It all starts with the Redskins winning out the rest of the way, ending the season with a 10-6 record. Since 2002, the year the league expanded to 32 teams, 162 clubs have won 10 or more games; 152 (94 percent) of those made the playoffs. Washington will have to produce road victories over the Dallas Cowboys and Los Angeles Chargers in Weeks 13 and 14, win at home against the Arizona Cardinals and Denver Broncos in Weeks 15 and 16 and end its season with another victory over the Giants at MetLife Stadium.
Using the same win probabilities that fuel our weekly power rankings, the most likely win is the game against the Denver Broncos (73.1 percent), who can’t seem to get consistent quarterback play from Trevor Siemian, Brock Osweiler or second-year pro Paxton Lynch, who left his debut Sunday due to injury. Overall, the Broncos are scoring 1.4 points per drive, the fifth-lowest mark this season, while allowing 1.9, the 13th most.
The hardest game remaining for Washington is against the Chargers (34.5 percent win probability). According to the game charters at Pro Football Focus, the Chargers have the second-best pass rush in the NFL. Edge rushers Joey Bosa and Melvin Ingram have combined for 113 total sacks, hits and hurries in 2017, with 52 stops at or behind the line of scrimmage.
Win those two games though and things take a very optimistic turn.
But the Redskins need help, too. A 10-6 record won’t win the Redskins the NFC East. It would put them at the mercy of the wild-card race which, if the season ended today, would be won by the Carolina Panthers and Atlanta Falcons. So in addition to Washington winning out, it will need the Falcons, plus the Seattle Seahawks, Detroit Lions and Green Bay Packers. to lose at least seven games. (Note: The Panthers can finish at 10-6 and would lose a tiebreaker to Washington based on conference winning percentage.) Here’s how that happens based on the most likely losses for each team over the remaining five weeks.
Baltimore Ravens defeat the Detroit Lions
Probability: 58 percent
Minnesota Vikings defeat the Atlanta Falcons
Probability: 61 percent
Philadelphia Eagles defeat the Seattle Seahawks
Probability: 71 percent
New Orleans Saints defeat the Carolina Panthers
Probability: 63 percent
Minnesota Vikings defeat the Carolina Panthers
Probability: 55 percent
Tampa Bay Buccaneers defeat the Detroit Lions
Probability: 38 percent
New Orleans Saints defeat the Atlanta Falcons
Probability: 57 percent
Jacksonville Jaguars defeat the Seattle Seahawks
Probability: 64 percent
Los Angeles Rams defeat the Seattle Seahawks
Probability: 57 percent
Tampa Bay Buccaneers defeat the Atlanta Falcons
Probability: 34 percent
Minnesota Vikings defeat the Green Bay Packers
Probability: 75 percent
Atlanta Falcons defeat the Carolina Panthers
Probability: 51 percent
If the above scenario happens, the Redskins would most likely finish as the first wild-card team ahead of the 10-6 Panthers, due to the third conference tiebreaker (best win-loss-tie percentage in conference games). If things play out in this fashion, Washington will have seven wins over NFC teams while the Panthers would only have six.
|NFC team||Projected wins in 2017||Playoff seed|
|Philadelphia Eagles||14||No. 1 seed|
|Minnesota Vikings||13||No. 2 seed|
|New Orleans Saints||12||No. 3 seed|
|Los Angeles Rams||12||No. 4 seed|
|Green Bay Packers||7|
|Tampa Bay Buccaneers||7|
|New York Giants||3|
|San Francisco 49ers||2|
Speaking of tiebreakers, for the purposes of this projection we didn’t incorporate them beyond the final application between the Redskins and Panthers. With so much still in flux, it’s virtually impossible to project all the permutations for some of the NFL’s tiebreakers like strength of victory and strength of schedule. But it is a little easier to calculate the first two conference tiebreakers — head-to-head victories and conference winning percentage.
The biggest notes there as far as the Redskins are concerned is that they hold the head-to-head tiebreaker over the Seahawks and that winning out will give them seven NFC victories. A 10-6 record by the Falcons would mean they’d have at least nine conference wins, giving them the tiebreaker over the Redskins. The Lions need four wins to get to 10-6, but if their one loss is to an NFC team they would tie the Redskins at seven conference wins and proceed further down the tiebreaker chain. Should the Panthers finish at 10-6, they would only have six NFC victories, putting Carolina behind Washington in a head-to-head tiebreaker. Things would get much, much more complicated should three teams finish at 10-6 for the wild-card slots.
But in the scenario presented above, Washington would make the playoffs. You’d be wise, however, not to get your hopes up. The odds of all 17 events occurring is approximately 22,000 to 1 — nearly twice the odds of you being injured by a toilet — but it gets the Redskins one of the two wild-card spots 98 percent of the time, with two-thirds of the simulations seeing Washington as the top wild-card team in the NFC.
So yes, I’m saying there is a chance; just don’t bet the ranch on it.
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