Does a 9-7 record get the No. 6 seed in the AFC?
After glancing at the Raiders’ remaining games, it feels like the best case scenario for this team would be to go 5-2 the rest of the way to reach 9-7 on the season. Oakland can beat Denver, Los Angeles and New York. The Raiders will need to take two of these four games to achieve 5-2: New England, Kansas City, Dallas and Philadelphia. Though not easy, it is possible.
So in essence, Oakland should be good enough to get to 7-9 or 8-8 this season, but will reaching for a 9-7 mark be enough to get the Raiders the No. 6 seed in the AFC? Right now, New England and Pittsburgh feel like locks to win their respective divisions. Either Jacksonville or Tennessee will win the AFC South. Both might get into the playoffs anyway.
Kansas City is the team to beat in the AFC West, but Oakland does hold the head-to-head tiebreaker over the Chiefs. However, 10-6 wins this division, not 9-7. For the sake of simplicity, let’s just say New England, Pittsburgh, Kansas City and the winner of Jacksonville at Tennessee in a few weeks are playoff teams. Who are the two other teams coming out of the AFC?
It seems that the AFC East could get a team in, as the worst team in that division is the Jets at 4-5. However, how much are you buying the Buffalo Bills? They need to force turnovers to win games, where are mostly random. The loser of Jacksonville at Tennessee probably gets in any way, but we’ll have to wait and see.
Losing the head-to-head tiebreaker to Baltimore really hurts the Raiders’ playoff chances. Baltimore could flounder down the stretch to finish at or below .500. Frankly, 9-7 should get the No. 6 seed. The only way it doesn’t happen is if the Bills and the second-place AFC South team are both 10-6. Is 11-5 what it takes to win the AFC South? Maybe.