MILWAUKEE | Monday night was a banner night for a couple of Milwaukee Bucks. Guard Malcolm Brogdon was named the 2017 NBA Rookie of the Year. Star forward Giannis Antetokounmpo was named the NBA’s Most Improved player.
Antetokounmpo beat out Utah Jazz center Rudy Gobert and Denver Nuggets center Nikola Jokic for the honor given during Monday’s NBA Awards. Brogdon, who was also selected to the NBA All-Rookie First Team earlier in the day, beat out Joel Embiid and Dario Saric of the Philadelphia 76ers. Brogdon becomes the first player who wasn’t a first-round pick to win NBA Rookie of the Year in the common-draft era.
Antetokounmpo was also named to the NBA’s All-Defensive Second Team earlier in the day.
Brogdon led rookies in assists (4.2 apg) and steals (1.12 spg) and ranked second in three-point field goal percentage (40.4) and free throw percentage (86.5). He was also third in field goal percentage (45.7) and fourth in scoring (10.2 ppg). A second round pick (36th overall) in the 2016 NBA Draft, Brogdon became one of just five rookies in NBA history to shoot 40.0 percent or better from 3-point range while averaging at least 4.0 assists per game. He recorded the first rookie triple-double in Bucks’ history, and the only one by a rookie during the 2016-17 season, when he scored 15 points with 12 assists and 11 rebounds at Chicago on Dec. 31.
Antetokounmpo averaged career highs in both blocks (1.9) and steals (1.6) per game during the 2016-17 season, which ranked sixth and 11th in the NBA respectively. He also ranked fifth in total blocks (151) and ninth in total steals (131). Antetokounmpo is the sixth Buck to be named to an NBA All-Defensive team and the first since Alvin Robertson was a First Team All-Defensive Team pick in 1990-91.