Jrue Holiday agrees to four-year contract extension with Milwaukee Bucks
Bucks guard's new deal worth at least $134 million
Holiday has been a key part of Milwaukee's success since arriving in 2020
The Milwaukee Bucks have officially acquired guard Jrue Holiday in a 4-team deal, the teams announced on Tuesday.
Holiday, 31, is a 6-foot-3 guard who has played 11 seasons in the NBA. He was originally drafted by the Philadelphia 76ers with the 17th overall pick in the 2009 NBA draft. Holiday played four seasons with Philadelphia before being traded to the New Orleans Pelicans in 2013. In 2020, he was traded to the Milwaukee Bucks.
Holiday has been a key part of Milwaukee's success since arriving in 2020. He helped the Bucks win the NBA championship in 2021 and was named to the NBA All-Defensive First Team in 2022. Holiday is averaging 18.4 points, 4.8 rebounds, and 6.8 assists per game this season.
Holiday's new contract extension is worth at least $134 million over four years, according to ESPN's Adrian Wojnarowski. The deal includes a player option for the 2025-26 season.
The Bucks are currently 30-17 this season and are in first place in the Eastern Conference. Holiday's new contract extension will help the Bucks remain a contender for the NBA championship.
The ultimate swing came with the acquisition of guard Jrue Holiday, who scored 27 points and added 9 rebounds and 9 assists on Saturday.
MILWAUKEE — When Bucks guard Jrue Holiday scores in home games, DJ Shawna, the Bucks in-arena DJ, plays a snippet of the chorus from "All of Me" by John Legend.
The song’s lyrics, which include the line "All of me loves all of you," are not only an anthem of affection, but an accurate description of how Bucks fans feel about Holiday.
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