Lesson 35Game Theory and Linear Programming

Math 20

December 14, 2007

Announcements

I Pset 12 due December 17 (last day of class)

I Lecture notes and K&H on website

I next OH Monday 1–2 (SC 323)

Outline

RecapDefinitionsExamplesFundamental TheoremGames we can solve so far

GT problems as LP problemsFrom the continuous to the discreteStandardizationRock/Paper/Scissors again

The row player’s LP problem

DefinitionA zero-sum game is defined by a payoff matrix A, where aij

represents the payoff to the row player if R chooses option i and Cchooses option j .

I The row player chooses from the rows of the matrix, and thecolumn player from the columns.

I The payoff could be a negative number, representing a netgain for the column player.

DefinitionA zero-sum game is defined by a payoff matrix A, where aij

represents the payoff to the row player if R chooses option i and Cchooses option j .

I The row player chooses from the rows of the matrix, and thecolumn player from the columns.

I The payoff could be a negative number, representing a netgain for the column player.

DefinitionA strategy for a player consists of a probability vector representingthe portion of time each option is employed.

I We use a row vector p for the row player’s strategy, and acolumn vector q for the column player’s strategy.

I A pure strategy (select the same option every time) isrepresented by a standard basis vector ej or e′j . For instance,if R has three choices and C has five:

e′2 =

010

e4 =(0 0 0 1 0

)I A non-pure strategy is called mixed.

DefinitionA strategy for a player consists of a probability vector representingthe portion of time each option is employed.

I We use a row vector p for the row player’s strategy, and acolumn vector q for the column player’s strategy.

I A pure strategy (select the same option every time) isrepresented by a standard basis vector ej or e′j . For instance,if R has three choices and C has five:

e′2 =

010

e4 =(0 0 0 1 0

)I A non-pure strategy is called mixed.

DefinitionThe expected value of row and column strategies p and q is thescalar

E (p, q) =n∑

i ,j=1

piaijqj = pAq

Probabilistically, this is the amount the row player receives (or thecolumn player if it’s negative) if players employ these strategies.

DefinitionThe expected value of row and column strategies p and q is thescalar

E (p, q) =n∑

i ,j=1

piaijqj = pAq

Probabilistically, this is the amount the row player receives (or thecolumn player if it’s negative) if players employ these strategies.

Rock/Paper/Scissors

Example

What is the payoff matrix for Rock/Paper/Scissors?

SolutionThe payoff matrix is

A =

0 −1 11 0 −1−1 1 0

.

Rock/Paper/Scissors

Example

What is the payoff matrix for Rock/Paper/Scissors?

SolutionThe payoff matrix is

A =

0 −1 11 0 −1−1 1 0

.

Example

Consider a new game: players R and C each choose a number 1,2, or 3. If they choose the same thing, C pays R that amount. Ifthey choose differently, R pays C the amount that C has chosen.What is the payoff matrix?

Solution

A =

1 −2 −3−1 2 −3−1 −2 3

Example

Consider a new game: players R and C each choose a number 1,2, or 3. If they choose the same thing, C pays R that amount. Ifthey choose differently, R pays C the amount that C has chosen.What is the payoff matrix?

Solution

A =

1 −2 −3−1 2 −3−1 −2 3

Theorem (Fundamental Theorem of Matrix Games)

There exist optimal strategies p∗ for R and q∗ for C such that forall strategies p and q:

E (p∗, q) ≥ E (p∗, q∗) ≥ E (p, q∗)

E (p∗, q∗) is called the value v of the game.

Theorem (Fundamental Theorem of Matrix Games)

There exist optimal strategies p∗ for R and q∗ for C such that forall strategies p and q:

E (p∗, q) ≥ E (p∗, q∗) ≥ E (p, q∗)

E (p∗, q∗) is called the value v of the game.

Reflect on the inequality

E (p∗, q) ≥ E (p∗, q∗) ≥ E (p, q∗)

In other words,

I E (p∗, q) ≥ E (p∗, q∗): R can guarantee a lower bound onhis/her payoff

I E (p∗, q∗) ≥ E (p, q∗): C can guarantee an upper bound onhow much he/she loses

I This value could be negative in which case C has theadvantage

Fundamental problem of zero-sum games

I Find the p∗ and q∗!I Last time we did these:

I Strictly-determined gamesI 2× 2 non-strictly-determined games

I The general case we’ll look at next.

Pure Strategies are optimal in Strictly-Determined Games

TheoremLet A be a payoff matrix. If ars is a saddle point, then e′r is anoptimal strategy for R and es is an optimal strategy for C. Alsov = E (e′r , es) = ars .

