MILWAUKEE, Wis. – 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.