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Milwaukee Bucks’ Jrue Holiday wins NBA Sportsmanship Award

Apr 27, 2021; Charlotte, North Carolina, USA; Milwaukee Bucks guard Jrue Holiday (21) brings the ball up court against the Charlotte Hornets in the second half at Spectrum Center. The Milwaukee Bucks won 114-104. Mandatory Credit: Nell Redmond-USA TODAY Sports
Credit: Nell Redmond-USA TODAY Sports

Milwaukee Bucks guard Jrue Holiday received the Joe Dumars Trophy on Friday as the winner of the 2020-21 NBA Sportsmanship Award.

The annual award is presented to the player who best represents the ideals of sportsmanship on the court, with each of the league’s 30 teams nominating one of its own players.

Holiday received 130 of 343 first-place votes and earned 2,752 points in voting by NBA players. Boston’s Kemba Walker, a two-time winner of the award, finished second with 2,474 points and Miami’s Bam Adebayo was third with 2,199.

Sacramento’s Harrison Barnes (2,008), San Antonio’s Derrick White (1,635) and Minnesota’s Josh Okogie (1,280) were also finalists for the award.

The six finalists were awarded 11 points for each first-place vote, nine for each second-place vote, seven for each third-place vote, five for each fourth-place vote, three for each fifth-place vote and one for each sixth-place vote.

Holiday, 30, averaged 17.7 points, 6.1 assists and 4.5 rebounds in 59 games (56 starts) during his first season with the Bucks and 12th season in the NBA.

A first-round pick by Philadelphia (17th overall) in 2009, Holiday has career averages of 16.0 points, 6.4 assists and 4.0 rebounds in 772 games (696 starts) with the 76ers (2009-13), New Orleans Pelicans (2013-20) and Bucks.

The award is named for Dumars, a Hall of Fame guard for the Detroit Pistons who was the inaugural Sportsmanship Award recipient in 1995-96.

Related: NBA games today – Full offseason schedule, including start of NBA free agency

–Field Level Media

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