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Clippers center Montrezl Harrell is defended by Nuggets center Mason Plumlee during Game 1 on Thursday
Reuters: Los Angeles Clippers forward Montrezl Harrell is the 2019-20 NBA Sixth Man of the Year, the league announced Friday.
Harrell received 58 first-place votes and 397 points in balloting conducted by a panel of 100 sportswriters and broadcasters. Oklahoma City Thunder guard Dennis Schroder (35 first-place votes, 328 points) was second and Clippers guard Lou Williams (seven, 127) was third.
Statistics toward consideration for the award were used through March 11, the night the NBA season was suspended after Rudy Gobert tested positive for COVID-19 prior to a game at Oklahoma City.
Harrell posted career highs with 18.6 points and 7.1 rebounds in 63 games (two starts) before the interruption of the season. He didn’t play in any regular season games after the resumption of play due to the death of his grandmother.
Harrell, 26, was a second-round pick by Houston in 2015 and played two seasons with the Rockets before joining Los Angeles as part of the June 2017 trade that sent All-Star guard Chris Paul to the Rockets.
A Clippers player has won the award in five of the past seven seasons, with Harrell joining Jamal Crawford (2013-14, 2015-16) and Williams (2017-18, 2018-19). Williams also won it with the Toronto Raptors in 2014-15. Harrell averaged a career-high 27.8 minutes and shot 58.0 percent from the field and 65.8 percent from the free-throw line in 2019-20.
Through five seasons, Harrell has averaged 12.7 points and 4.9 rebounds in 318 games (25 starts).
Detroit Pistons forward Christian Wood and Milwaukee Bucks guard George Hill tied for fourth place with 17 points.