Los Angeles Angels outfielder Mike Trout, a two-time American League Most Valuable Player, finished second in the voting for the 2018 award.
Boston’s Mookie Betts was the AL’s winner, the league announced on Thursday. Milwaukee’s Christian Yelich was close to a unanimous pick for the NL honor.
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Trout secured one first-place vote and 265 points. Trout tied the record of four second-place finishes shared by Stan Musial, Ted Williams and Albert Pujols. Trout won in 2014 and 2016, finished second in ’12, ‘13 and ’15, and was fourth in 2017.
Trout hit a career-high 39 home runs with 79 RBIs, a .312 batting average and a .460 on-base percentage.
— The Associated Press contributed to this report
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