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 '14 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. – PopularIsn'tRight Nov 24 '14 at 15:02

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.

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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.

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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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