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A suggestion for the NFL’s head-to-head tiebreaker James A. Swenson Dan Swenson UW-Platteville Black Hills State University [email protected] [email protected] 23 April 2016 Swenson/Swenson (UWP/BHSU) NFL Tiebreakers 23 April 2016 1 / 19

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Page 1: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 2: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 3: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 4: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 5: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 6: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 7: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 8: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 9: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 10: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 11: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 12: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 13: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 14: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 15: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 16: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 17: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 18: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 19: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 20: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 21: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 22: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 23: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 24: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 25: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 26: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

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

Page 27: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 28: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 29: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 30: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 31: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 32: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 33: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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

Page 34: A suggestion for the NFL's head-to-head tiebreakerpeople.uwplatt.edu/~swensonj/SwensonTiebreakers.pdf · 8 Best combined ranking among all teams in points scored and points allowed

“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.

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“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.

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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.

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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.

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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.

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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

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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

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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.

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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/

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2014 NFC Playoffs: What would have happened?

http://www.nj.com/super-bowl/index.ssf/2015/02/

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Thanks!

http://www.sportslogos.net/logos/list_by_team/172/

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