Pittsburgh Steelers outside linebacker T.J. Watt holds the second-best odds in the league to win NFL Defensive Player of the Year at 7/4, courtesy of BetOnline.
Watt’s nine sacks are tied for the third-most in the NFL, and he is tied for second with 14 tackles-for-loss. He also has 32 total tackles, an interception and a league-leading 30 hits on the quarterback.
Three other Steelers defenders are also listed as contenders for the award, with defensive end Cam Heyward (25/1), defensive end Stephon Tuitt (25/1) and safety Minkah Fitzpatrick (33/1) all trailing Watt.
The only defender with better odds than Watt is Pittsburgh native and Los Angeles Rams defensive tackle Aaron Donald, who is seeking his fourth Defensive Player of the Year award. Donald leads the NFL with ten sacks.
Watt finished tied for third in NFL Defensive Player of the Year voting in 2019 with Tampa Bay Buccaneers linebacker Shaw Barrett. Watt led the league with eight forced fumbles and recorded 14.5 sacks, but New England Patriots cornerback Stephon Gilmore ultimately took home the honor.
Watt has been selected to two Pro Bowls, and was named First-Team All-Pro in 2019. He has also received NFL Defensive Player of the Week honors three times.
His brother J.J. of the Houston Texans saw his odds shorten to 33/1 following a dominant showing on Thanksgiving against the Detroit Lions, where he recorded four tackles and a pick-six.
Watt will look to improve upon his candidacy when the Steelers host the Baltimore Ravens on Wednesday afternoon. He has 3.5 sacks and two forced fumbles in seven career games versus Baltimore.