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What is a football team's best mix of running and passing plays?
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What is a football team's best mix of running and passing plays?

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  • 1. Should a football team run or pass? A game theory approach Laura A. McLay (c) 2012 Based on Mathletics by Wayne Winston
  • 2. The problem The problem• An offense can run or pass the ball An offense can run or pass the ball• The defense anticipates the offense’s choice  and chooses a run or pass offense. and chooses a run or pass offense• Given this strategic interaction,  – what is the best mix of pass and run plays for the  offense? – what is the best mix of pass and run defenses?
  • 3. Idealized payoffs (yards) Idealized payoffs (yards) Run defense (x) Pass defense (1‐x) Offense runs (q) ‐5 5 Offense passes (1 q) Offense passes (1‐q) 10 0We consider a zero sum game. The offense wants the most yards. The defense wants the offense to have the fewest yards.A pureA pure strategy is deterministic: the offense or defense makes the same is deterministic: the offense or defense makes the same decision all the timeA mixed strategy is a random strategy that assigns probabilities to the available A mixed strategy is a random strategy that assigns probabilities to the availablechoices.
  • 4. Case 1: Defense chooses a pure  strategy• The offense chooses a mixed strategy The offense chooses a mixed strategy – Run with probability q – Pass with probability 1‐q Pass with probability 1 q• If a run defense is chosen the expected gain is: If a run defense is chosen, the expected gain is: q(‐5) + (1‐q)10 = 10‐15q• If a pass defense is chosen the e pected gain is If a pass defense is chosen, the expected gain is: q(5) + (1‐q) 0 = 5q
  • 5. Case 1: Defense chooses a pure  strategy• For any value of q chosen by the offense the For any value of q chosen by the offense, the  defense wants to minimize the yards: min{ 10‐15q, 5q } min{ 10 15q 5q }• The offense should choose q (0 < q < 1) that  maximizes the min{ 10‐15q, 5q }
  • 6. Case 1: Defense chooses a pure  strategyExpected payoff q The offense should run half the time, gaining 2.5  yards per attempt (on average). yards per attempt (on average)
  • 7. Case 2: Offense chooses a pure  strategy• The defense chooses a mixed strategy The defense chooses a mixed strategy – Run defense with probability x – Pass defense with probability 1 x Pass defense with probability 1‐x• If th ff If the offense runs, the expected gain is: th t d i i x(‐5) + (1‐x)(5) = 5 – 10x• If the offense passes, the expected gain is: x(10) + (1 x)(0)  10x x(10) + (1‐x)(0) = 10x 
  • 8. Case 2: Offense chooses a pure  strategy• For any value of x chosen by the defense the For any value of x chosen by the defense, the  offense wants to maximize the yards: max{ 5 – 10x, 10x } max{ 5 10x 10x }• The defense should choose x (0 < x < 1) that  minimizes the max{ 5 – 10x, 10x}
  • 9. Case 2: Offense chooses a pure  strategyExpected payoff x The defense should choose a run defense 1/4 of the time,  allowing 2.5 yards per attempt (on average). (The offense gain and defensive loss are always identical)
  • 10. Idealized payoffs (yards) Idealized payoffs (yards) Run defense (x) Pass defense (1‐x) Offense runs (q) r‐k r+k Offense passes (1 q) Offense passes (1‐q) p+mk p mk p‐mkSuppose 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 The correct choice on defense has m times more effect on passing as it does onrunning (range of 2mk vs. 2k)
  • 11. Idealized payoffs, cont d. Idealized payoffs, cont’d. Run defense (x) Pass defense (1‐x) Offense runs (q) r‐k r+k Offense passes (1 q) Offense passes (1‐q) p+mk p mk p‐mkSuppose the defense chooses a pure strategy.If a run defense is chosen, the expected gain is: q(r‐k) + (1‐q)(p+mk) = (p+mk) + (r‐k‐p‐mk)q q(r k) + (1 q)(p+mk) (p+mk) + (r k p mk)qIf a pass defense is chosen, the expected gain is: q(r+k) + (1‐q) (p‐mk) = (p‐mk)+(r+k‐p+mk)q
  • 12. Case 3: Idealized inputs Case 3: Idealized inputsExpected payoff q • q = m/(m+1) [Does not depend on r or p!] • Lik i Likewise, x = 1/2 + (r‐p)/(2km+m) for the defense 1/2 + ( )/(2k + ) f th d f
  • 13. Case 3: intuition Case 3: intuition• For m=1 For m=1 – Offense runs pass and run plays equally• For m>1 For m>1 – Offense runs more since the defensive call has  more of an effect on passing plays more of an effect on passing plays• For m<1 – Offense passes more since the defensive call has  less of an effect on passing plays