We don't know for sure where the Milwaukee Bucks will select in the first round of the June 25 draft, but after today's tiebreaker drawing we have a much better idea.
The Bucks lost a three-way tiebreaker with Sacramento (8th) and Denver (9th) on Friday afternoon, meaning that (sadly) they're all but assured to select 10th overall on draft night. The Bucks finished tied with the Kings and Nuggets at 33-49 in the NBA's final standings, resulting in Friday's drawing to determine both a) the final number of lottery combinations each team would have and b) (more importantly) the default draft order if the teams don't jump into one of the top three lottery combinations. Overall, there's an 84.6% likelihood the Bucks end up in tenth and less than 7% chance they end up in the top three of next month's lottery.
Normally, the 8-9-10 teams receive 28, 17 and 11 lottery combinations, respectively. But due to the tie, the total combinations were pooled and distributed as evenly as possible. Since 56 combinations can't be divided evenly among three teams, each team was guaranteed 18 combinations with the first two tiebreaker winning teams receiving one additional combination (19 + 19 + 18 = 56, get it?). As a result, the Kings and Nuggets will have a 1.9% chance of nabbing the top spot and 6.8% chance of landing a top three pick, while the Bucks will have a 1.8% chance of winning the draft lottery and 6.5% chance of leaping into the top three.