Mookie Betts was named the 2018 American League Most Valuable Player Thursday, easily beating out two-time winner Mike Trout to become the first Red Sox player since Dustin Pedroia in 2008 to capture the award.
Betts won 28 first-place votes to one for Trout and one for teammate J.D. Martinez. Cleveland's Jose Ramirez third in the voting. Trout beat out Betts for the same award in 2016. Trout has finished in the top five in balloting for the award every since 2012, his first full season and has finished either first or second in six of those seven seasons.
Betts finished with 410 points, followed by 265 for Trout, 208 for Ramirez and 198 for Martinez.
Xander Bogaerts finished 13th in the voting, with Chris Sale finishing 24th.
Betts led baseball in batting average, slugging percentage, runs scored and WAR. He posted a slash line of .346/.438/.640 while adding 32 homers and 30 stolen bases. He also won a Gold Glove in right field, while saving 20 defensive runs.
He becomes the 11th player in Red Sox history to win the award, following Tris Speaker, Jimmy Foxx, Ted Williams, Jackie Jensen, Carl Yastrzemski, Fred Lynn, Jim Rice, Roger Clemens, Mo Vaughn and Dustin Pedroia.
More to come
(Billie Weiss/Boston Red Sox/Getty Images)
Red Sox
Mookie Betts wins 2018 AL MVP, easily beating out Mike Trout
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