Milwaukee Bucks forward Giannis Antetokounmpo (34) blocks a shot by Detroit Pistons big man Thon Maker. Antetokounmpo becomes the second player in Bucks history to win the NBA's Defensive Player of the Year Award. File Photo by Tannen Maury/EPA-EFE
Aug. 25 (UPI) -- Milwaukee Bucks superstar Giannis Antetokounmpo was named the 2019-20 NBA Defensive Player of the Year on Tuesday.
Antetokounmpo, 25, beat out Los Angeles Lakers star Anthony Davis and Utah Jazz big man Rudy Gobert to earn the award for the first time in his seven-year career. He captured the award in a landslide, receiving 75 first-place votes and earning 432 total points.
The "Greek Freak" becomes the second player in Bucks history to win the award, joining two-time winner Sidney Moncrief (1983-84).
Davis finished second with 14 first-place votes (200 points) and Gobert -- who won the award in each of the past two seasons -- came in third with six first-place votes (187 points).
Ben Simmons (fourth), Bam Adebayo (fifth), Patrick Beverley (sixth), Marcus Smart (seventh), Andre Drummond (eighth), Kawhi Leonard (ninth), Brook Lopez (10th), Hassan Whiteside (11th) and Jarrett Allen (12th) also received votes.
Antetokounmpo, a two-time NBA All-Defensive selection, anchored the best defensive team in the NBA in the Bucks. He averaged 13.6 rebounds, one block and one steal in 30.4 minutes per game this season.
Antetokounmpo -- the reigning NBA Most Valuable Player -- also is considered the favorite to win the league's MVP award this year. He would join Michael Jordan (1987-88) and Hakeem Olajuwon (1993-94) as the only players in league history to win MVP and DPOY in the same season.