Bucks guard Jrue Holiday has been named the NBA’s Teammate of the Year for the 2021/22 season, the league announced today in a press release.
The Twyman-Stokes Teammate of the Year award, introduced in 2012/13, is voted on by NBA players after a panel of league executives selects 12 finalists (six from each conference). A total of 306 players submitted ballots this season.
The winner is deemed to be the best teammate based on his “selfless play, on- and off-court leadership as a mentor and role model to other NBA players, and commitment and dedication to team.”
As we relayed last month, Holiday, DeMar DeRozan, Rudy Gay, Jeff Green, Udonis Haslem, Andre Iguodala, Jaren Jackson Jr., Kevin Love, Boban Marjanovic, Chris Paul, Fred VanVleet, and Grant Williams were this year’s nominees.
Holiday received 39 first-place votes and 964 total points, narrowly edging out Marjanovic, who got 48 first-place votes but just 936 total points. DeRozan, Green, and Paul rounded out the top five.
It’s the second time Holiday has won the award, making him the first player to earn the honor more than once — he first won it in 2020 when he was a Pelican.
Holiday, 31, averaged 18.3 PPG, 6.8 APG, and 4.5 RPG on .501/.411/.761 shooting in 67 games (32.9 MPG) for the Bucks this season. He’s also a strong candidate to earn an All-Defensive nod this spring.
Ben Simmons robbed
He got screwed because him and Kyrie split the vote.
Westbrook enters the chat.
Grayson Allen got all the Bulls votes.
New HR poll: Worst teammate of the year? NBA won’t do it, we should.
Should we just call it the Ben Simmons Award?
I’m sorry, but Kevin Love was a finalist? The same Kevin Love who threw a temper tantrum in front of his team earlier this season?
Again?— or was that last year. This year he was planned as a backup from the gitgo.
Yeah but that was last year, this season he embraced his role and mentored the young guys.
Nobody beats boban