A coin flip will determine 2021 NBA Draft tiebreakers, including Oklahoma City and Cleveland’s tie for fourth, per a report from ESPN draft analyst Jonathan Givony.
Givony reported that the Draft order will be conducted on Tuesday, May 25.
Oklahoma City’s 117-112 win over the Clippers on Sunday cost a chance at the third-best odds, sending the Thunder back one spot into a tie with Cleveland.
The teams with the fourth and fifth best odds will both have the same chance at the second, third and fourth pick.
The winner of Oklahoma City and Cleveland’s coin flip will have a 12.5 percent chance at the top pick in the draft, opposed to a 10.5 percent chance should they lose the flip.
The winner will have a 25.7 percent chance at the sixth pick and a 16.7 percent chance at the seventh pick.
The loser of the flip will have a 2.2 percent chance at the fifth pick, 19.6 percent chance at the sixth, 26.7 percent chance ash the seventh, an 8.7 percent chance at the eighth and a 0.6 percent chance at the ninth.
There are six tiebreakers in the standings currently, most of which are a 50/50 split.