LaMelo Ball Is NBA’s 2020-2021 Top Rookie

LaMelo Ball
LaMelo Ball

This year’s Rookie of the year in the NBA has been awarded to LaMelo Ball, as per reports on Wednesday. LaMelo has received the award even though he has only been able to play 51 games because of injury. In the history of the award, it is the 4th-least number of games played.

LaMelo Ball Comes Out On Top Even With Fewer Appearances

The player to have won the award with the least number of appearances is Paul Hoffman during the season of 1947-48. He had played 37 matches. However, Hoffman’s team had played a total of 48 games in the year. In a usual season with 82 games, Vince Carter and Patrick Ewing both hold the record, having won the award after playing 50 games.

Kyrie Irving had been the only one below them with 51 games. Now LaMelo Ball has tied with him. Irving, similar to Ball, had his debut during a season that was shortened. For LaMelo Ball, the shortening was due to the ongoing pandemic. For Irving, the reason was the lockout during 2011.

Anthony Edwards, the Minnesota Timberwolves overall number 1 draft pick; and Tyrese Haliburton, of the Sacramento Kings, were ranked 2nd and 3rd respectively, for this year’s best rookie award.

LaMelo Ball was the third overall pick. However, his spectacular performances in the few matches he did play made the voters overlook the long absence due to injury. His scoring average was 15.7 with 5.9 rebounds and an average assists of 6.1 while playing with the Charlotte Hornets. He was the top assister among rookies, and in the others, he was among the best three. His highlights while passing are possibly unprecedented among rookies in the NBA.

Bell’s season had begun from the bench. However, his performances earned him a permanent starting position. He had even led Charlotte past the regular season, which is a first in 5 years.

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