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Bucks, Holiday agree to 4-year extension worth up to $160M

David Liam Kyle / National Basketball Association / Getty

The Milwaukee Bucks and star guard Jrue Holiday have agreed to a four-year, $135-million maximum contract extension, Holiday's agent, Jason Glushon, told ESPN's Adrian Wojnarowski.

Bonuses can make the contract worth up to $160 million. Holiday's deal contains a player option for the 2024-25 season, The Athletic's Shams Charania notes.

The two-time All-Defensive team member has enjoyed a strong debut season with the Bucks, averaging 17 points, 5.4 assists, 4.6 rebounds, and a career-high 1.8 steals over 38 games. Milwaukee traded Eric Bledsoe, George Hill, three future first-round picks, and two pick swaps to acquire Holiday from the New Orleans Pelicans as part of a four-team deal this past offseason.

The Bucks now have their three top players - Giannis Antetokounmpo, Khris Middleton, and Holiday - on the books through the end of the 2022-23 season. Middleton's deal contains a $40.3-million player option for the 2023-24 campaign, while Antetokounmpo and Holiday will still be under contract. Milwaukee currently sits third in the Eastern Conference with a 32-17 record.

Prior to suiting up for the Bucks and Pelicans, Holiday was drafted by the Philadelphia 76ers with the 17th overall pick in the 2009 draft. The 30-year-old owns career averages of 15.9 points, 6.3 assists, 3.9 rebounds, and 1.5 steals across 12 NBA seasons.

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