Should a football team go for a one or two point conversion? A dynamic programming approach

Should a football team go for a one
or two point conversion?
A dynamic programming approach
Dr. Laura Albert
University of Wisconsin-Madison
laura@engr.wisc.edu
@lauraalbertphd
(c) 2019

The problem
• When an NFL team scores a touchdown, they score six points
• They have two choices:
– kick a point after to score 1 extra point
– try for a 2 point conversion
• When should they try a 2 point conversion? When should
they kick an extra point?
– Solution approach: dynamic programming
– Model output: a series of decisions based on the score differential and
the remaining time/remaining possessions
(c) Laura Albert 2019 2Two point conversion DP

We make several assumptions
1. There are three possible outcomes for each team’s
possession:
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 depend
on the score or the time left.
3. If the game ends in a tie, each team’s probability of winning
is 0.5.
4. We know the number of remaining possessions.
5. The teams are equally matched
Note: we can easily lift this assumption

Our approach
• This decision depends on two factors:
– The point differential
– The remaining number of possessions (assumed to be known)
• The state of the system depends on these two factors
– The minimal amount of information needed to make our decision
• The success rate for two point conversions is 48% and the
success rate for extra points is 96%.

5
Let’s solve this with dynamic
programming (DP)
• We model a sequence of interrelated decisions for two teams.
• Each team seeks to maximize the probability they win the
game.
• DP is a general approach to problem solving, not a particular
model or algorithm.
• DP is smart enumeration. All possible outcomes are not
enumerated, but DP guarantees the optimal solution.
(c) Laura Albert 2019Two point conversion DP

6
Shortest Path Problem
INSTANCE: Directed graph 𝐺𝐺 = (𝑉𝑉, 𝐸𝐸), nonnegative
arc distances 𝑑𝑑𝑖𝑖𝑖𝑖 for each edge, a starting node 𝑠𝑠.
OBJECTIVE: Find the shortest path from node 𝑠𝑠 to
node 𝑡𝑡.
s
p
t

7
Shortest Path Problem, cont’d.
• Observation 1: If the shortest path from s to r passes through node p, the
subpaths (𝑠𝑠, 𝑝𝑝) and (𝑝𝑝. 𝑡𝑡) are shortest paths from 𝑠𝑠 to 𝑝𝑝 and 𝑝𝑝 to 𝑡𝑡,
respectively.
– If not, the shorter subpath would allow us to construct a shorter path from 𝑠𝑠
to 𝑡𝑡, leading to a contradiction.
• Observation 2: Let 𝑉𝑉𝑗𝑗 denote the length of a shortest path from 𝑗𝑗 to 𝑡𝑡,
then
𝑉𝑉𝑗𝑗 = min
𝑖𝑖∈𝑉𝑉𝑖𝑖{𝑗𝑗}
{𝑉𝑉𝑖𝑖 + 𝑑𝑑𝑑𝑑𝑑𝑑}
In other words, the lengths of the shortest paths from every neighbor of 𝑗𝑗
to 𝑡𝑡, then we can find the length of the shortest part from 𝑗𝑗 to 𝑟𝑟.
Note that this is not necessarily an algorithm for finding the shortest path.

8
Principle of Optimality
• Principle of Optimality: Observations 1 and 2 define the
optimal substructure of all DP models and algorithms. Each
solution is characterized by
– A sequence of decisions (more on this later),
– The optimal solution is composed of optimal solutions to
subproblems.
• In the Shortest Path problem, the DP gives the shortest paths
from all nodes to node 𝑡𝑡
– 𝑉𝑉𝑡𝑡 = 0.

9
Shortest Path DP algorithms
• Since DP algorithms are smart enumerations, good
DP algorithms are important to find.
• Let Vj
n = the shortest path from j to r at iteration n.
• We want to find the optimal distances Vj by initially
setting Vj
0 = large number (j ≠ r)
• The idea: iteratively loop over all nodes to find the
best edge to traverse
– Because of the principle of optimality, we only improve
(lower) Vj
n.

