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# Should a football team go for a one or two point conversion after scoring a touchdown?

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### Should a football team go for a one or two point conversion after scoring a touchdown?

1. 1. Should an NFL team go for a one or  two point conversion? two point conversion?A dynamic programming approach Laura A. McLay (c) 2012 Based on Mathletics by Wayne Winston
2. 2. The problem The problem• When an NFL team scores a touchdown they When an NFL team scores a touchdown, they  score six points• They have two choices: They have two choices: – kick a point after to score 1 extra point – try for a 2 point conversion f 2 i i• The decision depends on the score and the  remaining time left
3. 3. The assumptions The assumptions• We make several assumptions:1. There are three possible outcomes for each team’s  possession:  1. touchdown (19% of the time),  1 touchdown (19% of the time) 2. field goal (13% of the time), 3. no score (68% of the time).2. The probability of success of each outcome does not  h b bl f f h d depend on the score or the time left. q y3. The teams are equally matched.4. The success rate for two point conversions is 42% and  the success rate for extra points is 100%.
4. 4. Our approach Our approach• This decision depends on two factors: This decision depends on two factors: – The point differential – The remaining number of possessions (assumed The remaining number of possessions (assumed  to be known)• The state of the system depends on these two The state of the system depends on these two  factors – Th The minimal amount of information needed to  i i l t fi f ti d dt make our decision
5. 5. Dynamic programming Dynamic programmingLet et• Fn(p) = the probability that the team wins the  g game if they are p points ahead, they have just  y pp y j gotten the ball, and there are n remaining  possessions.• Gn(p) = the probability that the team wins the  game if they are p points ahead, their opponent  has just gotten the ball, and there are n  has just gotten the ball and there are n remaining possessions.• n=0 means the game is over n=0 means the game is over.
6. 6. Boundary conditions Boundary conditions• G0(p) = F0(p) = 1 for p > 0 (p) = F (p) = 1 for p > 0• G0(p) = F0(p) = 0 for p < 0• G0(0) 0(0) 0 [ i ] (0) = F (0) = 0.5 [a tie]• Goal: maximizing Fn(p) for some n and p – Find by conditioning. d by co d o g
7. 7. Law of total probability Law of total probability• Let X denote a random variable. – X is a binary variable denoting the outcome of  the game• We can find the probability of X given some  additional information Y – Y is a random variable denoting the outcome of  the current possession (touchdown, field goal, or  no score).) P[ X ]   P ( X | y ) P( y ) yY
