4 3 2 1 0 −15 4 3 2 1 0

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⋅

161616161616

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

€

= 16⋅ 1 1 1 1 1 1[ ] ⋅

−15

−9−33

9

15

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⋅ 16

€

= 0A “fair game” if the dice are fair.

Expected value with an unfair die

SupposeThen

€

p =1

10

1

10

1

5

1

5

1

5

1

5

⎡ ⎣ ⎢

⎤ ⎦ ⎥

€

E(p,q) =1

101

1015

15

15

15

⎡ ⎣ ⎢

⎤ ⎦ ⎥⋅

0 −1 −2 −3 −4 −51 0 −1 −2 −3 −42 1 0 −1 −2 −33 2 1 0 −1 2

4 3 2 1 0 −15 4 3 2 1 0

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⋅

161

6161

6161

6

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=1

10⋅ 1 1 2 2 2 2[ ] ⋅

−15

−9−33

9

15

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⋅ 16

=160

(−15) + (−9) + 2(−3) + 2 ⋅ 3 + 2 ⋅ 9 + 2 ⋅15( ) =2460

=25

StrategiesWhat if we could

choose a die to be as biased as we wanted?

In other words, what if we could choose a strategy p for this game?

Clearly, we’d want to get a 6 all the time!

C’s outcomes

1 2 3 4 5 6

R’s outcom

es

1 0 -1 -2 -3 -4 -5

2 1 0 -1 -2 -3 -4

3 2 1 0 -1 -2 -3

4 3 2 1 0 -1 -2

5 4 3 2 1 0 -1

6 5 4 3 2 1 0

Flu Vaccination

Suppose there are two flu strains, and we have two flu vaccines to combat them.

We don’t know distribution of strains

Neither pure strategy is the clear favorite

Is there a combination of vaccines that maximizes immunity?

Strain

1 2Vaccine

1 0.85 0.70

2 0.60 0.90

Fundamental Theorem of Zero-Sum Games

There exist optimal strategies p* for R and q* for C such that for all strategies p and q:

E(p*,q) ≥ E(p*,q*) ≥ E(p,q*)E(p*,q*) is called the value v of the gameIn other words, R can guarantee a lower

bound on his/her payoff and C can guarantee an upper bound on how much he/she loses

This value could be negative in which case C has the advantage

Fundamental Problem of Zero-Sum games

Find the p* and q*!In general, this requires linear

programming. Next week!There are some games in which we can

find optimal strategies now:Strictly-determined games2� 2 non-strictly-determined games

Network Programming

Suppose we have two networks, NBC and CBS

Each chooses which program to show in a certain time slot

Viewer share varies depending on these combinations

How can NBC get the most viewers?

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Payoff MatrixCBS shows

60 Minutes

Survivor

CSI

Everybody Loves Raymond

NBC

Shows

Friends 60 20 30 55

Dateline 50 75 45 60

Law & Order

70 45 35 30

NBC’s Strategy

NBC wants to maximize NBC’s minimum share

In airing Dateline, NBC’s share is at least 45

This is a good strategy for NBC

60 M

Surv

CSI

ELR

F 60 20 30 55

DL 50 75 45 60

L&O 70 45 35 30

CBS’s Strategy

CBS wants to minimize NBC’s maximum share

In airing CSI, CBS keeps NBC’s share no bigger than 45

This is a good strategy for CBS

60 M

Surv

CSI

ELR

F 60 20 30 55

DL 50 75 45 60

L&O 70 45 35 30

Equilibrium

(Dateline,CSI) is an equilibrium pair of strategies

Assuming NBC airs Dateline, CBS’s best choice is to air CSI, and vice versa

60 M

Surv

CSI

ELR

F 60 20 30 55

DL 50 75 45 60

L&O 70 45 35 30

Characteristics of an Equlibrium

Let A be a payoff matrix. A saddle point is an entry ars which is the minimum entry in its row and the maximum entry in its column.

A game whose payoff matrix has a saddle point is called strictly determined

Payoff matrices can have multiple saddle points

Pure Strategies are optimal in Strictly-Determined Games

If ars is a saddle point, then er

T is an optimal strategy for R and es is an optimal strategy for C.

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Proof

E(erT ,q) = ερ

Τ Αθ = αρ1 αρ2 Λ αρν[ ] ⋅

θ1

θ2

Μθν

= αρ1θ1 + αρ2θ2 +Λ + αρνθν

≥ αρσθ1 + αρσθ2 +Λ + αρσθν

= αρσ(θ1 +Λ + θν ) = αρσ = Ε(ερΤ , εσ)

Proof

E(p,es ) = πΑεσ = π1 π2 Λ πµ[ ] ⋅

α1σ

α2σ

Μαµ σ

= π1α1σ + π2α2σ +Λ + πµ αµ σ

≤ π1αρσ + π2αρσ +Λ + πµ αρσ

= (π1 + π2 +Λ + πµ )αρσ = αρσ = Ε(ερΤ , εσ)

Proof

So for all strategies p and q:E(er

T,q) ≥ E(erT,es) ≥ E(p,es)

Therefore we have found the optimal strategies

2x2 non-strictly determined

In this case we can compute E(p,q) by hand in terms of p1 and q1

€

E(p,q) = p1 p2[ ] ⋅a11 a12

a21 a22

⎡

⎣ ⎢

⎤

⎦ ⎥⋅

q1

q2

⎡

⎣ ⎢

⎤

⎦ ⎥

= p1a11q1 + p1a12q2 + p2a21q1 + p2a22q2

= p1a11q1 + p1a12(1−q1) + (1− p1)a21q1 + (1− p1)a22(1−q1)

= (a11 + a22 − a12 − a21)p1 − (a22 − a21)[ ]q1 + (a12 − a22)p1 + a22

Optimal Strategy for 2x2 non-SD

LetThis is between 0 and 1 if A has no

saddle pointsThen €

p1 = p1∗ a22 − a21

a11 + a22 − a12 − a21

; p2 =1− p1

€

E(p,q) =(a12 − a22)(a22 − a21)

a11 + a22 − a12 − a21

+ a22

=a11a22 − a12a21

a11 + a22 − a12 − a21

Optimal set of strategies

We have

p∗ =α22 − α21

α11 + α22 − α12 − α21

α11 − α12

α11 + α22 − α12 − α21

θ∗ =

α22 − α12

α11 + α22 − α12 − α21

α11 − α21

α11 + α22 − α12 − α21

ϖ=α11α22 − α12α21

α11 + α22 − α12 − α21

Flu Vaccination

Strain

1 2Vaccine

1 0.85 0.70

2 0.60 0.90

p1∗ =

.90 − .60.85 + .90 − .70 − .60

=.30.45

=23

π2∗ =

13

θ1∗ =

.90 − .70.85 + .90 − .70 − .60

=.20.45

=49

θ2∗ =

59

ϖ=(.85)(.90) − (.70)(.60).85 + .90 − .70 − .60

=.345.45

= .766Κ

Flu Vaccination

Strain

1 2Vaccine

1 0.85 0.70

2 0.60 0.90

So we should give 2/3 of the population vaccine 1 and 1/3 vaccine 2

The worst that could happen is a 4:5 distribution of strains

In this case we cover 76.7% of pop

Other Applications of GT

WarBattle of Bismarck

Sea

Business Product IntroductionPricing

Dating

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