Mookie Betts has been a better player than Mike Trout through the first 60% of the season. That’s right, we only have 40% of the season left because the season started in March and the All-Star game was later than usual. All Mookie has to do is maintain his insane stat line for the last 40% of the season. Here are the players stats next to each other:
Mookie – 78 G, 79 R (2nd), 108 H, 25 2B, 3 3B, 23 HR, 51 RBI, 18 SB, .359 AVG (1st), .691 SLG (1st), 1.139 OPS (1st)
Trout – 97 G, 71 R, 104 H, 18 2B, 3 3B, 25 HR, 50 RBI, 15 SB, .310 AVG, .606 SLG, 1.060 OPS
Now Mike Trout has played nineteen more games than Betts due to injury. That said, Mookie still has eight more runs, four more hits, seven more doubles, the same amount of triples, only two less homers, one more RBI, and three more stolen bases. His batting average is 49 points higher, his slugging percentage is 85 points higher, and his OPS is 79 points higher.
Difference: -19 G, 8 R, 4 H, 7 2B, Tie 3B, -2 HR, 1 RBI, 3 SB, .049 AVG, .085 SLG, 0.079 OPS
This makes it pretty clear who the American League MVP would be if it ended today. The numbers are close, but factoring in having nineteen less games should make the decision clear, Mookie Betts is the frontrunner for MVP. Also, Mookie has the edge on defense and baserunning. I love Mike Trout and hope he ends his career with an insane amount of MVP trophies, but as of now, the 2018 trophy is headed to Boston. Betts 1.139 OPS would be the 50th best single season all time.