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2019 NBA Draft Lottery odds for the Cleveland Cavaliers


The 2019 NBA Draft Lottery will take place in Chicago on Tuesday night, and the Cavs are in line for a top pick.

For the 14 teams that failed to reach the playoffs in the 2018-19 season, the lottery will be a chance at landing a talented young prospect. Most experts believe Duke’s Zion Williamson will be taken first, but which team will have that first pick?

The Cavs, who failed to reach the postseason for the first time in four consecutive seasons, have a 14 percent chance of landing the No. 1 overall selection in the upcoming lottery.

Two other teams share the top odds of getting the first pick: the New York Knicks and Phoenix Suns. The Chicago Bulls (12.5 percent chance) and Atlanta Hawks (10.5 percent) round out the top five of this year’s lottery hopefuls:

The Cavs will deploy their good luck charm at this year’s lottery. Nick Gilbert, son of team owner Dan Gilbert, will once again represent the team at the event:

Nick has represented Cleveland at the lottery on five separate occasions. In 2011, Cleveland won the No. 1 overall selection, which turned into Kyrie Irving. In 2013, the Cavs selected Anthony Bennett with the top pick. In 2014, the Cavs picked Andrew Wiggins No. 1 before trading him and Bennett for Kevin Love, though Nick Gilbert wasn’t at the lottery for that one.

Will the Cavs win the first pick again? And if so, who might they be interested in taking? Some say Zion is a lock, while others are interested in Ja Morant. Let’s see how the ping pong balls fall.

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