Optimal strategies in 2× 2 non-Strictly-Determined Games

Let A be a 2× 2 matrix with no saddle points. Then the optimalstrategies are

p =(a22 − a21

∆

a11 − a12

∆

)q =

a22 − a12

∆a11 − a21

∆

where ∆ = a11 + a22 − a12 − a21. Also

v =|A|∆

Outline

RecapDefinitionsExamplesFundamental TheoremGames we can solve so far

GT problems as LP problemsFrom the continuous to the discreteStandardizationRock/Paper/Scissors again

The row player’s LP problem

This could get a little weird

This derivation is not something that needs to be memorized, butshould be understood at least once.

Objectifying the problem

Let’s think about the problem from the column player’sperspective. If she chooses strategy q, and R knew it, he wouldchoose p to maximize the payoff pAq. Thus the column playerwants to minimize that quantity. That is, C ’s objective is realizedwhen the payoff is

E = minq

(max

ppAq.

)

This seems hard! Luckily, linearity, saves us.

Objectifying the problem

Let’s think about the problem from the column player’sperspective. If she chooses strategy q, and R knew it, he wouldchoose p to maximize the payoff pAq. Thus the column playerwants to minimize that quantity. That is, C ’s objective is realizedwhen the payoff is

E = minq

(max

ppAq.

)This seems hard! Luckily, linearity, saves us.

From the continuous to the discrete

LemmaRegardless of q, we have

maxp

pAq = max1≤i≤m

e′iAq

Here e′i is the probability vector represents the pure strategy ofgoing only with choice i .

The idea is that a weighted average of things is no bigger than thelargest of them. (Think about grades).

From the continuous to the discrete

LemmaRegardless of q, we have

maxp

pAq = max1≤i≤m

e′iAq

Here e′i is the probability vector represents the pure strategy ofgoing only with choice i .

The idea is that a weighted average of things is no bigger than thelargest of them. (Think about grades).

Proof of the lemma

Proof.We must have

maxp

pAq ≥ max1≤i≤m

e′iAq

(the maximum over a larger set must be at least as big). On theother hand, let q be C ’s strategy. Let the quantity on the right bemaximized when i = i0. Let p be any strategy for R. Notice thatp =

∑i pie

′i . So

E (p, q) = pAq =m∑

i=1

pie′iAq ≤

m∑i=1

pie′i0Aq

=

(m∑

i=1

pi

)e′i0Aq = e′i0Aq.

Thusmax

ppAq ≤ e′i0Aq.

The next step is to introduce a new variable v representing thevalue of this inner maximization. Our objective is to minimize it.Saying it’s the maximum of all payoffs from pure strategies is thesame as saying

v ≥ e′iAq

for all i . So we finally have something that looks like an LPproblem! We want to choose q and v which minimize v subject tothe constraints

v ≥ e′iAq i = 1, 2, . . . m

qj ≥ 0 j = 1, 2, . . . nn∑

j=1

qj = 1

Trouble with this formulation

I Simplex method with equalities?

I Not in standard form

Resolution:

I We may assume all aij ≥ 0, so v > 0

I Let xj =qj

v

Since we know v > 0, we still have x ≥ 0. Now

n∑j=1

xj =1

v

n∑j=1

qj =1

v.

So our problem is now to choose x ≥ 0 which maximizes∑

j xj .The constraints now take the form

v ≥ e′iAq ⇐⇒ 1 ≥ e′iAx,

for all i . Another way to write this is

Ax ≤ 1,

where 1 is the vector consisting of all ones.

Upshot

TheoremConsider a game with payoff matrix A, where each entry of A is

positive. The column player’s optimal strategy q isx

x1 + · · ·+ xn,

where x ≥ 0 satisfies the LP problem of maximizing x1 + · · ·+ xn

subject to the constraints Ax ≤ 1.

Rock/Paper Scissors

The payoff matrix is

A =

0 −1 11 0 −1−1 1 0

.

We can add 2 to everything to make

A =

2 1 33 2 11 3 2

.

Rock/Paper Scissors

The payoff matrix is

A =

0 −1 11 0 −1−1 1 0

.

We can add 2 to everything to make

A =

2 1 33 2 11 3 2

.

Convert to LP

The problem is to maximize x1 + x2 + x3 subject to the constraints

2x1 + x2 + 3x3 ≤ 1

3x1 + 2x2 + x3 ≤ 1

x1 + 3x3 + 2x3 ≤ 1.

We introduce slack variables y1, y2, and y3, so the constraints nowbecome

2x1 + x2 + 3x3 + y1 = 1

3x1 + 2x2 + x3 + y2 = 1

x1 + 3x3 + 2x3 + y3 = 1.

