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Highest record a nfl team can get and miss the playoffs

Sports Asked by That one person on December 18, 2020

What is the highest record a team can get and miss the playoffs? Assume the 2019 scheduling rules and playoff setup.

One Answer

Under the 2019 schedule, where the playoff structure was:

4 division winners + 2 wildcards from 4 4-team divisions per conference

And each team played a schedule as follows:

  1. 6 games against their division (2 versus each team)
  2. 4 games against another division in their conference (1 each)
  3. 2 games against two other teams in the other two divisions, based on last year's record
  4. 4 games against teams in the other conference

Then the best possible record to miss the playoff is 14-2.

First, to show that it is not possible to miss the playoffs at 15-1:

Choose two divisions who do not play each other in their entirety (2. above). Choose two teams in each division who do not play the other two teams in 3. above.

If this were 2019, we might choose the Baltimore Ravens and the Pittsburgh Steelers (AFC North), and the Tennessee Titans and the Jacksonville Jaguars (AFC South). (See here for the 2019 NFL season matchups.)

Those four teams could each go 15-1 at best (splitting the series against the other member of the pair, and winning all other games). Further, the division winners for the other two divisions could potentially both go 14-2; for example, the Buffalo Bills and the Kansas City Chiefs could have gone 14-2, only losing to the pair of AFC North and AFC South teams respectively. But, no team could be 15-1 and miss the playoffs; clearly, no team in the AFC East or West could (As they all must have two losses from the AFC North and South, respectively), and no third team in either the AFC North or AFC South could (as the other teams both have four losses to each of the first pair).

Three second place teams at 15-1 are impossible as well; that's because the opposite divisions (the East and the West here) the two top teams would each have 2 games against both of the pair in their paired division, plus 2 games against each other for each pair; that's six games, six losses, to split among three teams, so at best they're all 14-2.

Since 15-1 is out, let's consider 14-1-1 (or 13-0-3, which is effectively the same record). In the same setup as above, let's say the teams as above still split their series against each other, but then a third team is considered in the AFC North picture - the Browns. The Browns cannot be 13-0-3 or better, because they have four games against the Ravens and the Steelers, and those six games (Ravens-Steelers, Ravens-Browns, Steelers-Browns, two each) cannot result in fewer than six losses total for the three teams. The same logic applies to a team in the South, and to three second place teams as above.

That leads us to 14-2. As we noted above, the three teams in the AFC North can all go 14-2, if they each split the series against each other - Browns, Steelers, and Ravens. The AFC South then would not have two 15-1 teams, however; in that case, the Browns would have had a game against the Titans, and would have to win that to remain at 14-2. So, the final playoff standings would be:

Division Team 1 Team 2 Team 3
AFC North Ravens (14-2) Steelers (14-2) Browns (14-2)
AFC East Bills (12-4)
AFC South Jaguars (15-1) Titans (14-2)
AFC West Chiefs (13-3)

Then you would have a tiebreaker scenario that I won't get into that would determine which of the Ravens/Steelers/Browns/Titans did not make the playoffs.

Correct answer by Joe on December 18, 2020

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