Should an NFL team go for a one or two point conversion?two point conversion?

A dynamic programming approachLaura A. McLay

(c) 2012

Based on Mathletics by Wayne Winston

The problemThe problem

• When an NFL team scores a touchdown theyWhen 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

f 2 i i– try for a 2 point conversion

• The decision depends on the score and the remaining time left

The assumptionsThe 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).h b b l f f h d2. The probability of success of each outcome does not depend on the score or the time left.

3. The teams are equally matched.q y4. The success rate for two point conversions is 42% and 

the success rate for extra points is 100%.

Our approachOur 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 i i l t f i f ti d d t– The minimal amount of information needed to make our decision

Dynamic programmingDynamic programming

Letet• Fn(p) = the probability that the team wins the game if they are p points ahead, they have just g y p p y jgotten 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 nhas just gotten the ball, and there are n remaining possessions.

• n=0 means the game is overn=0 means the game is over.

Boundary conditionsBoundary conditions

• G0(p) = F0(p) = 1 for p > 0G0(p) = F0(p) = 1 for p > 0

• G0(p) = F0(p) = 0 for p < 0

G (0) (0) 0 [ i ]• G0(0) = F0(0) = 0.5 [a tie]

• Goal: maximizing Fn(p) for some n and p– Find by conditioning.d by co d o g

Law of total probabilityLaw 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).[ ] ( | ) ( )

y YP X P X y P y

y Y

Finding an expected valueFinding 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 thatThe expected value of I is the probability that the team wins

E[I] = P(I > 0)E[I]   P(I > 0)

• P(I > 0) = F (p)• P(I > 0)  = Fn(p)– This can be found by computing the expected value of an indicator variable E[I]value of an indicator variable E[I]

Law of total probabilityLaw of total probability• PTD = probability of a touchdown• PFG = probability of a field goal• P = probability of no score• PNS = probability of no score• PAT = probability of scoring a point after a touchdown• P2PT = probability of a two point conversion after a touchdown

Three outcomes of the current drive: TD, FG, or NS( PTD + PFG + PNS = 1)

Fn+1(p) = + PNSGn(p) No scoreNS n(p)+ PFGGn(p+3)  Field Goal+ PTD max{ PATGn(p+7) + (1‐PAT) Gn(p+6),      TD with AT

P G (p+8) + (1 P ) G (p+6) } TD with 2PTP2PTGn(p+8) + (1‐P2PT) Gn(p+6) }  TD with 2PT

Series of equationsSeries of equations• Fn+1(p) = PTD max{PATGn(p+7) + (1‐PAT) Gn(p+6), P G (p+8) + (1‐P ) G (p+6)} + P G (p+3) +P2PTGn(p+8) + (1‐P2PT) Gn(p+6)} + PFGGn(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) + 2PT n(p ) ( 2PT) n(p )} FG n(p )PNSFn(p)

• These are interdependent and can be solved iteratively

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 0 0.190 0.134 0.268 0.199 0.302 0.232 0.316 0.249 0.318 0.2565 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 0 660 0 500 0 620 0 493 0 584 0 477 0 547 0 454 0 510 0 4290 0.5 0.660 0.500 0.620 0.493 0.584 0.477 0.547 0.454 0.510 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 0 920 0 920 0 829 0 837 0 749 0 761 0 680 0 693 0 6199 1 1.000 0.920 0.920 0.829 0.837 0.749 0.761 0.680 0.693 0.619

10 1 1.000 0.920 0.920 0.840 0.843 0.765 0.770 0.696 0.703 0.634

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 0 0.000 0.134 0.095 0.199 0.148 0.233 0.179 0.251 0.197 0.2585 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 0 340 0 503 0 380 0 497 0 393 0 480 0 389 0 457 0 377 0 4310 0.5 0.340 0.503 0.380 0.497 0.393 0.480 0.389 0.457 0.377 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 0 822 0 833 0 738 0 754 0 667 0 684 0 606 0 6229 1 0.920 0.920 0.822 0.833 0.738 0.754 0.667 0.684 0.606 0.622

10 1 0.920 0.920 0.837 0.841 0.759 0.766 0.689 0.698 0.626 0.635

How do we determine whether to go ffor two points?

Test the condition in the max statement to seeTest the condition in the max statement to see what was selected:

For Fn+1(p):For Fn+1(p): If  ( P2PTGn(p+8) + (1‐P2PT) Gn(p+6) ) > ( P G (p+7) + (1‐P ) G (p+6) )( PATGn(p+7) + (1‐PAT) Gn(p+6) )Then go for two points. 

G f i t th iGo for one point otherwise.A similar approach works for Gn+1(p)

What should the team do if they score hda touchdown?

Decisions Possessions remaining nP 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 1 2 2 2 2 2 2 2 2 25 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 10 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 19 1 1 1 1 1 1 1 1 1 1

10 1 1 1 1 1 1 2 1 2 2

What should the opposing team do if h hdthey score a touchdown?

G_n(p) Possessions remaining nP 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 1 1 1 1 1 1 1 1 1 15 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 10 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 19 1 1 1 1 1 1 1 1 1 1

10 1 1 1 1 1 1 1 1 1 1