MILWAUKEE, Wis. — This week's NBA Eastern Conference Player of the Week is none other than Milwaukee Bucks guard Jrue Holiday!
The Bucks announced Holiday's recognition Monday, saying it's the second time he has earned Player of the Week honors in his career. He joins Giannis Antetokounmpo as Bucks to be named Player of the Week this season.
Holiday led the Bucks to a 2-1 record last week, averaging 33.3 points, 9.3 assists, 4.7 rebounds, and 1.7 steals per game. He shot 56.9% from the field and 47.8% from the three-point line.
The Bucks guard also set a new season-high in scoring last week. First, he scored 35 points in a win over the Pacers, and then beat that with 37 points on Tuesday. And if you still need proof that he deserves Player of the Week, Holiday also trallied two double-doubles last week.
It's Holiday's third season with the Bucks, and 14th in the NBA. Overall he is averaging 19.6 points, 7.5 assists, 4.9 rebounds and 1.4 steals in 32.7 minutes per game. He's shooting 46.6% overall, and 38.4% from the three-point line.
