Milwaukee Bucks guard Jrue Holiday has been named the Eastern Conference Player of the Week for Week 14 (Jan. 16-22), the NBA announced today.
Leading the Bucks to a 2-1 record last week, Holiday averaged 33.3 points, 9.3 assists, 4.7 rebounds and 1.7 steals per game while shooting 56.9% from the field and 47.8% from three. In back-to-back games, Holiday set a new season-high in scoring, first scoring 35 points in a win over the Pacers on Monday, before scoring a season-high 37 points in Tuesday’s win over the Raptors. This marked the first time in his career scoring 35+ points in consecutive games.
Holiday also tallied two double-doubles last week, dishing out 11 assists to go along with his 35 points on Monday against Indiana, before scoring 28 points with 10 assists on Saturday at Cleveland. Holiday has recorded a double-double in four of his last five games, pushing his season total to 11.
In his third season with the Bucks and 14th in the NBA, Holiday is averaging 19.6 points, 7.5 assists (T-8th in NBA), 4.9 rebounds and 1.4 steals in 32.7 minutes per game. He’s also shooting 46.6% overall and 38.4% from three, knocking down a career-high 2.4 threes a game.
This is the second time Holiday has earned Player of the Week honors in his career after previously doing so for Week 13 during the 2020-21 season. He joins Giannis Antetokounmpo (two times) as Bucks to be named Player of the Week this season.