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Pelicans’ Jordan wins NBA’s teammate of the year award

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New Orleans Pelicans forward Zion Williamson, center left, talks to center DeAndre Jordan, center right, while on the bench during the first half of an NBA basketball game against the San Antonio Spurs, Friday, Oct. 24, 2025, in New Orleans. (AP Photo/Peter Forest)

NEW YORK (AP) — New Orleans’ DeAndre Jordan won the Twyman-Stokes Teammate of the Year award Wednesday, edging Portland’s Jrue Holiday and Houston’s Jeff Green.

Jordan finished with 1,445 points in voting by 385 current NBA players. Holiday — a three-time winner — finished with 1,437 points and Green had 1,420.

A panel of league executives selected 12 finalists, six from each conference, to be considered for the honor. Jordan adds the award to a list of accolades that includes three All-NBA selections, two All-Defensive Team picks, one All-Star nod, an Olympic gold medal from the 2016 Rio de Janeiro Games and a championship with Denver in 2023.

The award pays tribute to the player deemed the league’s best teammate “based on selfless play, leadership on and off the court as a mentor and role model to other NBA players, and commitment to the team.”

It has been presented since 2013 and is named for Jack Twyman and Maurice Stokes, teammates on the Rochester/Cincinnati Royals from 1955-58. Stokes suffered a brain injury in the final game of the 1957-58 regular season, fell into a coma days later and was paralyzed. Twyman became Stokes’ legal guardian and advocate, and supported him for the rest of his life.

Other players nominated for the award this season were Toronto’s Garrett Temple, Charlotte’s Pat Connaughton, New York’s Jalen Brunson, Boston’s Jayson Tatum, San Antonio’s DeAaron Fox, Detroit’s Duncan Robinson, Oklahoma City’s Jaylin Williams, Orlando’s Desmond Bane and the Los Angeles LakersMarcus Smart.

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AP NBA: https://apnews.com/nba

The Associated Press