NBA insider Shams Charania revealed on Twitter that Holiday's agent Jason Glushon had confirmed the news to The Athletic. The deal will be worth a sum of $160 million, which will keep Jrue Holiday in Milwaukee for another four years.
The Milwaukee Bucks signed Jrue Holiday in the off-season from the New Orleans Pelicans before the start of the 2020-21 campaign. It was a four-team trade deal involving the Bucks, Pelicans, OKC Thunder and the Denver Nuggets.
Milwaukee Bucks and Jrue Holiday to carry on with their partnership for another four years
The max contract offered by the Milwaukee Bucks to Jrue Holiday comes as no surprise, as he has been stellar for them this season. Holiday has given the Bucks some much-needed star-depth along with Giannis and Khris Middleton. The three players have combined well for the Bucks this campaign and have helped them achieve a stellar 32-17 record.
The Milwaukee Bucks traded for the point guard because they wanted Giannis and Middleton to play more freely. The Bucks didn't want the duo to worry about extending their scope of play. But much to their surprise, Jrue has not only provided Giannis and Middleton valuable support, but he has also become a star in his own right.
Holiday is averaging 17 points, 5.4 assists, and 1.8 steals per game (2nd best in the NBA). The 30-year-old has been a solid upgrade on the likes of George Hill and Eric Bledsoe as he has been a terrific defender on the wings. Holiday's player efficiency rating is 22.95, which is the second-best on the Milwaukee Bucks roster.
The move seems like an ideal one for the Bucks. It will now keep them in high spirits as they prepare to match the likes of the Brooklyn Nets and Philadelphia 76ers in the post-season.
Considering how successful this campaign has been for the Milwaukee Bucks, it would be fair to say that the core of Giannis, Middleton and Holiday could land them an NBA championship soon.