Rhamondre Stevenson’s three-touchdown game earns league recognition

Rhamondre Stevenson ended his 2025 season on a high note, and the NFL took notice. On Wednesday afternoon, Stevenson was announced as the AFC Offensive Player of the Week for Week 18.

Against the Miami Dolphins in the regular season finale, Stevenson had one of the best games of his career. He carried the ball seven times for 131 yards and two touchdowns, and added another two catches for 22 yards and a score through the air. His performance helped key the Patriots' 38-10 win to close out the regular season.

Stevenson is the sixth Patriots player to win a Player of the Week award this season, and second on offense. He joins Antonio Gibson (special teams Week 2), Marcus Jones (special teams Week 4), K'Lavon Chaisson (defense Week 7), Andy Borregales (special teams Week 12), and Drake Maye (offense Week 13). Prior to this year, the Patriots hadn't won a Player of the Week award since 2022.

This is the first time the Patriots have won six or more Player of the Week awards in a season since 2017 (Tom Brady x3, Dion Lewis x2, Stephen Gostkowski, Rob Gronkowski), which was also the last time a Patriots running back (Lewis) won the award. It's the first time in franchise history that six different players have won the award in the same season.

While his big game against the Dolphins is getting this recognition, Stevenson has been playing some of the best football of his career since returning from a toe injury in late November. He finished the regular season with 603 rushing yards at 4.6 yards per carry, with another 32 catches for 345 yards, and nine total touchdowns.