8. 8. Finding an expected value Finding an expected value• Let I be an indicator variable if the team wins: Let I be an indicator variable if the team wins: – I = 1 if they win and I = 0 if they lose.• The expected value of I is the probability that The expected value of I is the probability that  the team wins E[I]  P(I > 0) E[I] = P(I > 0)• P(I > 0) = Fn(p) > 0)  = F – This can be found by computing the expected  value of an indicator variable  value of an indicator variable E[I]
9. 9. Law of total probability Law of total probability• PTD = probability of a touchdown• PFG = probability of a field goal• PNS = probability of no score = probability of no score• PAT = probability of scoring a point after a touchdown• P2PT = probability of a two point conversion after a touchdownThree outcomes of the current drive: TD, FG, or NS( PTD + PFG + PNS = 1)Fn+1(p) =  + PNSGn(p)  No score + PFGGn(p+3)   Field Goal + PTD max{ PATGn(p+7) + (1‐PAT) Gn(p+6),      TD with AT P2PTGn(p+8) + (1 P2PT) Gn(p+6) }  TD with 2PT (p+8) + (1‐P ) G (p+6) }  TD with 2PT
10. 10. Series of equations Series of equations• Fn+1(p) = PTD max{PATGn(p+7) + (1‐PAT) Gn(p+6),  P2PTGn(p+8) + (1‐P2PT) Gn(p+6)} + PFGGn(p+3) + (p+8) + (1‐P ) G (p+6)} + P (p+3) +  PNSGn(p)• Gn+1(p) = PTD min{PATFn(p‐7) + (1‐PAT) Fn(p‐6),  P2PTFn(p‐8) + (1‐P2PT) Fn(p‐6)} + PFGFn(p‐3) +  (p ) ( ) (p )} (p ) PNSFn(p)• These are interdependent and can be solved  iteratively
11. 11. What do the results look like? What do the results look like?F_n(p) Possessions remaining n P 0 1 2 3 4 5 6 7 8 9 10 ‐10 0 0.000 0.000 0.046 0.034 0.093 0.071 0.129 0.100 0.155 0.122 ‐9 0 0.000 0.000 0.063 0.046 0.119 0.090 0.158 0.122 0.183 0.143 ‐8 0 0.040 0.027 0.103 0.075 0.154 0.117 0.189 0.146 0.210 0.166 ‐7 0 0.095 0.065 0.158 0.116 0.201 0.154 0.229 0.180 0.246 0.196 ‐6 0 0.190 0.129 0.255 0.184 0.280 0.212 0.291 0.227 0.294 0.234 ‐5 5 0 0.190 0.134 0.268 0.199 0.302 0.232 0.316 0.249 0.318 0.256 ‐4 0 0.190 0.142 0.280 0.214 0.322 0.253 0.340 0.273 0.344 0.281 ‐3 0 0.255 0.198 0.361 0.285 0.399 0.320 0.406 0.330 0.399 0.328 ‐2 0 0.320 0.242 0.425 0.331 0.453 0.361 0.449 0.364 0.432 0.356 ‐1 0 0.320 0.250 0.434 0.345 0.468 0.378 0.466 0.382 0.451 0.375 0 0.5 05 0.660 0 660 0.500 0 500 0.620 0 620 0.493 0 493 0.584 0 584 0.477 0 477 0.547 0 547 0.454 0 454 0.510 0 510 0.429 0 429 1 1 1.000 0.743 0.790 0.621 0.677 0.550 0.603 0.501 0.548 0.462 2 1 1.000 0.743 0.790 0.624 0.679 0.557 0.609 0.510 0.557 0.472 3 1 1.000 0.787 0.829 0.677 0.727 0.608 0.655 0.557 0.597 0.513 4 1 1.000 0.843 0.869 0.738 0.771 0.660 0.693 0.596 0.627 0.543 5 1 1.000 0.845 0.869 0.741 0.772 0.664 0.695 0.602 0.631 0.549 6 1 1.000 0.845 0.895 0.765 0.803 0.694 0.725 0.632 0.658 0.577 7 1 1.000 0.910 0.920 0.821 0.832 0.737 0.751 0.665 0.680 0.602 8 1 1.000 0.920 0.920 0.826 0.832 0.741 0.751 0.668 0.681 0.606 9 1 1.000 1 000 0.920 0 920 0.920 0 920 0.829 0 829 0.837 0 837 0.749 0 749 0.761 0 761 0.680 0 680 0.693 0 693 0.619 0 619 10 1 1.000 0.920 0.920 0.840 0.843 0.765 0.770 0.696 0.703 0.634
