a suggestion for the nfl's head-to-head...
TRANSCRIPT
A suggestion for the NFL’s head-to-head tiebreaker
James A. Swenson Dan SwensonUW-Platteville Black Hills State University
[email protected] [email protected]
23 April 2016
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 1 / 19
Welcome!
Thanks for coming!
I hope you’ll enjoy the talk; please feel free to get involved!
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 2 / 19
Outline
1 What the NFL does now
2 What the NFL should do instead
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 3 / 19
The NFL playoffs
What are we doing?
There are 32 teams in the National Football League.
There are two conferences (AFC/NFC) of 16 teams each.
Each conference is divided into four divisions(North/South/East/West) of 4 teams each.
Each team plays a regular season of 16 games, including two gamesagainst each team in its division.After this, six teams from each conference compete in a playoff, endingwith the Super Bowl.In each conference, the four division winners and two “wild card” teamsqualify for the playoffs.
Images: http://www.nationalchamps.net/Helmet_Project/nfl.htm
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 4 / 19
The NFL playoffs
What are we doing?
There are 32 teams in the National Football League.
There are two conferences (AFC/NFC) of 16 teams each.
Each conference is divided into four divisions(North/South/East/West) of 4 teams each.
Each team plays a regular season of 16 games, including two gamesagainst each team in its division.
After this, six teams from each conference compete in a playoff, endingwith the Super Bowl.In each conference, the four division winners and two “wild card” teamsqualify for the playoffs.
Images: http://www.nationalchamps.net/Helmet_Project/nfl.htm
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 4 / 19
The NFL playoffs
What are we doing?
There are 32 teams in the National Football League.
There are two conferences (AFC/NFC) of 16 teams each.
Each conference is divided into four divisions(North/South/East/West) of 4 teams each.
Each team plays a regular season of 16 games, including two gamesagainst each team in its division.After this, six teams from each conference compete in a playoff, endingwith the Super Bowl.
In each conference, the four division winners and two “wild card” teamsqualify for the playoffs.
Images: http://www.nationalchamps.net/Helmet_Project/nfl.htm
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 4 / 19
The NFL playoffs
What are we doing?
There are 32 teams in the National Football League.
There are two conferences (AFC/NFC) of 16 teams each.
Each conference is divided into four divisions(North/South/East/West) of 4 teams each.
Each team plays a regular season of 16 games, including two gamesagainst each team in its division.After this, six teams from each conference compete in a playoff, endingwith the Super Bowl.In each conference, the four division winners and two “wild card” teamsqualify for the playoffs.
Images: http://www.nationalchamps.net/Helmet_Project/nfl.htm
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 4 / 19
The NFL Playoffs
Image: http://nittanysportshuddle.com/2016/01/smittys-picks-nfl-playoff-week-1/
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 5 / 19
NFL procedures
Breaking ties between 2 teams
Teams in same division:1 Head-to-head (best won-lost-tied percentage
in games between the clubs).
2 Best won-lost-tied percentage in gamesplayed within the division.
3 Best won-lost-tied percentage in commongames.
4 Best won-lost-tied percentage in gamesplayed within the conference.
5 Strength of victory.
6 Strength of schedule.
7 Best combined ranking among conferenceteams in points scored and points allowed.
8 Best combined ranking among all teams inpoints scored and points allowed.
9 Best net points in common games.
10 Best net points in all games.
11 Best net touchdowns in all games.
12 Coin toss
Teams in different divisions:1 Head-to-head, if applicable.
2 Best won-lost-tied percentage in gamesplayed within the conference.
3 Best won-lost-tied percentage in commongames, minimum of four.
4 Strength of victory.
5 Strength of schedule.
6 Best combined ranking among conferenceteams in points scored and points allowed.
7 Best combined ranking among all teams inpoints scored and points allowed.
8 Best net points in conference games.
9 Best net points in all games.
10 Best net touchdowns in all games.
11 Coin toss.
“NFL Tie-Breaking Procedures.” http://www.nfl.com/standings/tiebreakingprocedures
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 6 / 19
NFL procedures
Breaking ties among 3+ teams
Teams in same division:1 Head-to-head (best won-lost-tied percentage
in games among the clubs).
2 Best won-lost-tied percentage in gamesplayed within the division.
3 Best won-lost-tied percentage in commongames.
4 Best won-lost-tied percentage in gamesplayed within the conference.
5 Strength of victory.
6 Strength of schedule.
7 Best combined ranking among conferenceteams in points scored and points allowed.
8 Best combined ranking among all teams inpoints scored and points allowed.
9 Best net points in common games.
10 Best net points in all games.
11 Best net touchdowns in all games.
12 Coin toss
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Head-to-head sweep. (Applicable only if oneclub has defeated each of the others or if oneclub has lost to each of the others.)
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 Best won-lost-tied percentage in commongames, minimum of four.
5 Strength of victory.
6 Strength of schedule.
7 Best combined ranking among conferenceteams in points scored and points allowed.
8 Best combined ranking among all teams inpoints scored and points allowed.
9 Best net points in conference games.
10 Best net points in all games.
11 Best net touchdowns in all games.
12 Coin toss
(After finding a rule that eliminates a team, go back to the beginning.)
