The Kings won a key NBA draft tiebreaker that took place today, according to Shams Charania of The Athletic (Twitter link).
The league conducts tiebreakers when two teams finish the season with the same record. The winner of the tiebreaker gets the higher pick or the higher spot in the lottery standings.
In Sacramento’s case, the team entered the March 11 hiatus with a 28-36 record, identical to the Pelicans‘ 28-36 mark. Because the NBA is sorting its lottery standings based on non-playoff teams’ pre-hiatus records, the Kings and Pelicans had been tied for the 12th spot.
As a result of today’s tiebreaker, Sacramento will get the 12th pick in the draft, assuming neither of the two teams behind them jump into the top four; the Pelicans would receive the 13th pick in that scenario. The two teams’ draft lottery odds will be nearly identical, though Sacramento will have the slight edge, with the Kings receiving 13 of 1,000 available ping-pong ball combinations on lottery night, compared to 12 combinations for the Pelicans.
In the second round, the order will be flipped — the Pelicans will get the No. 42 pick and the Kings will be at No. 43.
Meanwhile, the Thunder finished the season tied with the Rockets and Jazz in the standings at 44-28, so a tiebreaker was required to determine their draft order.
As a result of today’s tiebreaker, the Oklahoma City pick will be No. 21, according to Charania, who adds that Houston’s pick will be No. 22, followed by Utah’s at No. 23.
Of those three teams, only the Jazz actually still own their 2020 first-rounder. The Sixers will receive OKC’s No. 21 pick, while the Nuggets will get Houston’s pick at No. 22. It’s an ideal outcome for Philadelphia, considering that Thunder pick was top-20 protected.
In the second round, Utah’s pick will be No. 51, Houston’s will be No. 52, and Oklahoma City’s will be No. 53. However, the Jazz and Rockets don’t own their second-rounders. Instead, it’ll be the Warriors drafting at No. 51 and the Kings at No. 52.