The Houston Rockets are going to look a bit different next season and they've added a new face into the mix on Sunday.
According to Chris Haynes of Yahoo Sports, Sterling Brown is electing to leave the Milwaukee Bucks for the Rockets in free agency.
One-year deal for Sterling Brown. https://t.co/OjwgcKiH8F
— Chris Haynes (@ChrisBHaynes) November 22, 2020
Brown is signing a one-year deal with the Rockets, giving him somewhere to call home for the upcoming season. Though, it remains to be seen what the roster around Brown looks like given the uncertainty surrounding James Harden and Russell Westbrook.
Prior to joining the Rockets, Brown spent his first three seasons in the NBA with the Bucks. However, he was drafted by the Philadelphia 76ers in the second round of the 2017 NBA Draft out of SMU. He would be traded to Milwaukee just a month after being drafted.
In his three years with the Bucks, Brown has served as a role player for Milwaukee, averaging only 15.7 minutes per game in his career thus far. During that span, he's compiled 5.2 points, 3.1 rebounds, and an assist per game.
Since the beginning of the offseason, the Bucks have added a few wing players to the roster, including the blockbuster trade for Jrue Holiday. Provided that, Brown was going to be an odd man out as Milwaukee wouldn't be able to give him many minutes on the floor.
On the other hand, he'll now get a chance to prove himself with a new team in the Rockets. The SMU product will likely compete for minutes with Danuel House, Eric Gordon, and Ben McLemore.
Follow Us
Subscribe