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Raiders DE Lamarr Houston wins AFC Defensive Player of the Week

The Raiders pulled out an overtime victory against the Jaguars on Sunday, and Lamarr Houston is getting recognized for his defensive efforts.


Oakland Raiders defensive end Lamarr Houston is the AFC Defensive Player of the Week for his performance against the Jacksonville Jaguars on Sunday, according to Vic Tafur of the San Francisco Chronicle. Houston helped the Raiders to a 26-23 victory in overtime over Jacksonville.

The third-year player out of Texas was all over the field for the Raiders. Houston recorded eight tackles, five of which were solo. He also posted one sack and forced a fumble in overtime that set the Raiders up for their game-winning field goal.

Sunday was the best statistical performance of Houston's career. The eight tackles were a career-high, and he had not recorded a sack and forced a fumble in the same game since Week 8 of his rookie season against the Seattle Seahawks.

The win put the Raiders at 2-4 on the season, just one game behind the Denver Broncos and San Diego Chargers for the AFC West lead.