Over the last few months of the season, debates raged over who should be the 2019 WNBA Rookie of the Year: Minnesota Lynx forward Napheesa Collier or Dallas Wings guard Arike Ogunbowale. Now, we have the answer.
Early Monday morning, the league announced that Collier won the honor, earning 29 out of 43 votes from the media panel. Ogunbowale received the other 14 votes, with no other rookie getting on the ballot.
Collier and Ogunbowale staged one of the best Rookie of the Year races in WNBA history, each making their case by getting even better as the season went on. In the end, Collier's consistency throughout the season, and impressive performance on the defensive end set her apart.
She finished the season averaging 13.1 points, 6.6 rebounds, 2.6 assists, 1.9 steals and 0.9 blocks while leading the league in minutes played. Her 1.9 steals per game were fourth among all players in the league, let alone rookies, and she finished just two blocks shy of averaging a steal and a block per game -- something only four players accomplished this season.
In winning the award, Collier is following in historic footsteps. Of the 21 previous Rookie of the Year Award winners, eight have gone on to win MVP, and she's the last Lynx player to win the honor since Maya Moore took it home in 2011.
Collier helped the Lynx make it back to the playoffs despite a huge roster overhaul, but they fell in the first round to the Seattle Storm. Collier finished that game with 19 points, 10 rebounds, three assists and two steals, joining Candace Parker and Tamika Catchings as the only rookies to have at least 18 points and 10 rebounds in a playoff game.
In addition to Rookie of the Year, Collier headlined the All-Rookie Team, which was also announced on Monday. Joining her on that team were Ogunbowale, Jackie Young, Brianna Turner and Teaira McCowan.