Free agent guard Shake Milton is leaving the Sixers, having agreed to a two-year, $10M deal with the Timberwolves, a source tells ESPN’s Adrian Wojnarowski (Twitter link).
The 54th pick of the 2018 draft after three seasons at SMU, Milton originally signed a two-way deal with Philadelphia, which was converted into a four-year, minimum-salary contract after his rookie season. That expired after 2022-23, making Milton an unrestricted free agent.
While Milton isn’t a great athlete or a defensive stopper, he has been quite productive when given a chance at a bigger offensive role throughout his pro career. For example, in 11 starts (38.3 minutes) last season, he averaged 20.8 points, 7.2 assists and 4.8 rebounds on a sparkling .518/.413/.970 shooting line.
His overall numbers have been more modest, with the 26-year-old averaging 9.7 points, 2.7 assists and 2.4 rebounds on .458/.368/.830 shooting in 234 games (45 starts, 21.4 MPG) over the past four seasons. Still, that’s solid production for a reserve.
Based on the contract value, it appears as though Milton will be getting part of Minnesota’s mid-level exception. The team also re-signed Nickeil Alexander-Walker and added Troy Brown, so the Wolves have been quite active in free agency, prioritizing young players to fill out their backcourt and wing depth.
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