No Runny Eggs

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Tiebreak THIS!, AFC 8-way 8-8 edition

by @ 22:58 on December 27, 2009. Filed under Sports.

NBC Sports’ Gregg Rosenthal challenged the Twittersphere to tiebreak a hypothetical 8-way 8-8 tie between the Miami Dolphins (7-8), New York Jets (8-7), Pittsburgh Steelers (8-7), Baltimore Ravens OldBrowns (8-7), Houston Texans (8-7), Jacksonville Jaguars (7-8), Tennessee Titans (7-8) and Denver Broncos (8-7) for the two AFC wild-card playoff spots. The NFL tiebreaking procedures state that ties within the division get broken first.

Let’s start with the easiest division, the AFC West. Denver would be the only team at 8-8, so they go on to the first wild-card tiebreaker

Next, let’s go to the AFC North. Baltimore split their season series with Pittsburgh, but they have a better division record (3-3 versus 2-4). Baltimore would go on to the first wild-card tiebreaker.

The AFC East is a bit easier as Miami swept New York, and they would go on to the first wild-card tiebreaker.

Finally, we get to the wild division, the AFC South. Jacksonville has the best head-to-head-to-head record of the three teams (3-1 versus 2-2 for Tennessee and 1-3 for Houston), so they would go on to the first wild-card tiebreaker.

Now, we can go to the conference-level tiebreakers between Denver, Baltimore, Miami and Jacksonville:

  • No team either beat or lost to all of the others, so the first tiebreaker (head-to-head sweep/swept) is out.
  • Each team would have a 6-6 conference record, so the second tiebreaker (conference record) is out.
  • The four teams do not have 4 games against common opponents (sharing only Indianapolis and New England), so the third tiebreaker (record against common opponents, minimum of 4 games apiece) is out.
  • That devolves to the 4th tiebreaker, strength of victory. Between the games through tonight and the games necessary to create the 8-8 tie, Denver would have a SoV of 66-59, Baltimore would have a SoV of 58-67, Miami would have a SoV of 55-69, and Jacksonville would have a SoV of 46-80.

Denver would get the first wild-card spot based on strength of victory (none of the other 3 teams can make up the 10+ games). If either Baltimore or Miami has the 2nd-best strength of victory outright, that team would get the second wild-card spot since Jacksonville would not be able to catch either team. However, a difference of 2 1/2 games going into tomorrow night’s game means that it is possible they would tie. In that case, they would go to a two-team tiebreaker:

  • Baltimore and Miami did not play each other this year, so the first tiebreaker (head-to-head) is out.
  • Once again, they would have identical 6-6 conference records, so the second tiebreaker (conference record) is out.
  • They would have identical 2-3 records against common opponents (Indianapolis, New England, Pittsburgh and San Diego), so the third tiebreaker (record against common opponents, minimum of 4 games apiece) is out.
  • We already stipulated that strength of victory would be identical, so the fourth tiebreaker is out.
  • Between the games through tonight and the games necessary to create the 8-team 8-8 tie, the two teams’ strength of schedule is an identical 138-111. If one team pulls ahead, they would get the last wild-card spot. Otherwise…
  • …Things devolve to best combined ranking among conference teams in points scored and points allowed. Miami has scored 336 points (8th in the AFC) and given up 360 points (14th in the AFC), for a combined 11th. Baltimore has scored 370 points (4th in the AFC) and given up 248 points (2nd in the AFC) for a combined 3rd, easily favoring Baltimore.
  • Revisions/extensions (11:20 pm 12/27/2009) - Things only start over if a tie remains after breaking up a 4-way tie. The post has been shortened to reflect that.

    R&E part 2 (11:25 pm 12/27/2009) - Corrected a rather-embarrassing typo. I originally listed Houston twice in the list of 8.

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