Is it possible for three NFL teams in different divisions in the same conference to all go 16-0 during the regular season? I'm not sure if this is possible with all the scheduling rules that they have.

If this is possible, then one of those 3 teams would lose the tie-breaker and have to play a wild card game. So they would have to play 4 playoff games to win the Super Bowl, and they would be 20-0 if they won the Super Bowl.

  • My comment pertains to Diggers3 last statement, "As an add on, there could be a total of 4 undefeated regular season teams in one year. 2 in each conference." This statement is incorrect. You can have two 16-0 teams in one conference and only one 16-0 team in the other (two AFC 16-0 and one NFC 16-0, or two NFC 16-0 and one AFC 16-0) because of the inter-conference play. Attached is one combination of teams that could have gone 16-0 for the 2014 season based on their opening day schedule, but a fourth team is not possible to add. – BrownBall Nov 22 '14 at 21:23
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    Are you sure this is true? For interconference play, each division will only play teams from one division in the other conference. So couldn't you in theory have an NFC East team go 16-0 (while playing against the AFC East), an NFC North team go 16-0 (while playing against the AFC North), an AFC South team go 16-0 (while playing against the NFC South), and an AFC West team go 16-0 (while playing against the NFC West)? – pacoverflow Nov 22 '14 at 22:00
  • I will check. I recall the rotation of intra-conference and inter-conference accounts for it not lining up cleanly to have 4 regular season undefeated teams. – BrownBall Nov 22 '14 at 23:23
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    That is incorrect. I checked before adding that to my answer. If you check the schedule and draw it out you can clearly see that it is possible to have 4 teams. Your example does not cover every case. It is a single example. – diggers3 Nov 23 '14 at 0:58

It is not. This is because of the regular reason scheduling:

  • Each team plays 16 games.

  • 6 games are in their own division (3 teams, each twice).

  • 4 games are against a division in the other conference.

  • 4 games are against a division in their own conference.

  • 2 remaining games are against the other 2 divisions in their conference based on final standings from the year before.

There are 4 divisions in each conference. So when the inter-conference divisions play each other, there can be a maximum of 2 undefeated teams in a conference. This also (obviously) means they didn't play each other in the 2 remaining games.

As an add on, there could be a total of 4 undefeated regular season teams in one year. 2 in each conference.


NO, 3 teams can't go 16-0 in a conference.

Each division in a conference plays all of the teams in their own division twice, another division in their conference, and the other two teams that finished in the same spot as them the prior year.

This means that at maximum the highest number of teams that can go 16-0 in a conference is 2. Say you have four teams that win all of their games except games head to head against another team in the group. Each is a team from the East, West, North and South, they manage some good luck and none of them play each other as their same position game. Say for this year the East plays the West and the North plays the South. Only two of the teams in this group will be 16-0 at years end as there will be 2 head to head games among the 4 teams that only lose to teams in the group.

  • "In the same spot as them" isn't quite correct, it's more complicated than that. Otherwise correct. – Joe Feb 13 '15 at 21:33

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