A suggestion for the NFL`s head-to-head tiebreaker

Transcription

A suggestion for the NFL’s head-to-head tiebreaker
James A. Swenson
UW-Platteville
[email protected]
Dan Swenson
Black Hills State University
[email protected]
23 April 2016
Swenson/Swenson (UWP/BHSU)
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Welcome!
Thanks for coming!
I hope you’ll enjoy the talk; please feel free to get involved!
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Outline
1
What the NFL does now
2
What the NFL should do instead
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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.
Images: http://www.nationalchamps.net/Helmet_Project/nfl.htm
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The NFL playoffs
What are we doing?
Each team plays a regular season of 16 games, including two games
against each team in its division.
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The NFL playoffs
What are we doing?
After this, six teams from each conference compete in a playoff, ending
with the Super Bowl.
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The NFL playoffs
What are we doing?
After this, six teams from each conference compete in a playoff, ending
with the Super Bowl.
In each conference, the four division winners and two “wild card” teams
qualify for the playoffs.
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The NFL Playoffs
Image: http://nittanysportshuddle.com/2016/01/smittys-picks-nfl-playoff-week-1/
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NFL procedures
Breaking ties between 2 teams
Teams in different divisions:
Teams in same division:
1
Head-to-head (best won-lost-tied percentage
in games between the clubs).
1
Head-to-head, if applicable.
2
Best won-lost-tied percentage in games
played within the conference.
2
played within the division.
3
3
Best won-lost-tied percentage in common
games.
games, minimum of four.
4
Strength of victory.
4
5
Strength of schedule.
6
Best combined ranking among conference
teams in points scored and points allowed.
5
6
7
7
Best combined ranking among all teams in
points scored and points allowed.
8
Best net points in conference games.
8
9
Best net points in all games.
9
Best net points in common games.
10 Best net touchdowns in all games.
11 Coin toss.
10 Best net points in all games.
12 Coin toss
“NFL Tie-Breaking Procedures.” http://www.nfl.com/standings/tiebreakingprocedures
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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
played within the division.
3
games.
4
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 one
club has defeated each of the others or if one
club has lost to each of the others.)
3
5
4
6
games, minimum of four.
7
5
6
8
7
9
Best net points in common games.
8
9
Best net points in conference games.
12 Coin toss
12 Coin toss
(After finding a rule that eliminates a team, go back to the beginning.)
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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/
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Example: NFC, 2014
Breaking ties among 3+
teams
Who’s #1?
Week 1: SEA 36, GB 16.
Week 6: DAL 30, SEA 23.
1
Then...
2
3
4
...
DAL vs. GB: no game.
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Example: NFC, 2014
teams
Who’s #1?
1
Then...
2
3
4
...
#1: SEA.
#2: GB.
NFL Tiebreakers
#3: DAL.
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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 GB
and SEA got a week off.
Second weekend:
SEA won a home game against
CAR, a team with a losing record.
GB got a controversial win at
home against DAL.
NFC Championship:
SEA scored two TDs in the last
2:09, then beat GB in OT.
Image: CBS Sports, via http://www.interbasket.net/news/16746/2014/12/print-nfl-bracket-2015-15-wildcards/
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Outline
1
What the NFL does now
2
What the NFL should do instead
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Modeling the regular season
Definition
Let x and y be teams. We say x ⊳ y ⇐⇒ x beat or tied y at least once.
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Definition
Definition
The transitive closure of a relation R is the smallest transitive relation
that contains R.
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Definition
Definition
The transitive closure of a relation R is the smallest transitive relation
that contains R.
Definition
We write ▸ for the transitive closure of ⊳. Thus x ▸ y when there is a
sequence of teams x = v0 ⊳ v1 ⊳ . . . ⊳ vn = y .
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Team ordering
Definition
Let x and y be teams. We say x > y ⇐⇒ (x ▸ y ) and not (y ▸ x).
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Team ordering
Definition
Lemma
The relation > is antisymmetric: If x > y , then y >/ x.
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Team ordering
Definition
Lemma
Lemma
The relation > is transitive: If x > y > z, then x > z.
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Team ordering
Definition
Lemma
Lemma
Proof.
Suppose x > y and y > z. Then x ▸ y ▸ z, so x ▸ z.
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Team ordering
Definition
Lemma
Lemma
Proof.
Sftsoc that z ▸ x. Also, x ▸ y , so z ▸ y . But then y >/ z. ⇒⇐
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Team ordering
Definition
Lemma
Lemma
Proof.
Sftsoc that z ▸ x. Also, x ▸ y , so z ▸ y . But then y >/ z. ⇒⇐
Theorem
> is a partial order.
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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.
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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} or
W = {x, y }. We should pick W = {x}, eliminating both y and z. In
general, we ought to take the smallest possible W .
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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 for
i ∈ {1, 2}:
Wi is non-empty;
If x ∈ Wi and y ∈ Li , then x > y .
Then W1 ⊆ W2 or W2 ⊆ W1 .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
If W1 = W2 , we’re done, so assume W1 ≠ W2 .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
If W1 = W2 , we’re done, so assume W1 ≠ W2 . Wlog, let z ∈ W2 ∖ W1 .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
Now z ∈ L1 ∩ W2 .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
Now z ∈ L1 ∩ W2 . We claim W1 ⊆ W2 . To see this, pick w ∈ W1 .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
Now z ∈ L1 ∩ W2 . We claim W1 ⊆ W2 . To see this, pick w ∈ W1 . Then
w > z.
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
w > z. By hypothesis, z >/ w .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
w > z. By hypothesis, z >/ w . Hence w ∈/ L2 .
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Theorem
[x > y ⇒ y >/ x] for all x, y ∈ X .
i ∈ {1, 2}:
Wi is non-empty;
Proof.
w > z. By hypothesis, z >/ w . Hence w ∈/ L2 . Thus w ∈ W2 .
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2014 NFC Playoffs: What would have happened?
teams
Who’s #1?
1
Then...
2
Apply the partition rule.
3
4
...
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teams
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
NFL Tiebreakers
1
Then...
2
3
4
...
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teams
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL ▸ SEA ▸ GB.
NFL Tiebreakers
1
Then...
2
3
4
...
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teams
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL ▸ SEA ▸ GB.
DAL > SEA > GB.
NFL Tiebreakers
1
Then...
2
3
4
...
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teams
Who’s #1?
SEA ⊳ GB
DAL ⊳ SEA
DAL ▸ SEA ▸ GB.
DAL > SEA > GB.
W = {DAL}, L = {SEA, GB}.
1
Then...
2
3
4
...
#1: DAL.
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teams
Who’s #1?
SEA ⊳ GB
SEA ▸ GB.
SEA > GB.
W = {SEA}, L = {GB}.
#1: DAL. #2: SEA.
NFL Tiebreakers
1
Then...
2
3
4
...
#3: GB.
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First weekend:
CAR won at home against ARI,
whose top 2 QBs were injured.
GB beats DET, while DAL and
SEA get a week off.
Second weekend:
DAL gets an easy win over CAR,
the worst division champion ever.
SEA, at home, beats GB, as they
really did twice in 2014.
NFC Championship:
DAL, at home, beats SEA, then
beats 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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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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A suggestion for the NFL`s head-to-head tiebreaker

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