Milwaukee Bucks point guard Jrue Holiday received the Sportsmanship of the Season award, becoming the first franchise player to earn the honor, the North American Basketball League (NBA) announced on Friday.
This award is presented annually to the player who best represents the ideals of sportsmanship on the field, as voted by current NBA players.
NBA executives selected six finalists from 30 nominees, one per franchise, and nearly 350 players voted on a weighting system, with 11 points for first place and one point for sixth.
The 30-year-old American won with 130 votes in first place for 2,752 points in total against 2,474 to his runner-up Kemba Walker, the leader of the Boston Celtics, already winner in 2017 and 2018, while playing in Charlotte . Miami Bam Adebayo’s interior completes the podium.
Holiday, a 12-year NBA veteran, last year won the Best Teammate of the Year award for his leadership, selfless play and dedication to his team.
Jrue Holiday, who arrived in Wisconsin last November, is averaging 17.7 points, 6.1 assists and 4.5 rebounds per game this season for the Bucks, against the Brooklyn Nets in the second round of the NBA play-offs. from Saturday.