Noticing a trend, yet? For the third week in a row, a Buffalo Bills player has been awarded with an AFC player of the week honor—and this time, it was rookie defensive end Greg Rousseau.
Rousseau won the award for his part in stomping the Kansas City Chiefs in a 38-20 victory. He had several impact plays in the win—a sack, a tackle-for-loss, five tackles, and the first interception of his career. Rousseau is the fifth Bills rookie to win a Player of the Week award, joining Cornelius Bennett, C.J. Spiller, Greg Bell, and Josh Allen.
On the season, Rousseau leads the Bills with three sacks. He has 18 total tackles and four TFLs, as well as a pass defense and the aforementioned interception.
Last week, Tremaine Edmunds was named AFC Defensive Player of the Week for his role in the shutout of the Houston Texans. The prior week, Josh Allen won Offensive Player of the Week for scoring five touchdowns in a rout of the Washington Football Team.
In case you were wondering how Allen stacked up for the offensive award this time around, he was beaten out by Lamar Jackson. Jackson completed 37/43 passes for 442 yards and four touchdowns, and rushed 14 times for 62 yards, in a 31-25 comeback victory over the Indianapolis Colts.