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Cleveland Cavaliers win lottery tiebreaker, can only pick as low as No. 6 in 2019 NBA Draft

The Cleveland Cavaliers have a 14 percent chance of landing the No. 1 pick and a 52.1 percent chance of their pick being in the top four.

The NBA Draft Lottery isn't until May 14. But the Cleveland Cavaliers have already enjoyed their first taste of lottery luck.

On Friday, the Cavs won a tiebreaker with the Phoenix Suns to determine which team will sit second and which will sit third in next month's draft lottery. Should both teams be jumped in the top four selections, which are drawn by ping pong balls, the Cavs will pick ahead of the Suns, meaning their lowest possible pick in the 2019 draft will be sixth overall.

Under the new NBA Draft Lottery format, the Cavs, Suns and New York Knicks will each have a 14 percent chance of landing the No. 1 overall pick and a 52.1 percent chance of having their picks land in the top four. The need for a tiebreaker between Cleveland and Phoenix came as a result of both teams finishing with a 19-63 record.

After winning Friday's drawing, the Cavs would only pick sixth should both they and New York be jumped in the drawing of the top four lottery selections.

Although Cleveland won its primary tiebreaker, it lost another one by way of the 2019 first-round pick its owed from the Houston Rockets. As a result, the Cavs' second first-round selection will be the No. 26 pick in the draft, which will be held on June 20 in Brooklyn, N.Y.

With the tiebreakers out of the way, all attention now turns toward next month's draft lottery in Chicago. Should the Cavs land the No. 1 overall pick, they would likely select Duke forward Zion Williamson, who is considered the consensus top prospect in the upcoming draft.

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