2017 NBA Awards show: Malcolm Brogdon wins NBA Rookie of the Year

The 2017 NBA Awards Show crowned the Rookie of the Year on Monday night with Milwaukee Bucks guard Malcolm Brogdon taking home the honors

Despite a prolific and successful college career for the Virginia Cavaliers, there wasn’t a great deal of hype about guard Malcolm Brogdon entering the 2016 NBA Draft. That’s why he slipped to the second round, where the Milwaukee Bucks eventually made him the 36th overall pick. From there, he outplayed his draft status from the onset.

Brogdon eventually went on to play in 75 games for the Bucks, starting in 28 games. He wasn’t just a role player, though, but a key contributor. Between his defensive prowess and ability as a combo guard, the UVA product was a key cog in helping Milwaukee reach the playoffs.

On Monday night at the 2017 NBA Awards Show, the Bucks guard was rewarded for his efforts. For the first award of the night, Brogdon was named the 2017 Kia NBA Rookie of the Year:

On the season, Brogdon averaged 10.2 points, 4.2 assists, 2.8 rebounds and 1.1 steals per game. What’s more, he was a highly efficient scorer with 45.7 percent overall shooting and 40.4 percent shooting from 3-point range.

The competition for the Rookie of the Year honors wasn’t exactly high profile, and was certainly weird. Both of the two other big contenders for the award, Joel Embiid and Dario Saric, play for the Philadelphia 76ers. Both had their own cases, sure. And had Embiid played a full season without getting injured, there’s a good chance that he would’ve won. The voters, instead, rewarded the Bucks guard for his ability to get on the floor in addition to playing well.

Next: Every NBA team's greatest playoff moment

Now the question for Brogdon will be what can he do with a Bucks team on the rise. Playing alongside Giannis Antetokounmpo, he’s going to be an integral piece moving forward in Milwaukee’s backcourt. If he can build off of a Rookie of the Year effort, though, he and the team will be just fine.