Bucks' Giannis Antetokounmpo Wins Defensive Player of the Year

Bucks star Giannis Antetokounmpo has won the NBA's Defensive Player of the Year award.

Bucks star Giannis Antetokounmpo has won the NBA's Defensive Player of the Year award, the NBA announced Tuesday. 

The anchor of Milwaukee's stellar unit had 75 first-place votes, 18 second-place votes and three third-place votes. Lakers big Anthony Davis finished second in the voting while Jazz center Rudy Gobert, 76ers guard Ben Simmons and Heat forward Bam Adebayo finished third, fourth and fifth, respectively. 

"Congrats to Giannis on being named Defensive Player of the Year," head coach Mike Budenholzer said on TNT's Inside the NBA. "His commitment to defending, his commitment to winning is beyond incredible. He impacts the game with his blocked shots, with his rebounding, with his ability to guard all five positions, his chase down blocks, his challenge to do everything defensively. His talent is beyond special and combined with his team, it's made for a very special moment and for us to recognize him as the Defensive Player of the Year."

In games played through March 11, Antetokounmpo led the Bucks to the NBA’s best defensive rating as they allowed 101.6 points per 100 possessions. From the start of the regular season through the March 11 stoppage, Milwaukee’s opponents also shot an NBA-low 41.3% from the field.

Antetokounmpo won the 2018–19 MVP award and is favored to take home the 2019–20 award. Only Michael Jordan and Hakeem Olajuwon won both the MVP and Defensive Player of the Year award in the same season. Just five players in league history have won both awards in the same career. 

The Bucks look to close out the Magic on Wednesday and advance to the Eastern Conference semifinals. Tipoff for Wednesday's game is at 4 p.m. ET.