Williamson's deal is reportedly among the highest guaranteed contracts Nike has handed out.

By Alaa Abdeldaiem
July 24, 2019

Zion Williamson's shoe deal with Jordan Brand extends five years and is worth $75 million, the richest annual rookie shoe deal in NBA history, The Action Network's Darren Rovell reported on Wednesday.

According to Rovell, Williamson's deal is among the highest guaranteed contracts Nike has agreed to. Michael Jordan, LeBron James, Kevin Durant and Cristiano Ronaldo are among a group of players who have larger deals.

Williamson, the No. 1 overall pick in the 2019 NBA draft, announced his decision to sign with Jordan Brand on Instagram on Tuesday.

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Let’s Dance #JUMPMAN

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"I feel incredibly blessed to be a part of the Jordan Brand family," Williamson said in a statement. "Since I was a kid, I dreamed of making it to the league and having the type of impact on the game Michael Jordan had and continues to have today. He was one of those special athletes I looked up to, and I really can't express how happy and excited I am for this journey."

According to ESPN's Adrian Wojnarowski, several shoe competitors made offers on totals that extended into the nine-figure range before Williamson committed to Jordan Brand. Williamson had other offers in excess of $10 million annually, according to reporting from ESPN shoe analyst Nick DePaula.

The 2018-19 Naismith winner averaged 22.6 points and 8.9 rebounds per game at Duke last season and led the Blue Devils to the Elite Eight.

Before Williamson's deal, James had the largest annual rookie shoe contract in history.

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