Although neither the Texans nor Washington finished the 2015 season at or below .500, both will host teams in the wild-card round having better records. In Houston, the 11-5 Chiefs visit the 9-7 Texans. In Washington, the home team is 9-7 . . . and the Packers are 10-6.
It’s not an uncommon phenomenon. In all, it has happened eight times from 2010 through 2014.
Last year, the 7-8-1 Panthers hosted the 11-5 Cardinals. The team with the lesser record won.
In 2013, the 12-4 49ers had to travel to Green Bay, to face the 8-7-1 Packers. The 49ers won, narrowly.
For each of the three years before that, each conference had a road team with a better record in the wild-card round. Baltimore, which eventually won the Super Bowl, started the trek as a 10-6 home team against the 11-5 Colts. Washington, also 10-6 that year, lost to the 11-5 Seahawks in the RGIII blown-out-knee game.
In 2011, the 9-7 Giants dismantled the 10-6 Falcons, 24-2 — and the 8-8 Broncos unleashed Tebowmania on the 12-4 Steelers, winning 29-23 in overtime.
In 2010, the 10-6 Chiefs lost soundly at home to the 12-4 Ravens. That same year, however, the 7-9 Seahawks upended the 11-5 Saints, thanks to the original Beastquake run.
Is it fair for a team to host an opponent with a better record? Fair or not, it’s been happening for years, thanks to the preferred position on the playoff tree that goes to a team that wins a division, regardless of record.
And it’s unlikely to change, in part because the NFL places heavy emphasis on winning a division. And in part because the teams realize that sometimes they benefit from this quirk in the rules — and sometimes they don’t.
For most fans, it’s only an issue when the home team finishes at or below .500. This year, it seemed inevitable for weeks that the champions of both the AFC South and NFC East would land into that category.