As the 2025 NBA playoffs roll along, the league continues to announce winners for several of its annual awards. On Thursday, it announced that Boston Celtics guard Jrue Holiday won the 2024-25 NBA Sportsmanship Award. Shai Gilgeous-Alexander finished in fourth place.
Presented annually since the 1995-96 season, the NBA Sportsmanship Award honors a player who best represents the ideals of sportsmanship on the court. Each NBA team nominated one of its players for the award. From the list of 30 team nominees, a panel of league executives selected six finalists, one from each division. Current NBA players selected the winner from the list of six finalists.
Gilgeous-Alexander tallied 2,270 voting points. He had 61 first-place votes, 36 second-place votes, 80 third-place votes, 87 fourth-place votes, 75 fifth-place votes and 55 sixth-place votes. The full voting results can be viewed below:
The Oklahoma City Thunder await their Round 2 opponent. They’ll enjoy over a week off as the Denver Nuggets and LA Clippers fight it out to determine who advances from their Round 1 playoff series.
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