Wolves center Rudy Gobert named NBA Defensive Player of the Year
Rudy Gobert #27 of the Minnesota Timberwolves reacts during the second half of game three of the Western Conference First Round Playoffs at Footprint Center on April 26, 2024 in Phoenix, Arizona. The Timberwolves defeated the Suns 126-109. (Christian Petersen / Getty Images)
MINNEAPOLIS (FOX 9) - Minnesota Timberwolves center Rudy Gobert has been named as the 2024 Defensive Player of the Year, the NBA announced Tuesday.
This is Gobert's fourth time taking home the Hakeem Olajuwon Trophy. For the voting, he received 72 first-place votes out of 99 possible.
Gobert was the anchor of the NBA's best defense. The Wolves ranked first in the league in team defensive rating for the regular season.
Individually, Gobert averaged 2.1 blocks and 12.9 rebounds per game (second-best in the NBA) for the season. He also led the league in defensive win shares -- an advanced stat that factors in a player's individual defensive performance and how it impacts the team's defense. For context, Gobert's 2024 defensive win shares rating of 5.8 is the second-best of his career. (Fun fact: Anthony Edwards finished third in the NBA in defensive win shares as well).
Gobert beat out rookie sensation Victor Wembanyama (19 first-place votes) and Heat center Bam Adebayo (three first-place votes) for the award. He was also one of three Timberwolves players who were named finalists for end-of-season awards, alongside coach Chris Finch and Naz Reid. Reid took home the Sixth Man of the Year trophy, while Finch finished third in Coach of the Year voting.
The Wolves currently lead the Nuggets 2-0 in the second round of the NBA playoffs, spurred in part by their suffocating defense. However, Gobert missed Game 2 of the series after returning to Minnesota for the birth of his first child.