Report: Malcolm Brogdon leaving Bucks for Pacers on 4-year, $85 million deal

Milwaukee Bucks restricted free agent Malcolm Brogdon has agreed to a deal with the Indiana Pacers, The Athletic’s Shams Charania reports.

ESPN’s Adrian Wojnawrowski reports that the deal is for four years and $85 million and that the Pacers are sending the Bucks a first-round and two second-round picks in return for Brogdon.

Scroll to continue with content
Malcolm Brogdon is reportedly leaving Milwaukee to join the Indiana Pacers. (Getty)
Malcolm Brogdon is reportedly leaving Milwaukee to join the Indiana Pacers. (Getty)

Bucks couldn’t keep everyone

Brogdon is the second player the Bucks have lost from the roster that produced the NBA’s best regular-season record last season after Nikola Mirotic left Milwaukee and the NBA to join EuroLeague club Barcelona.

[Free agency updates: Keep track of the moves, rumors, cap space and more]

The Bucks reportedly retained All-Star guard Khris Middleton on a 5-year, $178 million deal and center Brook Lopez on a four-year, $52 million deal.

While they retained those key pieces around league MVP Giannis Antetokounmpo, Brogdon was a luxury the salary cap would not allow the Bucks to afford.

Brogdon will have chance to shine in Indiana

Brogdon, the 2017 Rookie of the Year, averaged 15.6 points, 4.5 rebounds and 3.2 assists over 64 starts last season, his third in the NBA. In the process, he shot 50.5 percent from the field, 42.6 percent from 3-point distance and 92.8 percent from the free throw line, joining the NBA’s elusive 50/40/90 club.

After playing a secondary role behind Antetokounmpo and Middleton in Milwaukee, Brogdon will have a chance to be a featured player in Indiana.

The Pacers have also reportedly agreed to a three-year, $31.5 million deal with Jeremy Lamb while watching Bojan Bogdanovic leave for the Utah Jazz and Thaddeus Young go to the Chicago Bulls in free agency.

More from Yahoo Sports:

What to Read Next