Should a football team run or pass?

A game theory approach

Laura Albert McLayBadger Bracketology

@lauramclay@badgerbrackets

http://bracketology.engr.wisc.edu/

© 2015

The problem

• An offense can run or pass the ball

• The defense anticipates the offense’s choice and

chooses a run or pass offense.

• Given this strategic interaction,

o what is the best mix of pass and run plays for the offense?

o what is the best mix of pass and run defenses?

The solution:

Linear programming!

Definitions

• Players

• Actions

• Information

• Strategies

• Payoffs

• Equilibria

Eventually we’ll relate this to linear programming!

Definitions

• Players: we have 2

• Actions: discrete actions of actions available to each player and when they are available (order of play)

• Information: what each player knows about variables at each point in time

• Strategies: a rule that tells each player which action to choose at each decision point

• Payoffs: the expected utility/reward each player receives as a function of every players’ decisions

• Equilibria: strategy profiles consisting of best strategies for each of the players in the game

Payoff matrix

• A two person game is between a row player R and a

column player C

• A zero-sum game is defined by a � × �payoff matrix

� where ��� is the payoff to C if C chooses action and R chooses action o R chooses from the rows ∈ {1,… ,�}o C chooses from the columns ∈ {1,… , �}o Note: deterministic strategies can be bad!

• Zero-sum: my gain is your loss. Examples?

Rock-Paper-Scissors

• Payoff matrix?

� =� � �

���

Rock-Paper-Scissors

Payoff matrix?

� =� � �

���

0 1 −1−1 0 11 −1 0

Why is this zero sum?

Strategies:

• Deterministic: pure strategies

• Random/stochastic: mixed strategies

Strategies

Payoff matrix �• A strategy for a player is a probability vector

representing the portion of time each action is used

o R chooses with probability ��� = ��, ��, … , �� �

o C chooses with probability ��� = ��, �� , … , �� �

o We have: �� ≥ 0, = 1,… , �∑ ��� = 1

Payoffs

Expected payoff from R to C:

�, � =!!���������

= ����

Note:

• � and � are our variables

Problem:

• We want to solve this as a linear program but �, � is a

quadratic function with two players with opposing goals.

Solution

Game theory to the rescue!

Theorem

Expected payoff from R to C:

�, � =!!���������

= ����

Theorem:

There exist optimal strategies �∗ and �∗such that for all strategies � and �:

�, �∗ ≤ �∗, �∗ ≤ [�∗, �]

Note we call �∗, �∗ the value of the

game.

Hipster mathematician

Reflect on the inequality

�, �∗ ≤ �∗, �∗ ≤ [�∗, �]

• �∗, �∗ ≤ �∗, � : C guarantees a lower bound

(worst−case) on his/her payoff

• �, �∗ ≤ �∗, �∗ : R guarantees an upper bound

(worst-case) on how much he/she loses

• Fundamental problem: finding �∗ and �∗

Both R and C play

optimal strategies

C plays optimal,

R plays suboptimal

R plays optimal,

C plays suboptimal

Objective function analysis

• Suppose C adopts strategy �• Then, R’s best strategy is to find the � that minimizes

����:min* ����

• And therefore, C should choose the � that maximizes

these possibilities:

max- min* ����This will give us �∗ and �∗.This is hard!

Useful result

• Let’s focus on the inner optimization problem: min* ����o This is easy since it treats � as “fixed” so we have a linear

problem.

Lemma: min* ���� = min� /����where /�� is the pure vector of only selecting action (e.g., /�� = [100 …0])

Idea: a weighted average of things is no bigger than the largest of them.

Put it together

We now have:

max- min� /����subject to ∑ ��� = 1

�� ≥ 0, = 1,2, … , �

This is a linear program!!

Reduction to a linear program

• Now introduce a scalar 1 representing the value of

the inner minimization (min� /����):

max2,3 1subject to 1 ≤ /����, = 1,2, … ,�

∑ ��� = 1�� ≥ 0, = 1,2, … , �1 free

Reduction to a linear program

Matrix-vector notation

max11/ − �� ≤ 0/�� = 1� ≥ 0

/ is the vector of all 1’s

Block matrix form

max 01

� �1

−� //� 0

�1≤=

01

� ≥ 01 free

Now do the same from R’s perspective

Everything is analogous to what we did before!

