The 2018 Major League Baseball season has come to an end. The Boston Red Sox won the World Series two weeks ago. It is now time for the major awards: Rookie of the Year, Manager of the Year, Cy Young Award, and Most Valuable Player.
On Thursday, the American League MVP will be announced. The finalists are Los Angeles Angel Mike Trout, Boston Red Sox Mookie Betts, and Cleveland Indian Jose Ramirez.
Mike Trout
.312/.460/.628/1.088, 39 HR, 79 RBI
The Angel outfielder led the league with a .460 on-base percentage, 1.088 OPS, and 122 walks. He finished in the top 5 in batting average and home runs, as well as leading all of baseball in offensive WAR for the 6th time in his career. Trout injured his wrist in August, causing him to miss 3 weeks, otherwise his numbers would be even better.
Trout is a 2-time AL MVP and the 2012 AL Rookie of the Year. He has made the All-Star team in each of his 7 full seasons. Last week Trout won his 6th Silver Slugger.
Mookie Betts
.346/.438/.640/1.078, 32 HR, 80 RBI
The Red Sox outfielder had a phenomenal season for the high flying Red Sox. He led all of baseball with a .346 batting average, .640 slugging percentage, and 129 runs scored. He became just the second Red Sox in the 30/30 clube with 32 home runs and 30 stolen bases. Betts hit over .370 in May and in September.
Betts is a 3-time All-Star that received his third Gold Glove and second Silver Slugger last week.
Jose Ramirez
.270/.387/.552/.939, 39 HR, 105 RBI
The Indian third baseman finished in the top 5 in the American League in home runs, RBI, runs scored, and stolen bases this season. Ramirez set career highs with 39 home runs and 34 stolen bases, becoming just the third Indian to join the 30/30 club.
The 2-time All-Star received his second Silver Slugger last week.
Who Will Win?
Betts had an incredible season from start to finish, getting on and scoring runs almost at will. He will be the American League MVP.
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