should a football team run or pass? a linear programming approach to game theory
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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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