Gobert named NBA Defensive Player of the Year
NEW YORK: Minnesota Timberwolves center Rudy Gobert won the 2024 NBA Defensive Player of the Year award on Tuesday (Wednesday in Manila), matching a league record by taking the honor for the fourth time.
The 31-year-old Frenchman previously captured the award in 2018, 2019 and 2021 while with the Utah Jazz and this season was a key player on a T-Wolves squad that was the NBA’s top-ranked defensive club.
Gobert matched the all-time record for four Defensive Player of the Year awards won by Detroit’s Ben Wallace (2002, 2003, 2005 and 2006) and Dikembe Mutombo in 1995, 1997, 1998 and 2001 with Denver, Atlanta and Philadelphia.
The 7-foot-1 (2.16 m) star missed Minnesota’s playoff victory at Denver on Monday so he could be with girlfriend Julia Bonilla, who gave birth to their son earlier in the day.
“Romeo is the name, and he’s doing great,” Gobert told US NBA telecaster TNT. “A lot of blessings. Just really grateful.”
Gobert received 72 first-place votes from a media panel with fellow Frenchman Victor Wembanyama, the 20-year-old San Antonio Spurs prodigy named Monday as the NBA Rookie of the Year, second with 19 first-place votes and Miami’s Bam Adebayo third with three.
Gobert averaged 14.0 points, 12.9 rebounds, 2.1 blocked shots and 1.3 assists a game this season for the Timberwolves.
“Rudy has driven the defensive culture here,” Timberwolves coach Chris Finch said. “It’s a testament to his impact, his presence and what he has infused into the team of how important defense is and how great it can be when we play it.”
Gobert said the secret to that success was building a group of players determined to dominate defensively.