10
Shortest Path DP algorithm 1
Bellman’s Algorithm
1. Let Vr
0=0 and Vj
0 = M, j ≠ r (M is a large number) and let
n=1.
2. For all j ∈ V{r}
Vj
n=mini{dji+Vi
n-1}
3. If Vj
n < Vj
n-1 for any i, let n=n+1 and return to step 2. Else
stop.
Note: only consider nodes where an edge from i to j exists.
Problems: In step 2, most of the time we have to consider
edges with distance M. A better algorithm only
considers nodes with paths less than M

11
Characteristics of DP Problems
1. Problem divided into stages with a policy decision required at
each stage.
Shortest path = which node to visit next
2. Each stage has a number of states associated with it.
3. The effect of each policy decision in each state (at each stage)
transforms the current state into a state in the next stage
(deterministic or stochastic).
4. The solution procedure is designed to find the optimal policy.
Prescription for finding the best decision at each stage for each
state (state-action pairs)

12
Characteristics of DP Problems
5. Given current state, an optimal policy for the remaining stages is
independent of the policy in the previous stages.
This is the principle of optimality for dynamic programming
6. The solution procedure begins by finding the optimal policy for the
last stage.
This is usually trivial, such as 𝑉𝑉𝑡𝑡 = 0
7. A recursive relationship identifies the optimal policy for stage 𝑛𝑛,
given an optimal policy for stage 𝑛𝑛 + 1
Called the value functions or optimality equations or Bellman
equations or Hamilton-Jacobi-Bellman equations

Dynamic Programs as shortest or
longest paths
• Almost any DP problem can be formulated as a shortest or longest
path (by maximizing) in a graph with directed arcs (digraph)
• Nodes of the digraph correspond to states in an incomplete
solution
• Arcs of the digraph correspond to decisions. They link each state to
a subsequent state to which the decision leads.
Implication for our football problem:
• The nodes represent the triplet: (1) who has the possession, (2) the
current score differential, (3) how many possessions are left.
• We solve a longest path problem to maximize win probability (not
distance)
13

Dynamic programming
Let
• 𝑉𝑉𝑛𝑛
𝐴𝐴
(𝑝𝑝) = the probability that team A wins the game if they are
𝑝𝑝 points ahead, they have just gotten the ball, and there are 𝑛𝑛
remaining possessions.
• 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝) = the probability that team A wins the game if they are
𝑝𝑝 points ahead, their opponent team B has just gotten the
ball, and there are 𝑛𝑛 remaining possessions.
• 𝑛𝑛 = 0 means the game is over.

Boundary conditions
• 𝑉𝑉0
𝐴𝐴
𝑝𝑝 = 𝑉𝑉0
𝐴𝐴
𝑝𝑝 = 1 for 𝑝𝑝 > 0
• 𝑉𝑉0
𝐴𝐴
𝐴𝐴
𝑝𝑝 = 0 for 𝑝𝑝 < 0
• 𝑉𝑉0
𝐴𝐴
𝐴𝐴
𝑝𝑝 = 0.5 for 𝑝𝑝 = 0 [a tie]
• Goal: maximizing 𝑉𝑉𝑛𝑛
𝐴𝐴(𝑝𝑝) for some 𝑛𝑛 and 𝑝𝑝
– Find by conditioning on the possession outcome and
applying the law of probability

Law of total probability
• Let 𝑋𝑋 denote a random variable.
– 𝑋𝑋 is a binary variable denoting the outcome of the game
(win/loss)
• We can find the probability of 𝑋𝑋 given some
additional information 𝑌𝑌
– 𝑌𝑌 is a random variable denoting the outcome of the
current possession (touchdown, field goal, or no score).
[ ] ( | ) ( )
y Y
P X P X y P y
∈
= ∑

Finding an expected value
• Let 𝐼𝐼 be an indicator variable if the team wins:
– 𝐼𝐼 = 1 if they win and 𝐼𝐼 = 0 if they lose.
• The expected value of 𝐼𝐼 is the probability that the team wins
𝐸𝐸[𝐼𝐼] = 𝑃𝑃(𝐼𝐼 > 0)
• 𝑃𝑃(𝐼𝐼 > 0) = 𝑉𝑉𝑛𝑛
𝐴𝐴
(𝑝𝑝)
– This can be found by computing the expected value of an
indicator variable 𝐸𝐸[𝐼𝐼]
– This is the solution to a longest path problem!

Law of total probability
• 𝑃𝑃𝑇𝑇𝑇𝑇 = probability of a touchdown
• 𝑃𝑃𝐹𝐹𝐹𝐹 = probability of a field goal
• 𝑃𝑃𝑁𝑁𝑁𝑁 = probability of no score
• 𝑃𝑃𝐴𝐴𝐴𝐴 = probability of scoring a point after a touchdown
• 𝑃𝑃2𝑃𝑃𝑃𝑃 = probability of a two point conversion after a touchdown
Three outcomes of the current drive: TD, FG, or NS
( 𝑃𝑃𝑇𝑇𝑇𝑇 + 𝑃𝑃𝑃𝑃𝑃𝑃 + 𝑃𝑃𝑃𝑃𝑃𝑃 = 1)
𝑉𝑉𝑛𝑛+1
𝐴𝐴
𝑝𝑝 =
𝑃𝑃𝑁𝑁𝑁𝑁 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝)  No score
+ 𝑃𝑃𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝 + 3)  Field Goal
+ 𝑃𝑃𝑃𝑃𝑃𝑃 max{𝑃𝑃𝐴𝐴𝐴𝐴 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝 + 7) + (1 − 𝑃𝑃𝐴𝐴𝐴𝐴) 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝 + 6) , TD with AT
𝑃𝑃2𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝 + 8) + (1 − 𝑃𝑃2𝑃𝑃𝑃𝑃) 𝑉𝑉𝑛𝑛
𝐵𝐵
(𝑝𝑝 + 6) }  TD with 2PT
Note: define 𝑉𝑉𝑛𝑛+1
𝐵𝐵
𝑝𝑝 in an analogous way

