Star Red Sox outfielder Mookie Betts has been named American League MVP.
Major League Baseball announced Thursday that Betts had been voted Most Valuable Player for the 2018 season. He received 28 first-place votes out of a possible 30.
Los Angeles Angels center fielder Mike Trout finished second in the balloting with one first-place vote. Cleveland Indians infielder Jose Ramirez came in third.
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One first-place vote went to Betts' teammate, Red Sox designated hitter J.D. Martinez, who finished fourth.
Betts, who turned 26 in October, has proven to be a force since ascending to the big leagues in June of 2014, but 2018 has been, far and away, his best season to date. He posted career highs in homers (32), steals (30), batting average (.346), on-base percentage (.438), slugging average (.640) and on-base plus slugging (1.078). And he did so while providing outstanding defense in right field.
The award caps off a memorable season for Betts, who helped the Red Sox win a franchise-record 108 games in the regular season on their way to a World Series championship.
"It means a lot," Betts said on MLB Network while cradling his newborn daughter, who stayed asleep as his friends and family members cheered. "It's definitely a special award, and something that I'll cherish, but I think the most important thing is, that you know, we won a World Series and got to bring a trophy back to Boston."
Betts has received MVP votes in all four of his full seasons, serving as runner-up to Trout in 2016. He has been an All-Star in each of the last three seasons.
Milwaukee Brewers outfielder Christian Yellich took home the NL MVP Award in his first season with the club following a trade from the Miami Marlins.