Successfully reported this slideshow.
×
1 of 35

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

3

Share

Laura Albert McLay's slides from Oberlin College (2015)

See all

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

1. 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. 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?
3. 3. The solution: Linear programming!
4. 4. Definitions • Players • Actions • Information • Strategies • Payoffs • Equilibria Eventually we’ll relate this to linear programming!
5. 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. 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?
7. 7. Rock-Paper-Scissors • Payoff matrix? =
8. 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. 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. 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. 11. Solution Game theory to the rescue!
12. 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. 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. 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. 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. 16. Put it together We now have: max - min / subject to ∑ = 1 ≥ 0, = 1,2, … , This is a linear program!!
17. 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. 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. 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. 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. 21. OK, so now we have two ways to solve the same problem Let’s examine how these solutions are related.
22. 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
23. 23. We did it!
24. 24. Let’s work on an example Example from Mathletics by Wayne Winston (2009), Princeton University Press, Princeton, NJ.
25. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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