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Should a football team run or pass?
A game theory approach
Laura Albert McLay
Badger 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 1
1 −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.,
/ = [1	0	0	 … 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	/ 	 ):
max
2,3
	1
subject to 1 ≤ / 	 , = 1,2, … ,
∑ = 1
≥ 0, = 1,2, … ,
1 free
Reduction to a linear program
Matrix-vector notation
max	1
1/ − ≤ 0
/ = 1
≥ 0
/ is the vector of all 1’s
Block matrix form
max
0
1 1
− /
/ 0 1
≤
=
0
1
≥ 0
1 free
Now do the same from R’s perspective
Everything is analogous to what we did before!
• R solves this problem:
min	
*
max
-
	 	
• Lemma: max
2
	 	 = 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
min	4
4/ − ≥ 0
/ = 1
≥ 0
/ is the vector of all 1’s
Block matrix form
min
0
1 4
− /
/ 0 4
≥
=
0
1
≥ 0
4 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:
max
2
∗
	 	 = min
*
	 ∗
Proof:
From strong duality, we have 4∗ = 1∗. Also
1∗ = min / 	 ∗ = min
*
	 ∗ from C’s problem
4∗ = max
2
∗ 	/ 	 = max
2
∗ 	 	 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:
max 1	
subject to
1 ≤ −5	 + 10	
1 ≤ 5	
+ = 1
, ≥ 0
Case 1: Look at the offense
We know that = 1 − , which simplifies the
formulation to:
max 1	
subject 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
Run	defense
10 − 15	
Pass	defense
5
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:
min 4	
subject 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
Run	offense
5 − 10
Pass	offense
10
, 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
min 4	
subject to
4 ≥ (I − J)	 + (I + J)	
4 ≥ (K + J)	 + (K − J)	
+ = 1
, ≥ 0
max 1	
subject 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
Run	defense
K + J + (I − K − ( + 1)J)	
Pass	defense
K − 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 = 1
o Offense runs pass and run plays equally
• For > 1
o Offense runs more since the defensive call has more of an
effect on passing plays
• For < 1
o Offense passes more since the defensive call has less of an
effect on passing plays
Related blog posts
• Happiness is assuming the world is linear
• Why the Patriots’ decision to let the
Giants score a touchdown makes sense
• Introducing Badger Bracketology 1.0
• Some thoughts on the College Football
Playoff

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

  • 1. Should a football team run or pass? A game theory approach Laura Albert McLay Badger Bracketology @lauramclay @badgerbrackets http://bracketology.engr.wisc.edu/ © 2015
  • 2. 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?
  • 4. Definitions • Players • Actions • Information • Strategies • Payoffs • Equilibria Eventually we’ll relate this to linear programming!
  • 5. 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
  • 6. 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?
  • 8. Rock-Paper-Scissors Payoff matrix? = 0 1 −1 −1 0 1 1 −1 0 Why is this zero sum? Strategies: • Deterministic: pure strategies • Random/stochastic: mixed strategies
  • 9. 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
  • 10. 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.
  • 11. Solution Game theory to the rescue!
  • 12. 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
  • 13. 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
  • 14. 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!
  • 15. 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., / = [1 0 0 … 0]) Idea: a weighted average of things is no bigger than the largest of them.
  • 16. Put it together We now have: max - min / subject to ∑ = 1 ≥ 0, = 1,2, … , This is a linear program!!
  • 17. Reduction to a linear program • Now introduce a scalar 1 representing the value of the inner minimization (min / ): max 2,3 1 subject to 1 ≤ / , = 1,2, … , ∑ = 1 ≥ 0, = 1,2, … , 1 free
  • 18. Reduction to a linear program Matrix-vector notation max 1 1/ − ≤ 0 / = 1 ≥ 0 / is the vector of all 1’s Block matrix form max 0 1 1 − / / 0 1 ≤ = 0 1 ≥ 0 1 free
  • 19. Now do the same from R’s perspective Everything is analogous to what we did before! • R solves this problem: min * max - • Lemma: max 2 = 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 / ):
  • 20. Reduction to a linear program Matrix-vector notation min 4 4/ − ≥ 0 / = 1 ≥ 0 / is the vector of all 1’s Block matrix form min 0 1 4 − / / 0 4 ≥ = 0 1 ≥ 0 4 free
  • 21. OK, so now we have two ways to solve the same problem Let’s examine how these solutions are related.
  • 22. Minimax Theorem • Let ∗ denote C’s solution to the max-min problem • Let ∗ denote R’s solution to the min-max problem • Then: max 2 ∗ = min * ∗ Proof: From strong duality, we have 4∗ = 1∗. Also 1∗ = min / ∗ = min * ∗ from C’s problem 4∗ = max 2 ∗ / = max 2 ∗ from R’s problem
  • 24. Let’s work on an example Example from Mathletics by Wayne Winston (2009), Princeton University Press, Princeton, NJ.
  • 25. 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)
  • 26. 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: max 1 subject to 1 ≤ −5 + 10 1 ≤ 5 + = 1 , ≥ 0
  • 27. Case 1: Look at the offense We know that = 1 − , which simplifies the formulation to: max 1 subject to 1 ≤ −5 + 10 (1 − ) = 10 − 15 1 ≤ 5 , ≥ 0 Let’s solve the problem visually.
  • 28. 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 Run defense 10 − 15 Pass defense 5
  • 29. 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: min 4 subject to 4 ≥ −5 + 5 = 5 − 10 4 ≥ 10 + = 1 , ≥ 0
  • 30. 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 Run offense 5 − 10 Pass offense 10 , proportion of time defense prepares for run
  • 31. 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)
  • 32. Football example #2: offense vs. defense Offense problem Defense problem min 4 subject to 4 ≥ (I − J) + (I + J) 4 ≥ (K + J) + (K − J) + = 1 , ≥ 0 max 1 subject to 1 ≤ (I − J) + (K + J) 1 ≤ (I + J) + (K − J) + = 1 , ≥ 0
  • 33. 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 Run defense K + J + (I − K − ( + 1)J) Pass defense K − J + (I − K + ( + 1)J) K + J K + J
  • 34. Intuition The correct choice on defense has times more effect on passing as it does on running • For = 1 o Offense runs pass and run plays equally • For > 1 o Offense runs more since the defensive call has more of an effect on passing plays • For < 1 o Offense passes more since the defensive call has less of an effect on passing plays
  • 35. Related blog posts • Happiness is assuming the world is linear • Why the Patriots’ decision to let the Giants score a touchdown makes sense • Introducing Badger Bracketology 1.0 • Some thoughts on the College Football Playoff