FILE – Jrue Holiday #21 of the Milwaukee Bucks handles the ball during a game against the Boston Celtics at Fiserv Forum on March 24, 2021 in Milwaukee, Wisconsin. Stacy Revere/Getty Images/AFP
NEW YORK – Milwaukee Bucks guard Jrue Holiday was named the 2021 winner of the NBA Sportsmanship Award, the league announced Friday, becoming the club’s first player to receive the honor.
The award is presented annually to the player who best represents the ideals of sportsmanship on the court as voted upon by current NBA players.
NBA executives selected six finalists from 30 nominees, one per club, and nearly 350 players voted in a weighted system with 11 points for first place to one point for sixth.
Holiday won with 130 first-place votes and 2,752 total points to 2,474 for runner-up Kemba Walker of the Boston Celtics, who won the award in 2017 and 2018 while with Charlotte.
Miami’s Bam Adebayo was third, followed by Sacramento’s Harrison Barnes, San Antonio’s Derrick White and Minnesota’s Josh Okogie.
Holiday, a 12-year NBA veteran, won last year’s NBA Teammate of the Year Award for leadership, unselfish play and commitment to his club.
The 30-year-old American, named to the NBA’s All-Defensive Team in 2018 and 2019, has averaged 17.7 points, 6.1 assists and 4.5 rebounds a game this season for the Bucks, who are set to face the Brooklyn Nets in the second round of the NBA playoffs starting Saturday.
Milwaukee obtained Holiday from New Orleans as part of a four-team trade last November.