An easy initial basic solution is to let x = 0 and y = 1. The initialtableau is therefore

x1 x2 x3 y1 y2 y3 z valuey1 2 1 3 1 0 0 0 1y2 3 2 1 0 1 0 0 1y3 1 3 2 0 0 1 0 1z −1 −1 −1 0 0 0 1 0

Which should be the entering variable? The coefficients in thebottom row are all the same, so let’s just pick one, x1. To find thedeparting variable, we look at the ratios 1

2 , 13 , and 1

1 . So y2 is thedeparting variable.We scale row 2 by 1

3 :

x1 x2 x3 y1 y2 y3 z valuey1 2 1 3 1 0 0 0 1y2 1 2/3 1/3 0 1/3 0 0 1/3

y3 1 3 2 0 0 1 0 1z −1 −1 −1 0 0 0 1 0

Then we use row operations to zero out the rest of column one:

x1 x2 x3 y1 y2 y3 z valuey1 0 −1/3 7/3 1 −2/3 0 0 1/3

x1 1 2/3 1/3 0 1/3 0 0 1/3

y3 0 7/3 5/3 0 −1/3 1 0 2/3

z 0 −1/3 −2/3 0 1/3 0 1 1/3

We can still improve this: x3 is the entering variable and y1 is thedeparting variable. The new tableau is

x1 x2 x3 y1 y2 y3 z valuex3 0 −1/7 1 3/7 −2/7 0 0 1/7

x1 1 5/7 0 −1/7 3/7 0 0 2/7

y3 0 18/7 0 −5/7 1/7 1 0 3/7

z 0 −3/7 0 2/7 1/7 0 1 3/7

Finally, entering x2 and departing y3 gives

x1 x2 x3 y1 y2 y3 z valuex3 0 0 1 7/18 −5/18 1/18 0 1/6

x1 1 0 0 1/18 7/18 −5/18 0 1/6

x2 0 1 0 −5/18 1/18 7/18 0 1/6

z 0 0 0 1/6 1/6 1/6 1 1/2

So the x variables have values x1 = 1/6, x2 = 1/6, x3 = 1/6.Furthermore z = x1 + x2 + x3 = 1/2, so v = 1/z = 2. This alsomeans that p1 = 1/3, p2 = 1/3, and p3 = 1/3. So the optimalstrategy is to do each thing the same number of times.

Outline

RecapDefinitionsExamplesFundamental TheoremGames we can solve so far

GT problems as LP problemsFrom the continuous to the discreteStandardizationRock/Paper/Scissors again

The row player’s LP problem

Now let’s think about the problem from the column player’sperspective. If he chooses strategy p, and C knew it, he wouldchoose p to minimize the payoff pAq. Thus the row player wantsto maximize that quantity. That is, R’s objective is realized whenthe payoff is

E = maxp

minq

pAq.

LemmaRegardless of p, we have

minq

pAq = min1≤j≤n

pAej

The next step is to introduce a new variable v representing thevalue of this inner minimization. Our objective is to maximize it.Saying it’s the minimum of all payoffs from pure strategies is thesame as saying

v ≤ pAej

for all j . Again, we have something that looks like an LP problem!We want to choose p and v which maximize v subject to theconstraints

v ≤ pAej j = 1, 2, . . . n

pi ≥ 0 i = 1, 2, . . . mm∑

i=1

pi = 1

As before, we can standardize this by renaming

y =1

vp′

(this makes y a column vector). Then

m∑i=1

yi =1

v,

So maximizing v is the same as minimizing 1′y. Likewise, theequations of constraint become v ≤ (vy′)Aej for all j , or y′A ≥ 1′,or (taking transposes) A′y ≥ 1. If all the entries of A are positive,we may assume that v is positive, so the constraints p ≥ 0 aresatisfied if and only if y ≥ 0.

Upshot

TheoremConsider a game with payoff matrix A, where each entry of A is

positive. The row player’s optimal strategy p isy′

y1 + · · ·+ yn,

where y ≥ 0 satisfies the LP problem of minimizingy1 + · · ·+ yn = 1′y subject to the constraints A′y ≥ 1.

The big idea

The big observation is this:

TheoremThe row player’s LP problem is the dual of the column player’s LPproblem.

The final tableau in the Rock/Paper/Scissors LP problem was this:

x1 x2 x3 y1 y2 y3 z valuex3 0 0 1 7/18 −5/18 1/18 0 1/6

x1 1 0 0 1/18 7/18 −5/18 0 1/6

x2 0 1 0 −5/18 1/18 7/18 0 1/6

z 0 0 0 1/6 1/6 1/6 1 1/2

The entries in the objective row below the slack variables are thesolutions to the dual problem! In this case, we have the samevalues, which means R has the same strategy as C . This reflectsthe symmetry of the original game.

Example

Consider the game: players R and C each choose a number 1, 2,or 3. If they choose the same thing, C pays R that amount. Ifthey choose differently, R pays C the amount that C has chosen.What should each do?

Answer.

Choice R C

1 54.5% 22.7%

2 27.3% 36.3%

3 18.2% 40.1%

The expected payoff is 2.71 to the column player.

Example

Consider the game: players R and C each choose a number 1, 2,or 3. If they choose the same thing, C pays R that amount. Ifthey choose differently, R pays C the amount that C has chosen.What should each do?

Answer.

Choice R C

1 54.5% 22.7%

2 27.3% 36.3%

3 18.2% 40.1%

The expected payoff is 2.71 to the column player.