It's impossible for all the teams in a division to have losing records in their divisional games--every time someone loses a game, someone else in the division wins it. But the home team could win every game, and everyone would be 3-3 in divisional play. But as far as I can tell there's no reason why the teams in this unhappy division couldn't lose all ten of their other games. Is this true--is it mathematically possible for a three win team to go to the playoffs?

Naturally, this question is motivated by the fact that there's serious talk of a six win team representing the NFC South.

  • For this unlikely scenario to be resolved, it likely would be the fifth tie breaker to break the logjam. - Strength of victory. nfl.com/standings/tiebreakingprocedures
    – BrownBall
    Nov 19, 2014 at 18:40
  • If you include ties, a team could make the playoffs with even fewer than 3 wins. It's very unlikely, but in theory a division could have all four teams end up at 0-10-6, and one would have to make the playoffs. Nov 24, 2014 at 15:02

3 Answers 3


Technically yes. One of those 3-3 teams will win the division by tie-breaker and, strangely enough, get a home game as the #4 seed against the top wild card team.


I suppose in theory a team could get to the playoffs without winning a regular season game.

If all divisional games finish as ties, and all teams lose every other game, they'll all end up with 0-10-6 records. Whoever comes on top after the tiebreakers will win the division and get the #4 seed.

In practice having 12 ties for the divisional games is highly unlikely of course, so is each team losing their other 10 games.


Yes. The Division champion will make it to the playoffs no matter what the team's record is and will host a wildcard team (which will have a better record!). That being said, it sounds extremely unlikely to have a team holding the best record in a division with only 3 wins.


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