Series of equations
Team A:
𝑉𝑉𝑛𝑛+1
𝐴𝐴
𝑝𝑝 =
𝑃𝑃𝑇𝑇𝑇𝑇 max{ 𝑃𝑃𝐴𝐴𝐴𝐴 𝑉𝑉𝑛𝑛
𝐵𝐵
𝑝𝑝 + 7 + 1 − 𝑃𝑃𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵
𝑝𝑝 + 6 , 𝑃𝑃2𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵(𝑝𝑝 +

Team A’s win probability
Note: the points P are before the touchdown is scored
V^A_n(p) Possessions remaining n
P 0 1 2 3 4 5 6 7 8 9 10
-16 0 0.000 0.000 0.003 0.002 0.014 0.010 0.030 0.022 0.047 0.036
-15 0 0.000 0.000 0.006 0.004 0.020 0.014 0.038 0.029 0.058 0.044
-14 0 0.000 0.000 0.015 0.010 0.035 0.025 0.056 0.042 0.077 0.059
-13 0 0.000 0.000 0.024 0.016 0.050 0.036 0.075 0.056 0.097 0.075
-12 0 0.000 0.000 0.026 0.018 0.057 0.042 0.086 0.066 0.112 0.088
-11 0 0.000 0.000 0.035 0.026 0.074 0.056 0.108 0.084 0.136 0.108
-10 0 0.000 0.000 0.045 0.034 0.092 0.070 0.131 0.102 0.161 0.129
-9 0 0.000 0.000 0.062 0.046 0.119 0.090 0.162 0.126 0.193 0.154
-8 0 0.046 0.031 0.108 0.079 0.159 0.122 0.198 0.155 0.226 0.182
-7 0 0.091 0.062 0.154 0.113 0.199 0.153 0.234 0.186 0.260 0.212
-6 0 0.186 0.127 0.252 0.183 0.283 0.216 0.303 0.240 0.318 0.258
-5 0 0.190 0.135 0.271 0.202 0.311 0.242 0.334 0.268 0.350 0.288
-4 0 0.190 0.141 0.282 0.218 0.332 0.265 0.362 0.296 0.382 0.319
-3 0 0.255 0.198 0.362 0.288 0.409 0.332 0.430 0.356 0.441 0.371
-2 0 0.320 0.243 0.427 0.336 0.465 0.378 0.480 0.398 0.485 0.410
-1 0 0.320 0.251 0.440 0.357 0.489 0.405 0.509 0.430 0.518 0.444
0 0.5 0.660 0.500 0.622 0.500 0.601 0.500 0.588 0.500 0.579 0.500
1 1 1.000 0.749 0.805 0.643 0.714 0.594 0.667 0.569 0.639 0.555
2 1 1.000 0.757 0.817 0.662 0.736 0.621 0.695 0.600 0.670 0.588
3 1 1.000 0.803 0.849 0.713 0.774 0.668 0.733 0.644 0.708 0.628
4 1 1.000 0.859 0.896 0.782 0.832 0.734 0.789 0.703 0.758 0.681
Team A’s last
possession
Team A’s second to
last possession(c) Laura Albert 2019 20Two point conversion DP
Boundary
conditions

How do we determine whether to go
for two points?
Test the condition in the max statement to see what was
selected:
For Team A who gets the ball with 𝑛𝑛 + 1 possessions to go and is
up by 𝑝𝑝 points 𝑉𝑉𝑛𝑛+1
𝐴𝐴
𝑝𝑝
𝑃𝑃𝐴𝐴𝐴𝐴 𝑉𝑉𝑛𝑛
𝐵𝐵
𝑝𝑝 + 7 + 1 − 𝑃𝑃𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵
𝑝𝑝 + 6 <
𝑃𝑃2𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵
𝑝𝑝 + 8 + 1 − 𝑃𝑃2𝑃𝑃𝑃𝑃 𝑉𝑉𝑛𝑛
𝐵𝐵
𝑝𝑝 + 6
Then go for two points is better than going for one point.
A similar approach works for Team B by solving 𝑉𝑉𝑛𝑛+1
𝐵𝐵
𝑝𝑝

