Liberty Win the W.N.B.A. Draft Lottery Again

The Liberty will pick No. 1 for a second consecutive year in the W.N.B.A. draft, the league announced on Friday night. This time, the leading candidate to be the first pick is not nearly as clear as the last time, when the Liberty chose Oregon’s Sabrina Ionescu.

“We don’t know who will be at the top,” Liberty General Manager Jonathan Kolb said of the draft, which will be held in April. “Which players will rise through their play? For us, it’s wide open.”

Dallas will pick second, with Atlanta third and Indiana fourth. The W.N.B.A. determines the chances of winning the lottery based on the cumulative two-year records of the four participants. The Liberty, who struggled last season after losing Ionescu to an ankle injury, had an aggregate record of 12-44 and the best chance to win the lottery at 44.2 percent. Atlanta (15-41) was second at 27 percent, Dallas (18-38) third at 17 percent and Indiana (19-37) fourth at 10 percent.

There are several talented seniors in the college game, including Louisville’s Dana Evans, U.C.L.A.’s Michaela Onyenwere, Arizona’s Aari McDonald and Tennessee’s Rennia Davis, but the decisions could be complicated by the effects of the coronavirus pandemic on college sports. Because of the uncertainty surrounding winter sports seasons, the N.C.A.A. has granted all basketball players an extra year of eligibility, which means several elite prospects may wait to enter the draft.

Article source: https://www.nytimes.com/2020/12/05/sports/liberty-wnba-draft-lottery.html