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 7 / 19
Example: NFC, 2014
Who’s #1?
Week 1: Seattle 36, Green Bay 16.
Week 6: Dallas 30, Seattle 23.
Dallas vs. Green Bay: no regular-season game.
Image: http://espn.go.com/nfl/standings/_/season/2014/
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 8 / 19
Example: NFC, 2014
Who’s #1?
Week 1: SEA 36, GB 16.
Week 6: DAL 30, SEA 23.
DAL vs. GB: no game.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Head-to-head sweep. (Applicable only if oneclub has defeated each of the others or if oneclub has lost to each of the others.)
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: SEA. #2: GB. #3: DAL.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 9 / 19
Example: NFC, 2014
Who’s #1?
Week 1: SEA 36, GB 16.
Week 6: DAL 30, SEA 23.
DAL vs. GB: no game.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Head-to-head sweep. (Applicable only if oneclub has defeated each of the others or if oneclub has lost to each of the others.)
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: SEA. #2: GB. #3: DAL.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 9 / 19
2014 NFC Playoffs: What happened next?
First weekend:
CAR won at home against ARI,whose top 2 QBs were injured.
DAL had to play DET, while GBand SEA got a week off.
Second weekend:
SEA won a home game againstCAR, a team with a losing record.
GB got a controversial win athome against DAL.
NFC Championship:
SEA scored two TDs in the last2:09, then beat GB in OT.
Image: CBS Sports, via http://www.interbasket.net/news/16746/2014/12/print-nfl-bracket-2015-15-wildcards/
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 10 / 19
Outline
1 What the NFL does now
2 What the NFL should do instead
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 11 / 19
Modeling the regular season
Definition
Let x and y be teams. We say x ⊳ y ⇐⇒ x beat or tied y at least once.
Definition
The transitive closure of a relation R is the smallest transitive relationthat contains R.
Definition
We write ▸ for the transitive closure of ⊳. Thus x ▸ y when there is asequence of teams x = v0 ⊳ v1 ⊳ . . . ⊳ vn = y .
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 12 / 19
Modeling the regular season
Definition
Let x and y be teams. We say x ⊳ y ⇐⇒ x beat or tied y at least once.
Definition
The transitive closure of a relation R is the smallest transitive relationthat contains R.
Definition
We write ▸ for the transitive closure of ⊳. Thus x ▸ y when there is asequence of teams x = v0 ⊳ v1 ⊳ . . . ⊳ vn = y .
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 12 / 19
Modeling the regular season
Definition
Let x and y be teams. We say x ⊳ y ⇐⇒ x beat or tied y at least once.
Definition
The transitive closure of a relation R is the smallest transitive relationthat contains R.
Definition
We write ▸ for the transitive closure of ⊳. Thus x ▸ y when there is asequence of teams x = v0 ⊳ v1 ⊳ . . . ⊳ vn = y .
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 12 / 19
Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y) and not (y ▸ x).
Lemma
The relation > is antisymmetric: If x > y , then y /> x .
Lemma
The relation > is transitive: If x > y > z , then x > z .
Proof.
Suppose x > y and y > z . Then x ▸ y ▸ z , so x ▸ z .
Sftsoc that z ▸ x . Also, x ▸ y , so z ▸ y . But then y /> z . ⇒⇐
Theorem
> is a partial order.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 13 / 19
Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y) and not (y ▸ x).
Lemma
The relation > is antisymmetric: If x > y , then y /> x .
Lemma
The relation > is transitive: If x > y > z , then x > z .
Proof.
Suppose x > y and y > z . Then x ▸ y ▸ z , so x ▸ z .
Sftsoc that z ▸ x . Also, x ▸ y , so z ▸ y . But then y /> z . ⇒⇐
Theorem
> is a partial order.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 13 / 19
Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y) and not (y ▸ x).
Lemma
The relation > is antisymmetric: If x > y , then y /> x .
Lemma
The relation > is transitive: If x > y > z , then x > z .
Proof.
Suppose x > y and y > z . Then x ▸ y ▸ z , so x ▸ z .
Sftsoc that z ▸ x . Also, x ▸ y , so z ▸ y . But then y /> z . ⇒⇐
Theorem
> is a partial order.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 13 / 19
Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y) and not (y ▸ x).
Lemma
The relation > is antisymmetric: If x > y , then y /> x .
Lemma
The relation > is transitive: If x > y > z , then x > z .
Proof.
Suppose x > y and y > z . Then x ▸ y ▸ z , so x ▸ z .
Sftsoc that z ▸ x . Also, x ▸ y , so z ▸ y . But then y /> z . ⇒⇐
Theorem
> is a partial order.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 13 / 19
Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y) and not (y ▸ x).
Lemma
The relation > is antisymmetric: If x > y , then y /> x .
Lemma
The relation > is transitive: If x > y > z , then x > z .
Proof.
Suppose x > y and y > z . Then x ▸ y ▸ z , so x ▸ z .Sftsoc that z ▸ x . Also, x ▸ y , so z ▸ y . But then y /> z . ⇒⇐
Theorem
> is a partial order.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 13 / 19
Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y) and not (y ▸ x).
