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Raiders Have 7% Chance of Making Playoffs. Here's How.



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The heart of playoff football in Oakland is still beating.
Barely, but the Raiders head to the regular-season finale at Denver on Sunday alive for a playoff berth.
Many things have to happen for them to play another week. Namely: the 7-8 Raiders must win, the Colts (7-8) must win, and Pittsburgh (8-7) and Tennessee (8-7) have to lose. More games have to break Oakland’s way. But, in theory, it could happen. It did in Week 16 to get us to this point. Here’s a look at what needs to happen for the Raiders to enter the AFC playoffs as the No. 6 seed.
The New York Times puts the Raiders’ chances of qualifying for the postseason at 7%.

Titans Take Three-Way vs. Raiders, Steelers

Instead of an extra game between tied teams like baseball uses, the NFL has a series of tiebreaking formulas.
Thus, the Raiders would have to be 8-8, along with the other teams it’s chasing.
The first step to break ties for the Wild Card is to find the higher ranked team in each division. Then, the highest team in each division are compared to each other. If the Raiders, Pittsburgh, and Tennessee are 8-8, the criteria of head-to-head sweep — meaning one team beat the other two, or one team lost to the other two — does not apply. So, even though the Titans beat the Raiders in Week 14, because the Steelers aren’t in the equation, it doesn’t count.
The three teams would also tie in the next step, conference record at 6-6 each. The following step, record in common games, does not apply since the three teams would not have a minimum of four common games on the schedule.
Thus, the strength of victory tiebreaker would be used. SOV is the combined record of the teams defeated. Here, Tennessee would win out.
Currently, the Titans have 54 SOV wins. The Steelers have 37.5, with the Raiders at 34.5.
Thus, the Titans would win the three-team tiebreaker and advance.
However, if the Colts and Titans are tied at 8-8, Indianapolis would win the tiebreaker to determine the highest ranked team from the AFC South. The teams split their two games this year, but the Colts would have a record in the next tiebreaker — division record (4-2 vs. 2-4).

Team Record Conference Record Strength of Victory wins
Oakland 7-8 5-6 34.5
Pittsburgh 8-7 6-5 37.5
Tennessee 8-7 6-5 54

Raiders Take Three-Way vs. Steelers, Colts

However, if it is down to the Raiders, Steelers, and Colts, Oakland has a chance.
Back to three-team Wild Card tiebreaker No. 2: head-to-head sweep. The Colts lost to both Oakland and Pittsburgh, so they are out.
Now, instead of moving to the next three-team criteria, we go to the two-team Wild Card tiebreaker system. Since the Raiders and Steelers did not play this year, and the teams would have a tied conference record (6-6) and common game record (4-1), we go to the fourth criteria, SOV.
In the current standings, it appears the Raiders are ahead based on SOV percentage, .329 to .313. But this is misleading. Oakland’s .329 is based on 34.5 wins on 105 games (7 wins vs. 15 games each team played) vs. the Steelers 37.5 wins on 120 games (8 wins vs. 15 games played). So, using the raw SOV wins is the best.
If Oakland defeats Denver, that would add six more wins to its SOV, bumping it to 40.5. And, since Arizona is at the Rams, and Pittsburgh beat both teams, that is a guaranteed one more SOV win.
Because both teams defeated the Colts, and for this scenario to work the Colts need to win, that’s another SOV win for each team (now Oakland 41.5 to the Steelers 39.5).
Thus, in five other games, Oakland would need to retain its two-game advantage.
The chart of net SOV wins based on those games:

Game Pittsburgh Oakland
Ari at LAR +1 no matter what DNP
Ind at Jax If Ind wins*: +1 If Ind wins*: +1
Cin at Cle If Cin wins: +2
If Cle wins: +1
Tie: +1.5
If Cin wins: +1
If Cle wins: 0
Tie: +0.5
LAC at KC If SD wins: +1
If KC wins: 0
Tie: +0.5
If SD wins: +2
If KC wins: 0
Tie: +1
GB at Det DNP If Det wins: +1
If GB wins: 0
Tie: +0.5
Chi at Min DNP If Chi wins: +1
If Min wins: 0
Tie: +0.5
Mia at NE If Mia wins: +1
If NE wins: 0
Tie: +0.5

* Indianapolis must win in this theoretical scenario

What If?

What if the SOV is still tied (such as the Chargers, Lions and Bears lose, and Dolphins win)?
The next tiebreaker is Strength of Schedule, the record of all teams on the 16-game schedule of an opponent.
Through Week 16, the Raiders and Steelers are tied at with 107.5 SOS wins each.
Mathematically, Pittsburgh would win. Because of the final week schedule, 12 of their opponents play each other, meaning it will be a net-0 wins to the SOS.
For the other four games, in the scenario for the Raiders, Denver (the Week 16 opponent) would lose. The remaining two teams would be Cincinnati and the Jets, for a maximum net gain of two SOS wins.
For the Steelers, in the scenario, they would lose to their Week 17 opponent, Baltimore, actually giving them two net-SOS wins (since they played the Ravens twice). The other two teams are the Chargers and the Colts. Since in this whole scenario the Colts have to win, that would be at least +3 net SOS wins for the Steelers, and the tiebreaker.


For the Raiders to make the playoffs:
Raiders defeat Denver; AND
Pittsburgh loses at Baltimore; AND
Tennessee loses at Houston; AND
Indianapolis wins at Jacksonville
Miami loses/ties at New England
OR (if Miami wins)
Los Angeles Chargers win/tie at Kansas City; OR Chicago wins/ties at Minnesota; OR Detroit wins/ties vs. Green Bay.

David Taub has spent most of his career in journalism behind the scenes working as a TV assignment editor and radio producer. For more than a decade, he has worked in the Fresno market with such stops at KSEE-24, KMJ and Power Talk 96.7. Taub also worked the production and support side of some of TV sports biggest events including the Super Bowl, the NBA Finals and NASCAR to name a few. Taub graduated from the University of Michigan with dual degrees in communications and political science. You can contact David at 559-492-4037 or at Send an Email