DeAndre Jordan Wins 2025-26 NBA Teammate of the Year Award over Jayson Tatum, Jalen Brunson

New Orleans Pelicans center DeAndre Jordan has been named the recipient of the Twyman-Stokes Teammate of the Year Award for the 2025-26 season.

The NBA announced the winner on Wednesday, with Jordan narrowly finishing ahead of Jrue Holiday of the Portland Trail Blazers and Jeff Green of the Houston Rockets.

Jordan led the way with 66 first-place votes and 1,445 total points. Holiday finished second with 1,437 points, while Green had 1,420 points.

Other finalists included Jalen Brunson, Jayson Tatum, De'Aaron Fox and Desmond Bane.

Originally established during the 2012-13 season, the Teammate of the Year Award recognizes the player who is deemed to be the best teammate based on their selfless play, leadership on and off the court, being a mentor and role model to other players and commitment to their team.

Unlike the NBA's other awards that have three finalists, there are a total of 12 finalists for the Teammate of the Year Award. Each team nominates one player, which is then whittled down to 12 by a panel of league executives.

The winner is determined by a vote from their NBA fellow players. Holiday is the most decorated teammate since the award's inception, having won three times in his career. Mike Conley Jr., who won in 2019 and '24, is the only other player who was won multiple times.

Jordan is a first-time winner after being named a 2024-25 finalist. Stephen Curry won the award last year.