Through nearly three months in the 2018 MLB campaign, we have seen a surge of players making a strong case for the MVP award. The two standouts in the American League that I consider as the front runners are Boston Red Sox’s Mookie Betts and the Los Angeles Angels’ Mike Trout. Here is the case for each candidate:
Making The Case: Mike Trout
The Angels superstar has been nothing short of impressive in his career with Los Angeles, powering 220 career home runs, as well as maintaining a .306 batting average over the course of the last eight seasons. This year alone, Trout has 19 home runs, a .306 batting average, and an on-base percentage of .439. He has been phenomenal, and the Angels are only 37-28 at the time of this writing. Without Trout, Los Angeles would struggle substantially, and that is the definition of valuable.
Defensively, Trout has been extremely effective, putting out 131 batters on 134 chances without a single error on the season. Efficiency is his strength, and Mike Trout has made a strong case for the MVP award in the American League.
Making The Case: Mookie Betts
Mookie Betts in 2018 has been nothing short of electric. Coming off of the 10-day DL, Betts has shown to be this team’s franchise player of the future. With the bat, Betts has posted a league-high .359 batting average, launching 17 home runs. To boost these already impressive statistics, Betts also has an OBP of .437.
Defensively, Betts has been just as impressive as Trout. On 176 chances, he has put out 170 batters. The fielding percentage of Betts is just short of perfect with a .992 percentage. With a strong second half of the season, Mookie Betts also has a strong case for the MVP award.
Both of these players have made a viable case for the AL MVP, but I would choose Boston’s Mookie Betts this year. In addition, The Athletes Hub recently held a poll in which 56% of our readers chose Betts over Trout in 278 total votes.
Photo Source : Boston.CBSlocal.com