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One of the NBA’s most underrated players is getting paid.
Jrue Holiday and the Milwaukee Bucks have agreed to a four-year max contract extension that could be worth up to $160 million, Holiday’s agent Jason Glushon told Shams Charania of The Athletic. The fourth year of the contract is reportedly a player option.
Holiday, 30, has only been with the Bucks since November. His previous team, the New Orleans Pelicans, sent him to Milwaukee as part of a four-team trade. The Bucks sent George Hill, Eric Bledsoe, the draft rights to 2020 first round pick R.J. Hampton, two first-round draft picks, and two pick swaps to various teams, which was quite a lot to give up for Holiday.
But Holiday hasn’t disappointed. He’s averaging 17.0 points, 5.4 assists and 4.6 rebounds per game this year, but both he and the Bucks have really picked things up lately. Holiday is averaging 20 points over the last month, and the team has been on a tear since mid-February. Since February 19 the Bucks have won 16 games and lost just four, and they’re up to third place in the Eastern Conference.
Bucks core will be together for several years
The Bucks gave up a lot for Holiday, so it’s not surprising that they’d sign him to an extension to make the most of their investment. But he’s part of a core of players that will now be with the Bucks for several years as they continue to chase that elusive championship.
Before Holiday, the Bucks signed their most important player, Giannis Antetokounmpo, to the largest extension in NBA history. In December the two parties inked a five-year, $228 million contract extension, keeping Antetokounmpo in Milwaukee through at least 2025. And before Antetokounmpo, the Bucks signed Khris Middleton to a five-year, $178 million extension following the 2019 season.
The Bucks know that their best chance at winning a championship is to keep their core of Antetokounmpo-Middleton-Holiday together for as long as possible. Now that they’ve done that, they just have to get that ring.
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