Jrue Holiday wins 2021 NBA Sportsmanship Award

Bucks guard Jrue Holiday and Heat center Bam Adebayo
Stacy Revere/Getty Images

Jrue Holiday is an excellent player, which is why the Bucks signed him to a massive contract extension.

But Holiday brings far more than just his basketball talent to games.

He competes hard on both ends of the floor. He supports his teammates. He treats opponents – even locally loathed opponents – with respect.

Which is why – last season’s Teammate of the Year – Holiday has now won the NBA’s Sportsmanship Award.

NBA release:

Milwaukee Bucks guard Jrue Holiday has received the Joe Dumars Trophy as the winner of the 2020-21 NBA Sportsmanship Award, the NBA announced today.  The annual award is designed to honor a player who best represents the ideals of sportsmanship on the court.

Each NBA team nominated one of its players for the NBA Sportsmanship Award.  From the list of 30 team nominees, a panel of league executives selected one finalist from each of the NBA’s six divisions.  Current NBA players selected the winner from the list of six finalists, with nearly 350 players submitting their votes through confidential balloting conducted by the league office.

Full voting results with first-, second-, third-, fourth-, fifth- and sixth-place votes and total voting points:

1. Jrue Holiday (Bucks): 130-71-56-38-26-23-2,752

2. Kemba Walker (Celtics): 74-89-67-46-47-19-2,474

3. Bam Adebayo (Heat): 48-73-76-63-42-41-2,199

4. Harrison Barnes (Kings): 49-49-62-62-81-41-2,008

5. Derrick White (Spurs): 31-21-54-79-88-68-1,635

6. Josh Okogie (Timberwolves): 11-40-28-55-59-151-1,280