LaMelo Ball wins NBA rookie of year honor
LaMelo Ball of the Charlotte Hornets heads for the net in the second quarter at Madison Square Garden on May 15, 2021 in New York City. (Getty Images via AFP)
NEW YORK – Charlotte Hornets guard LaMelo Ball was named the NBA’s Rookie of the Year on Thursday, beating out competition from Minnesota’s Anthony Edwards and Sacramento Kings guard Tyrese Haliburton.
Ball, 19, was chosen by the Hornets with the third overall pick of the 2020 draft after spending the 2019-2020 season with the Illawara Hawks of Australia’s National Basketball League.
The teenager averaged 15.7 points, 6.1 assists, 5.9 rebounds and 1.59 steals in 51 games for Charlotte, leading all rookies in assists and steals.
Ball received 84 first-place votes from a global panel of 99 sportswriters and broadcasters who decided the award, for a total points tally of 465.
Timberwolves rookie Edwards received 15 first-place votes and finished second with 309 points. Haliburton earned 114 points.
Under the preferential voting system, players were awarded five points for each first-place vote, three points for each second-place vote and one point for each third-place vote.
Ball’s impressive maiden season included a triple-double for Charlotte in a game against the Atlanta Hawks in January, making him the youngest player ever to bag a triple in the NBA.
Other notable performances included a 34-point display in a game against the Utah Jazz in February.
Ball is the youngest of three basketball-playing brothers who rose to prominence during their high school careers in California.
Eldest brother Lonzo Ball, 23, was the second overall pick in the 2017 draft, chosen by the Los Angeles Lakers before being traded to the New Orleans Pelicans.
LiAngelo Ball, 22, is an undrafted free agent yet to break into the NBA.
