T.J. Watt seems unlikely to play in Sunday’s Steelers season finale against the Cleveland Browns, but he remains the favorite to become the NFL’s Defensive Player of the Year after this season.
Watt did not practice all week, along with starting quarterback Ben Roethlisberger and center Maurkice Pouncey. The Steelers have already announced that Roethlisberger will not play against the Browns and the Post-Gazette has reported that Pouncey and Watt will join him on the sidelines in Cleveland.
Watt, the Steelers’ MVP for the second straight season, leads the NFL in sacks, tackles for loss and quarterback hits entering the final week of the season, and his hold on at least two of those marks seems fairly secure.
It’s on that resume that SportsBetting.com lists Watt as the favorite to win Defensive Player of the Week entering the final week of the regular season. Watt is a 1/2 favorite, following by two-time winner Aaron Donald of the Los Angeles Rams at 2/1. Xavier Howard (8/1) and Myles Garrett (20/1) remain on the board as long shots.
The award has been given by the Associated Press to the top defensive player in the league since 1971. The Steelers have won the award seven times before, the most of any other franchise, and have done so with six different winners, most recently Troy Polamalu. Mel Blount, James Harrison, Jack Lambert and Rod Woodson also each won the award once while Joe Greene is one of seven two-time winners.
Donald and Watt’s brother J.J. Watt are also multi-time recipients of the award. J.J. Watt won in 2012, 2014 and 2015. Donald won in 2017 and 2018. New England Patriots cornerback Stephon Gilmore won in the 2019.