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Giannis Antetokounmpo chosen NBA defensive player of the year

Bucks forward Giannis Antetokounmpo (34) blocks a shot by Magic guard D.J. Augustin during a playoff game Aug. 24, 2020.
Bucks forward Giannis Antetokounmpo (34) blocks a shot by Magic guard D.J. Augustin during a playoff game Monday.
(Ashley Landis / Associated Press)

Milwaukee star Giannis Antetokounmpo claimed his first piece of hardware this season, winning the NBA’s defensive player of the year award in a runaway vote. It’s likely not his last major honor — Antetokounmpo is widely expected to be named the league’s most valuable player for a second consecutive season.

To earn the top defensive honor, Antetokounmpo beat out Lakers star Anthony Davis and two-time winner Rudy Gobert of Utah. Antetokounmpo received 75 of the possible 100 first-place votes from selected media and broadcasters.

Clippers guard Patrick Beverley received one of the other first-place votes and finished sixth. Clippers forward Kawhi Leonard received five third-place votes and finished ninth.

The Bucks finished with the most efficient defense in the league this season, holding teams to 102.5 points per 100 possessions, more than 2.0 points better than any other team. The Lakers, anchored by Davis’ defense, held teams to 106.1 points per 100 possessions, third in the league.

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