Bucks’ Jrue Holiday wins NBA teammate of year award again

Dallas Mavericks v Milwaukee Bucks
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MILWAUKEE — Milwaukee Bucks guard Jrue Holiday has been named the NBA’s teammate of the year for the second time in three seasons, the league announced Tuesday.

A panel of league executives selected 12 finalists for the honor, which is officially named the Twyman-Stokes Teammate of the Year Award. More than 300 current NBA players voted on the winner from that list of 12 finalists.

Dallas Mavericks center Boban Marjanovic finished second in the voting and Chicago Bulls guard/forward DeMar DeRozan was third. Holiday received 39 first-place votes and 964 total points. Marjanovic had more first-place votes – 48 – and received 936 total points. DeRozan had 34 first-place votes and 898 points.

Players got 10 points for a first-place vote, seven for second place, five for third, three for fourth and one for fifth.

Holiday also won this award in 2019-20 when he was with the New Orleans Pelicans. He won the NBA Sportsmanship Award last season, when he also was a finalist for the inaugural Kareem Abdul-Jabbar Social Justice Champion Award.

The Twyman-Stokes Teammate of the Year Award recognizes a player deemed the best teammate based on selfless play and dedication to his team, as well as leadership on and off the court as a mentor and role model to other players.