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AFC Playoff Picture: New York Jets Tie Most Wins Since 2013

Brad Penner-USA TODAY Sports

It's a beautiful day in the neighborhood. The New York Jets are 8-5, tying the most wins they've had in a season since 2013, when they went 8-8.

In the AFC East, we have the following:

1. New England Patriots (11-2)

2. New York Jets (8-5)

3. Buffalo Bills (6-7)

4. Miami Dolphins (5-7)

Here's the current playoff teams:

1. New England Patriots (11-2)

2. Cincinnati Bengals (10-3)

3. Denver Broncos (10-3)

4. Indianapolis Colts (6-7)

5. Kansas City Chiefs (8-5)

6. New York Jets (8-5)

If the Jets, Chiefs, and Pittsburgh Steelers (who are also 8-5) win out the three remaining games in the season, the Chiefs and Steelers will make it to the playoffs, as they will have the tiebreakers over the Jets. It's important the Steelers lose at least once in the final three games. As a result, since the Jets play the Dallas Cowboys, a NFC team, next week, while the Steelers play an AFC opponent, a Steelers loss is more important than a Jets win for tiebreaking purposes.

According to one website, if the Jets win the three remaining games, their odds are roughly 95% in favor of making the playoffs. If they go 2-1, the odds are 43%. If they go 1-2, it's 9%. In short... win. And the rest will take care of itself.