Breaking down stretch run for AFC playoff contenders, Raiders place in it

It’s crazy to think that 11 games into the season and Oakland is engulfed in playoff talk — but is it safe to assume the Raiders are in? At 9-2, the Raiders are tied for the best record in the AFC and are the No. 2 seed due to tie-breakers.

In the AFC West, the Raiders hold a one-game lead over the Chiefs and a two-game lead over the Broncos. At the moment, Kansas City holds the tie-breaker over Oakland and Denver, while the Raiders hold the tie-breaker over Denver — but all this could change as all three teams match-up against one another down the stretch.

While many assume the AFC West (and the three teams mentioned above) will claim three playoff spots, there are a number of scenarios in which that doesn’t happen. At the moment, 10 teams are either in the playoffs or within 1.5 games of the final wild card spot.

With that in mind, let’s look at what the Raiders are up against down the stretch compared to the competition both within the division and for the wildcard...

Patriots* (9-2): LA, BAL, @DEN, NYJ, @MIA (.491 opponent win percentage)

Raiders* (9-2): BUF, @KC, @SD, IND, @DEN (.564 opponent win percentage)

Chiefs (8-3): @ATL, OAK, TEN, DEN, @SD (.607 opponent win percentage)

Broncos (7-4): @JAX, @TEN, NE, @KC, OAK (.607 opponent win percentage)

Dolphins (7-4): @BAL, ARI, @NYJ, @BUF, NE (.509 opponent win percentage)

Ravens* (6-5): MIA, @NE, PHI, @PIT, @CIN (.555 opponent win percentage)

Steelers (6-5): NYG, @BUF, @CIN, BAL, CLE (.425 opponent win percentage)

Texans* (6-5): @GB, @IND, JAX, CIN, @TEN (.363 opponent win percentage)

Bills (6-5): @OAK, PIT, CLE, MIA, @NYJ (.446 opponent win percentage)

Titans (6-6): BYE, DEN, @KC, @JAX, HOU (.522 opponent win percentage)

*team is currently leading their division

And in order of opponent winning percentage:

1. Chiefs (8-3): @ATL, OAK, TEN, DEN, @SD (.607 opponent win percentage)

1a. Broncos (7-4): @JAX, @TEN, NE, @KC, OAK (.607 opponent win percentage)

3. Raiders* (9-2): BUF, @KC, @SD, IND, @DEN (.564 opponent win percentage)

4. Ravens* (6-5): MIA, @NE, PHI, @PIT, @CIN (.555 opponent win percentage)

5. Titans (6-6): BYE, DEN, @KC, @JAX, HOU (.522 opponent win percentage)

6. Dolphins (7-4): @BAL, ARI, @NYJ, @BUF, NE (.509 opponent win percentage)

7. Patriots* (9-2): LA, BAL, @DEN, NYJ, @MIA (.491 opponent win percentage)

8. Bills (6-5): @OAK, PIT, CLE, MIA, @NYJ (.446 opponent win percentage)

9. Steelers (6-5): NYG, @BUF, @CIN, BAL, CLE (.425 opponent win percentage)

10. Texans* (6-5): @GB, @IND, JAX, CIN, @TEN (.363 opponent win percentage)

A few interesting things to note about the remaining schedule before we dive into what all of this means for the Raiders. First: what you see above is good news for football fans — 66% of the games listed above (29/44) are between two teams currently in playoff contention, many of which feature two of the AFC teams listed above. At the very least the next five weeks will be exciting.

But what does it mean for the teams involved?

The good news for the Raiders of course is that they face the easiest schedule amongst the three teams vying for the AFC West crown according to opponent win percentage. While there aren’t any Clevelands gimmes, they do get both Indianapolis and Buffalo at home, who are 9th and 11th in the conference, and another match-up against 12th place San Diego.

This, of course, is a bit misleading, because on the flip side, Oakland’s two most important games (KC and Denver) are both on the road. While their overall schedule is ‘easier’, I’m sure they’d take the tougher schedule if it meant getting their divisional rivals at home.

From Oakland’s perspective, I think two wins guarantees them a playoff spot for the following reasons...

For Miami (currently second in wild card), it would require going 4-1 and overcoming Oakland’s advantage in tie-breakers. As it stands, Oakland is 6-1 in the AFC, while Miami is 5-3 (this metric is used as the primary tie-breaker in instances in which there is no head-to-head matchup). Should inter-conference record end in a tie (unlikely given who Miami would have to beat, but who knows), the tie-breaker would move to record against common opponents — where the Raiders also hold an advantage at the moment (3-0, compared to 2-1). Remember, though, lots of games still left to play.

Outside of Miami, two Oakland wins would require everyone else below Miami to go undefeated and would automatically eliminate Baltimore (head-to-head loss), Tennessee (already at six losses) and Houston (head-to-head loss) from wild card contention thanks to tie-breakers (although they could still win their division).

Basically: win twice and a lot would have to go perfectly wrong for Oakland to stay home from the playoffs. Win three times? Denver and Miami would have to go undefeated and earn all tie-breakers, while Kansas City would need to go at least 4-1.

So with all these hypotheticals in mind, let’s ask one last question: what’s most likely to happen?

As it stands, the Raiders are likely to be favored against Buffalo, San Diego and Indianapolis, and underdogs in Kansas City and Denver. Three wins from that bunch (while far from guaranteed) is well within the realm of possibilities. The more surprising outcome would actually be Oakland losing three of those games and putting their playoff spot in jeopardy.

As far as the division goes, things get complicated thanks to all the head-to-head stuff left to happen. Wins against Denver and Kansas City on the road would (obviously) all but guarantee Oakland the division crown barring a let down in their other three games. That said, Oakland could afford to stumble once or twice and still win the division if the chips fall the right way. But why risk it? The easiest way to lock this thing up is simple: win ‘em all.