WNBA Draft Lottery — Who will win the #1 overall pick?

The 2019 WNBA Draft Lottery will take place tonight during half time of the WNBA semi final game 2 (Washington vs Atlanta). The four worst teams in the league this season have the opportunity to grab a great college player in yet another deep WNBA draft class.

The four worst teams this season were the Indiana Fever, Chicago Sky, Las Vegas Aces and the New York Liberty. The lottery rules have shifted and instead of the 1 year record, it now accounts for 2 years (or seasons). So, this means that the Fever and the Sky have the best odds of landing a top 2 pick. The Las Vegas Aces improved significantly this season but can still land a top 3 pick. The “worst” odds out of all the teams belongs to the New York Liberty. The Liberty were great in 2017, and that great year will affect them for this draft lottery. However, it should be noted that finishing fourth in this year’s lottery is not that bad. There are a ton of talented players available and the Liberty should fare well.

Keeping all that in mind, I believe that lottery drawings are a luck of the draw. The Indiana Fever REALLY need this top pick and would be thrilled to have it. However, I do not think the Fever will win tonight. I have a feeling it will be the Chicago Sky who come away with the #1 overall pick. I think the Fever will pick #2, the Aces at #3 and the Liberty at #4 overall. (I could be totally wrong, and it could just be the Fever who finally manage to win the #1 overall pick)

Whatever team will end up winning the #1 overall pick in 2019, they will have plenty of talented senior college players to choose from. Players such as Teaira McCowan, Asia Durr, Arike Ogunbowale, Katie Lou Samuelson, Napheesa Collier, Kalani Brown and more. I tracked all of the college players for two years running and here was the latest on who the top overall junior player was for 2018.

This could change a WNBA teams fortune. So, who do you think will win the #1 overall pick? Tune in tonight to find out!

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