Lemma
The relation > is antisymmetric: If x > y , then y /> x .
Lemma
The relation > is transitive: If x > y > z , then x > z .
Proof.
Suppose x > y and y > z . Then x ▸ y ▸ z , so x ▸ z .Sftsoc that z ▸ x . Also, x ▸ y , so z ▸ y . But then y /> z . ⇒⇐
Theorem
> is a partial order.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 13 / 19
Tiebreaker proposal
Idea
Let X be a set of teams. Suppose X =W ∪ L, where:
W ≠ ∅;
W ∩ L = ∅;
∀x ∈W ,∀y ∈ L, x > y .
Then the league should rank teams in W above teams in L.
What if there’s a choice?
For example, suppose x > y , x > z , and y > z . We could take W = {x} orW = {x , y}. We should pick W = {x}, eliminating both y and z . Ingeneral, we ought to take the smallest possible W .
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 14 / 19
Tiebreaker proposal
Idea
Let X be a set of teams. Suppose X =W ∪ L, where:
W ≠ ∅;
W ∩ L = ∅;
∀x ∈W ,∀y ∈ L, x > y .
Then the league should rank teams in W above teams in L.
What if there’s a choice?
For example, suppose x > y , x > z , and y > z . We could take W = {x} orW = {x , y}. We should pick W = {x}, eliminating both y and z . Ingeneral, we ought to take the smallest possible W .
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 14 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2.
Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2.
Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.
Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2.
We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1.
Thenw > z . By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z .
By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w .
Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w . Hence w /∈ L2.
Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
“The smallest possible W ” is well-defined!
Theorem
Let X be a set, and let > be a binary relation on X such that[x > y ⇒ y /> x] for all x , y ∈ X .Suppose {W1,L1} and {W2,L2} are partitions of X such that fori ∈ {1,2}:
Wi is non-empty;
If x ∈Wi and y ∈ Li , then x > y .
Then W1 ⊆W2 or W2 ⊆W1.
Proof.
If W1 =W2, we’re done, so assume W1 ≠W2. Wlog, let z ∈W2 ∖W1.Now z ∈ L1 ∩W2. We claim W1 ⊆W2. To see this, pick w ∈W1. Thenw > z . By hypothesis, z /> w . Hence w /∈ L2. Thus w ∈W2.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 15 / 19
2014 NFC Playoffs: What would have happened?
Who’s #1?
Week 1: SEA 36, GB 16.
Week 6: DAL 30, SEA 23.
DAL vs. GB: no game.
DAL > SEA > GB.W = {DAL},L = {SEA,GB}.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Apply the partition rule.
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: DAL. #2: SEA. #3: GB.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 16 / 19
2014 NFC Playoffs: What would have happened?
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL > SEA > GB.W = {DAL},L = {SEA,GB}.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Apply the partition rule.
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: DAL. #2: SEA. #3: GB.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 16 / 19
2014 NFC Playoffs: What would have happened?
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL ▸ SEA ▸GB.
DAL > SEA > GB.W = {DAL},L = {SEA,GB}.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Apply the partition rule.
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: DAL. #2: SEA. #3: GB.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 16 / 19
2014 NFC Playoffs: What would have happened?
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL ▸ SEA ▸GB.
DAL > SEA > GB.
W = {DAL},L = {SEA,GB}.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Apply the partition rule.
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: DAL. #2: SEA. #3: GB.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 16 / 19
2014 NFC Playoffs: What would have happened?
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL ▸ SEA ▸GB.
DAL > SEA > GB.W = {DAL},L = {SEA,GB}.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Apply the partition rule.
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: DAL.
#2: SEA. #3: GB.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 16 / 19
2014 NFC Playoffs: What would have happened?
Who’s #1?
SEA ⊳ GB
SEA ▸GB.
SEA > GB.W = {SEA},L = {GB}.
Breaking ties among 3+teams
Teams in different divisions:1 Apply division tie breaker to eliminate all but
the highest ranked club in each division.Then...
2 Apply the partition rule.
3 Best won-lost-tied percentage in gamesplayed within the conference.
4 . . .
#1: DAL. #2: SEA. #3: GB.
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 16 / 19
2014 NFC Playoffs: What would have happened?
First weekend:
CAR won at home against ARI,whose top 2 QBs were injured.
GB beats DET, while DAL andSEA get a week off.
Second weekend:
DAL gets an easy win over CAR,the worst division champion ever.
SEA, at home, beats GB, as theyreally did twice in 2014.
NFC Championship:
DAL, at home, beats SEA, thenbeats NE in Super Bowl XLIX.
AP Photo/Brandon Wade, via http://www.bostonherald.com/sports/patriots_nfl/nfl_coverage/2014/09/
Charlie Riedel/AP, via http://www.si.com/nfl/2014/09/29/
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 17 / 19
2014 NFC Playoffs: What would have happened?
http://www.nj.com/super-bowl/index.ssf/2015/02/
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 18 / 19
Thanks!
http://www.sportslogos.net/logos/list_by_team/172/
Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 19 / 19