For the second year in a row, Wizards second-round draft pick Aaron White will play overseas, league sources tell CSNmidatlantic.com on Friday, but this time it’ll be in Russia.
White, who was drafted 49th overall in 2015, signed a two-year deal –- the second is a team option and there is an NBA buyout clause –- with Zenit St. Petersburg.
White played for the Wizards at Las Vegas summer league and started all five games. Projected to be a stretch option as a 6-8 forward, he averaged 7.2 points and only shot 29.4% from three-point range.
He bypassed competing for a roster spot with the Wizards last season when there weren’t any roster spots up for grabs and went to Germany to begin his pro career. The Wizards have two spots open as they remake the roster this year, but Jarell Eddie (non-guaranteed) and Danuel House (undrafted free agent) are in a better position entering camp to gain those spots.
If White decided to attend camp with the Wizards he could force them to cut him which would make him a free agent and he could try to play in on another NBA team. But if he did that, those international offers would be off the table since camp doesn’t open until Sept. 27. CSN reported a week ago that White was likely to spend this season abroad, too.
By taking the offer with St. Petersburg, White will earn more than he would even if he made the Wizards playing for the rookie minimum ($543,471).
"I'm really excited to continue my career in Europe and play at this high level for St. Petersburg," White told CSN via text message.
"Zenit is a very successful team and I look forward to contributing to them."