The Redskins sit at 2-2 with their season a quarter of the way over. They are tied for second place in the NFC East with the Dallas Cowboys. The Eagles lead the way at 3-1 and the New York Giants are bringing up the rear at 0-4.
Unless Philadelphia runs away with it, this should be a competitive division down to the wire. It seems likely that 10 wins will be enough to win it, as long as you have the right mix of tiebreakers. Doing the math, that means that the Redskins will need to go 8-4 the rest of the way to take the NFC East and get the home playoff game that goes with a division title.
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At first glance that looks to be a daunting task, winning at least two games for every loss. Let’s take a look at a formula that might get them where they want to be.
Beat the Eagles—They already have lost to the Eagles once. A second loss in Philly on Monday night in Week 6 would put them in a bad spot in tiebreakers.
Win three of four against the Giants and Cowboys—In a scheduling quirk, the Redskins will be done with one team in the division before facing either of the other two. Both teams made the playoffs in 2016 but both have proven to be very vulnerable this year. Taking three of four would probably consist of sweeping the Giants and splitting with Dallas but the Redskins will take it any way they can get it.
Don’t let an inferior team beat you—The way things look right now, Washington should be a solid favorite against the 49ers and Cardinals at home and at the Chargers. The home game against the Vikings may fall into this category as well. They need to take three out of these four games.
Pull off an upset—It looks like the Redskins will be underdogs on the road against the Saints and Seahawks and at home against the Broncos, although who really knows what either team might have to play for in Week 16. If they get the four division wins and three of the four against likely underdogs, they will need to take one of those three games to get to 10 wins.
Do you think they can do it? Vote in our poll.