Indians’ Shane Bieber unanimous winner of AL Cy Young Award

Cleveland Indians right-hander Shane Bieber won the 2020 American League Cy Young Award on Wednesday, becoming the AL's first unanimous choice since Justin Verlander in 2011 by receiving all 30 first-place votes from the Baseball Writers' Association of America.

Bieber went 8-1 with a 1.63 ERA and 122 strikeouts in 77⅓ innings, helping the Indians reach the postseason for the fourth time in five years. In all, Bieber led the majors in wins, ERA, strikeouts, FIP (2.07) and K/9 (14.2) and is the first pitcher since 2011 to win the Triple Crown (ERA, wins, strikeouts).

Kenta Maeda of the Minnesota Twins and the Toronto Blue Jays' Hyun-jin Ryu finished second and third, respectively, in the AL voting.

His 14.2 strikeouts per nine innings is the highest single-season ratio in history among players to have qualified for the ERA title – albeit in a shortened season.

Bieber had a 1.63 ERA in 77.1 innings in 2020. (Photo: David Richard, USA TODAY Sports)

Bieber struck out at least 10 batters in eight of his 12 starts and was unbeaten until his 10th outing of the season. Opposing hitters had a .167 average against Bieber, the best mark in the AL.

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The 25-year-old becomes the fifth Indians pitcher to win the award, joining Gaylord Perry (1972), CC Sabathia (2007), Cliff Lee (2008) and Corey Kluber (2014, 2017).

Bieber was a fourth-round pick in 2016 and made his big-league debut in 2018, before hitting his stride as a sophomore in 2019. He was named MVP of the All-Star Game in Cleveland and finished fourth in Cy Young voting, going 15-8 with a 3.28 ERA and 259 strikeouts.

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