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Giannis Antetokounmpo, Eric Bledsoe named NBA All-Defensive First Team

Milwaukee Bucks forward Giannis Antetokounmpo and guard Eric Bledsoe have both been named to the 2018-19 NBA All-Defensive First Team, the league announced today. This is the first time since 1990-91 (Alvin Robertson) that the Bucks have had a player selected to the All-Defensive First Team and just the third time in franchise history that Milwaukee has had two players selected to the All-Defensive First Team in the same season.

Led by Antetokounmpo and Bledsoe, the Bucks were the top-rated defense in the NBA this season and held opponents to a league-low 43.3 field goal percentage. Antetokounmpo averaged 1.5 blocks per game – the second-highest mark of his career – while Bledsoe added 1.5 steals per contest.

For Antetokounmpo, this is his first selection to the All-Defensive First Team after being named Second Team All-Defense in 2016-17. This is the first time Bledsoe has been selected to an NBA All-Defensive Team in his career. Antetokounmpo and Bledsoe are the fifth and sixth players in franchise history to be named First Team All-Defense.

Antetokounmpo and Bledsoe are joined on the All-Defensive First Team by Rudy Gobert (Utah Jazz), Paul George (Oklahoma City Thunder) and Marcus Smart (Boston Celtics). The NBA All-Defensive Teams were selected by a global panel of 100 sportswriters and broadcasters. Players were awarded two points for each First Team vote and one point for each Second Team vote. Antetokounmpo received a total of 145 points, including 94 First Team votes, while Bledsoe tallied 100 points with 36 First Team votes.

Antetokounmpo is also a finalist for the Kia NBA Defensive Player of the Year Award, which will be revealed at the 2019 NBA Awards presented by Kia on Monday, June 24 in Los Angeles. 

Photo: Getty Images

via Bucks press release

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