• R solves this problem:

min* max- ����• Lemma: max2 ���� = max� ���/�• That gives us the following linear program:

min* max� ���/�subject to ∑ ��� = 1

�� ≥ 0, = 1,2,… ,�• Introduce a scalar 4 representing the value of the inner

maximization (max� ���/� ):

Reduction to a linear program

Matrix-vector notation

min44/ − ��� ≥ 0/�� = 1� ≥ 0

/ is the vector of all 1’s

Block matrix form

min 01

� �4

−�� //� 0

�4≥=

01

� ≥ 04 free

OK, so now we have two ways to solve

the same problem

Let’s examine how these solutions are related.

Minimax Theorem

• Let �∗ denote C’s solution to the max-min problem

• Let �∗ denote R’s solution to the min-max problem

• Then:

max2 �∗��� = min* ����∗

Proof:

From strong duality, we have 4∗ = 1∗. Also

1∗ = min� /����∗ = min* ����∗ from C’s problem

4∗ = max2 �∗��/� = max2 �∗��� from R’s problem

We did it!

Let’s work on an example

Example from Mathletics by Wayne Winston (2009), Princeton University Press, Princeton, NJ.

Football example: offense vs. defense

5(7, 8) Offense runs (7:) Offense passes (;< = 1 − �:)

Run defense (��) -5 10

Pass defense (�� = 1 − ��) 5 0

The offense wants the most yards. The defense wants the offense to

have the fewest yards. This is a zero sum game.

Using this information, answer the following two questions:

(1) What fraction of time should the offense run the ball?

(2) If they adopt this strategy, how many yards will they achieve per

play on average?

Idealized payoffs (yards)

Case 1: Look at the offense

• The offense chooses a mixed strategy

o Run with probability ��o Pass with probability �� = 1 − ��

• Solve the linear program:

max1subject to

1 ≤ −5�� + 10��1 ≤ 5��

�� + �� = 1��, �� ≥ 0

Case 1: Look at the offense

We know that �� = 1 − ��, which simplifies the

formulation to:

max1subject to

1 ≤ −5�� + 10(1 − ��) = 10 − 15��1 ≤ 5����, �� ≥ 0

Let’s solve the problem visually.

Case 1: Look at the offense

We want the largest value of 1 that is “under” both lines.

This happens when �� = 1/2 (and �� = 1/2): run half the time,

pass half the time.

1∗ = min� /����∗ = 2.5 yards per play, on average.

Expected payoff

��, proportion of time offense runs the ball

Rundefense10 − 15��

Passdefense5��

Case 2: Look at the defense

• We still do not know the optimal defensive strategy.

• The defense chooses a mixed strategy

o Run defense with probability ��o Pass defense with probability �� = 1 − ��

• Solve the linear program:

min4subject to

4 ≥ −5�� + 5�� = 5 − 10��4 ≥ 10���� + �� = 1��, �� ≥ 0

Case 2: Look at the defense

We want the smallest value of 4 that is “over” both lines.

This happens when �� = 1/4 (and �� =3/4): prepare for run a quarter of the

time, prepare for a pass three quarters of the time.

This yields 4∗ = 2.5 yards per attempt (on average). The offense gain and

defensive loss are always identical!

Expected payoff

Runoffense5 − 10��

Passoffense10��

��, proportion of time defense prepares for run

Football example #2:

offense vs. defense

5(7, 8) Offense runs (7:) Offense passes (;< = 1 − �:)

Run defense (��) I − J K +�J

Pass defense (�� = 1 − ��) I + J K −�J

Suppose the defense chooses run and pass defenses with equal

likelihoods.

The offense would gain r yards per run, on average.

The offense would gain p yards per pass, on average.

The correct choice on defense has m times more effect on passing

as it does on running (range of 2�J vs. 2J)

Idealized payoffs (yards)

Football example #2:

offense vs. defense

Offense problem Defense problem

min4subject to

4 ≥ (I − J)�� + (I + J)��4 ≥ (K +�J)�� + (K −�J)��

�� + �� = 1��, �� ≥ 0

max 1subject to

1 ≤ (I − J)�� + (K +�J)��1 ≤ (I + J)�� + (K −�J)��

�� + �� = 1��, �� ≥ 0

Football example #2:

offense problem

After a lot of algebra…

�� = �/(� + 1)[Does not depend on I or K!]

Likewise, �� = 1/2 +(I − K)/(2J� +�)for the defense

Expected

payoff

��, proportion of time offense runs the ball

RundefenseK + �J + (I − K − (� + 1)J)��

PassdefenseK − �J + (I − K + (� + 1)J)��

K + �J

K +�J

Intuition

The correct choice on defense has � times more effect on passing as it does on running

• For � = 1o Offense runs pass and run plays equally

• For � > 1o Offense runs more since the defensive call has more of an

effect on passing plays

• For � < 1o Offense passes more since the defensive call has less of an

effect on passing plays

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