Bucks' Jrue Holiday named NBA teammate of year for 2nd consecutive season
Jrue Holiday and Khris Middleton (Photo by John Fisher/Getty Images)
MILWAUKEE - Milwaukee Bucks guard Jrue Holiday has been selected the NBA’s teammate of the year for the second straight year and third time in the last four seasons.
The NBA announced on Thursday the voting results for the Twyman-Stokes Teammate of the Year Award, which recognizes the player deemed the best teammate based on selfless play, leadership and dedication to a team.
Current players voted on the winner from a list of 12 finalists. The finalists were determined by a panel of league executives.
RELATED: Check out the new and improved FOX Sports app
Brooklyn’s Mikal Bridges finished second and Golden State’s Stephen Curry was third in the voting. The other finalists, in order of finish, were New York’s Derrick Rose, Miami’s Udonis Haslem, Portland’s Damian Lillard, Sacramento’s Harrison Barnes, Phoenix’s Damion Lee, Cleveland’s Darius Garland, Boston’s Grant Williams, Memphis’ Jaren Jackson Jr. and Denver’s Aaron Gordon.
Lillard won the award in 2021, the only time in the last four seasons it hasn’t gone to Holiday. Holiday won the award as a member of the New Orleans Pelicans in 2020 before earning it again with the Bucks the last two seasons.
The award is named after Jack Twyman and Maurice Stokes, who were Cincinnati Royals teammates from 1955-58. Stokes suffered an injury during the 1957-58 season that led to him falling into a coma, becoming permanently paralyzed and being diagnosed with post-traumatic encephalopathy, a brain injury. Twyman became his legal guardian and advocate.