12. 12. What do the results look like? What do the results look like?G_n(p) Possessions remaining n P 0 1 2 3 4 5 6 7 8 9 10 ‐10 0 0.000 0.000 0.000 0.034 0.026 0.071 0.054 0.101 0.078 0.123 ‐9 0 0.000 0.000 0.000 0.046 0.034 0.090 0.068 0.123 0.094 0.144 ‐8 0 0.000 0.027 0.018 0.075 0.055 0.117 0.089 0.147 0.114 0.167 ‐7 0 0.000 0.065 0.044 0.116 0.085 0.155 0.118 0.181 0.142 0.198 ‐6 0 0.000 0.129 0.088 0.184 0.133 0.212 0.160 0.228 0.177 0.235 ‐5 5 0 0.000 0.134 0.095 0.199 0.148 0.233 0.179 0.251 0.197 0.258 ‐4 0 0.000 0.142 0.105 0.216 0.165 0.255 0.200 0.275 0.220 0.283 ‐3 0 0.000 0.198 0.152 0.285 0.223 0.321 0.255 0.332 0.268 0.329 ‐2 0 0.000 0.242 0.182 0.335 0.259 0.364 0.288 0.367 0.296 0.358 ‐1 0 0.000 0.250 0.193 0.349 0.275 0.382 0.307 0.386 0.315 0.378 0 0.5 05 0.340 0 340 0.503 0 503 0.380 0 380 0.497 0 497 0.393 0 393 0.480 0 480 0.389 0 389 0.457 0 457 0.377 0 377 0.431 0 431 1 1 0.680 0.743 0.551 0.622 0.487 0.552 0.447 0.503 0.415 0.464 2 1 0.680 0.743 0.553 0.624 0.493 0.558 0.456 0.512 0.427 0.475 3 1 0.745 0.794 0.622 0.684 0.556 0.614 0.511 0.561 0.474 0.516 4 1 0.810 0.845 0.697 0.741 0.619 0.662 0.560 0.599 0.511 0.545 5 1 0.810 0.845 0.699 0.742 0.623 0.665 0.565 0.603 0.518 0.551 6 1 0.810 0.883 0.732 0.785 0.663 0.705 0.603 0.638 0.552 0.581 7 1 0.905 0.920 0.810 0.826 0.722 0.740 0.648 0.667 0.585 0.604 8 1 0.920 0.920 0.817 0.826 0.727 0.741 0.651 0.668 0.589 0.606 9 1 0.920 0 920 0.920 0 920 0.822 0 822 0.833 0 833 0.738 0 738 0.754 0 754 0.667 0 667 0.684 0 684 0.606 0 606 0.622 0 622 10 1 0.920 0.920 0.837 0.841 0.759 0.766 0.689 0.698 0.626 0.635
13. 13. How do we determine whether to go  for two points?Test the condition in the max statement to see Test the condition in the max statement to see what was selected:For F (p): For Fn+1(p):If  ( P2PTGn(p+8) + (1‐P2PT) Gn(p+6) ) >  ( P ( PATGn(p+7) + (1‐PAT) Gn(p+6) ) (p+7) + (1‐P ) G (p+6) ) Then go for two points. Go for one point otherwise.G f i t th iA similar approach works for Gn+1(p)
14. 14. What should the team do if they score Decisions a touchdown? hd Possessions remaining n P 0 1 2 3 4 5 6 7 8 9 10 ‐10 1 1 1 1 1 1 1 1 1 1 ‐9 1 1 1 1 1 1 1 1 1 1 ‐8 2 2 2 2 2 2 2 2 2 2 ‐7 1 1 1 1 1 1 1 1 1 1 ‐6 1 1 1 1 1 1 1 1 1 1 ‐5 5 1 2 2 2 2 2 2 2 2 2 ‐4 1 1 1 1 1 1 1 1 1 1 ‐3 1 1 1 1 1 1 1 1 1 1 ‐2 1 1 2 2 2 2 2 2 2 2 ‐1 1 2 1 2 1 2 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1 2 1 1 3 1 1 1 1 1 1 1 1 1 1 4 1 1 1 1 2 2 2 2 2 2 5 1 1 1 2 1 2 1 2 1 2 6 1 1 1 1 1 1 1 1 1 1 7 1 1 1 1 1 1 2 1 2 2 8 1 1 1 1 1 2 1 2 1 2 9 1 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 2 1 2 2
15. 15. What should the opposing team do if  they score a touchdown?G_n(p) h hd Possessions remaining n P 0 1 2 3 4 5 6 7 8 9 10 ‐10 1 1 1 1 1 1 1 1 1 1 ‐9 1 1 1 1 1 1 1 1 1 1 ‐8 1 1 1 1 1 1 1 1 1 1 ‐7 1 1 1 1 1 1 1 1 1 1 ‐6 1 1 1 2 2 2 2 2 2 2 ‐5 5 1 1 1 1 1 1 1 1 1 1 ‐4 1 1 1 1 1 1 1 1 1 1 ‐3 1 1 1 1 1 1 1 1 1 1 ‐2 1 1 1 1 1 1 1 1 1 1 ‐1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 4 1 1 1 1 2 2 2 2 2 2 5 1 2 2 2 2 2 2 2 2 2 6 1 1 1 1 1 1 1 1 1 1 7 1 1 1 1 1 1 1 2 2 2 8 2 2 2 2 2 2 2 2 2 2 9 1 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 1