Milwaukee Bucks Land Bogdan Bogdanovic from Kings and Jrue Holiday from Pelicans

Bogdan Bogdanovic with Serbia in 2019. Photo by Nikola Krstic/Shutterstock (10359146t)

After a disappointing playoff exit in 2019/20, Milwaukee Bucks started working on ways to improve their roster in order to impress reigning MVP Giannis Antetokounmpo and convince him to ink a long-term extension with the team. And it appears they are on a great track to pull that off.

According to ESPN’s Adrian Wojnarowski, Bucks acquired Sacramento Kings’ sharpshooter Bogdan Bogdanovic and New Orleans Pelicans’ guard Jrue Holiday in separate trades on Monday.

Bogdanovic was an impending restricted free agent, so Bucks are acquiring the Serbian in a sign-and-trade type of deal. Kings will receive Donte DiVincenzo, D.J. Wilson, and Ersan Ilyasova in return.

The package that Milwaukee gave up for Holiday is even bigger. Pelicans will receive guards Eric Bledsoe and George Hill as well as significant draft compensation.

Wojnarowski reports that Bucks will send Indiana Pacers’ first-round pick in 2020 as well as unprotected first-round picks in 2025 and 2027 to New Orleans. As part of the deal, the Pelicans will also have the option to swap picks in 2024 and 2026.

The two trades significantly add to Bucks’ already talented squad while addressing some of their biggest needs.

Bogdan Bogdanovic is a proven shooter in the league, having the ability to create a shot for himself while being extremely efficient from catch-and-shoot threes in recent seasons with Kings. Bogdanovic averaged career-high 15.1 points in 2019/20 while shooting 37.2 percent from deep and 44.0 from the field.

Jrue Holiday, on the other hand, comes with a reputation as one of the NBA’s best defenders and a player who has consistently produced during his time in the league. Holiday, who has one All-Star appearance to his name, finished the past season with 19.1 points, 4.8 rebounds, 6.7 assists, and 1.6 steals per game.