When should team A go for 2 if they score a
touchdown
Note: the points P are before the touchdown is scored
Possessions remaining n
P 0 2 2 3 4 5 6 7 8 9 10
-10 1 1 1 1 2 2 2 2 2 2 1
-9 1 1 1 1 1 1 2 2 2 2 1
-8 1 1 1 1 2 2 2 2 2 2 1
-7 1 1 1 1 2 2 2 2 2 2 1
-6 1 1 1 1 1 1 1 1 1 1 1
-5 1 1 1 1 2 2 2 2 2 2 1
-4 1 1 1 1 2 2 2 2 2 2 1
-3 1 1 1 1 2 2 2 2 2 2 1
-2 1 1 2 2 2 2 2 2 2 2 1
-1 1 1 1 1 2 2 2 2 2 2 1
0 1 1 2 2 2 2 2 2 2 2 1
1 1 1 1 1 1 1 1 1 1 1 1
2 1 1 1 1 1 1 1 1 1 2 1
3 1 1 2 2 2 2 2 2 2 2 1
4 1 1 1 1 1 1 1 1 1 1 1
5 1 1 1 1 1 1 1 1 1 1 1
6 2 2 2 2 2 2 2 2 2 2 2
7 1 1 1 1 2 1 2 1 2 1 1
8 1 1 1 1 1 1 1 1 1 1 1
9 1 2 2 2 2 2 2 2 2 2 1
10 1 1 2 2 2 2 2 2 2 2 1
Team A’s last
possession
Team A’s second to
last possession(c) Laura Albert 2019 22Two point conversion DP

When should team A go for 2 if they score
a touchdown
Two point conversion DP (c) Laura Albert 2019 23
Team A’s last
possession
Team A’s
second to
last
possession

When should Team A go for 2?
On their last possession (from 𝑉𝑉1
𝐴𝐴
𝑝𝑝 ):
When 𝑝𝑝 = −8
i.e., when Team A is down 8 and then scores a touchdown
Team A would be tied or down 2 after this possession.
On their second to last possession (from 𝑉𝑉3
𝐴𝐴
𝑝𝑝 ):
When 𝑝𝑝 = −16, −14, −11, −8, −5, −4, −2, −1
(Or 𝑝𝑝 = −10, −8, −3, −2, 1, 2, 4, 5 including the 6 points from
the touchdown)
Note: the points 𝑝𝑝 are before the touchdown is scored

When should team A go for 2 if they score
a touchdown
Go for it
when down 2
Go for it
when down 8
Go for it
when up
1, 2, 4, 5
Go for 1
here,
otherwise
the other
team could
win with a FG

When should Team A go for 2?
When we set 𝑃𝑃𝐴𝐴𝐴𝐴 = 0.95 (from 0.96) the results change slightly:
• Team A should go for 2 if they score a touchdown when down 15 before
the touchdown
• Note: the win probability does not change very much
On their second to last possession (from 𝑉𝑉3
𝐴𝐴
𝑝𝑝 ):
When 𝑝𝑝 = −16, −𝟏𝟏𝟏𝟏, −14, −11, −8, −5, −4, −2, −1
Note: the points 𝑝𝑝 are before the touchdown is scored
How much does going for 2 improve the win probability?

How much does going for 2 improve
the win probability?
Answer: not by much.
Rationale: if you are down by 15, you have to score two touchdowns while
keeping your opponent from scoring.
• If down by 8 (before the 6 TD points) with 1 or 3 possessions to go, the
strategy improves the win probability by 4.6% or 4.5%.
• If down by 1, 5, or 11 (before the 6 TD points), the strategy improves the win
probability by 0.6 – 1.8% depending on the remaining number of possessions
• If down by 14, 15, or 16 (before the 6 TD points), the strategy improves the
win probability by <1%
In some scenarios, the win probability gets worse (!)
• Your opponent is also finding the best strategy
The win probability can be further improved by improving other decisions
• E.g., Going for it on fourth down, better play calling

Why would a team go for 2 when they
are down by 8?
Let’s say a team is down by 14 and scores a touchdown, putting
them down by 8. Let’s say there will be 4 possessions left after
this possession.
• Traditional decision: go for 1 point.
If they succeed (with probability 0.96), they will have a 11.3% probability
of winning.
If they do not succeed, they will have a 7.9% probability of winning
Overall: 11.2% win probability
• Going for a two point conversion
If they succeed (with probability 0.48), they will have a 18.3% probability
of winning.
If they do not succeed, they will have a 7.9% probability of winning
Overall: 12.9% win probability

Should a football team go for a one or two point conversion? A dynamic programming approach

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