Outfielders Mike Trout of the Los Angeles Angels and Mookie Betts of the Boston Red Sox are both having great seasons for their respective teams. Betts leads the MLB in batting, hitting .359, as of June 12. Meanwhile, Trout leads in on base percentage (.435) and is second in home runs. The two players’ success raises an important question: which of them most deserves to be the American League Most Valuable Player for 2018?
Trout is widely considered to be the best player in baseball. In eight seasons with the Angels, he has won two AL MVPs and made the All-Star game twice. With 21 home runs in 67 games so far this season, he is currently on pace to hit 51 home runs, more than his previous MVP-season home run totals.
Trout plays a different role than Betts. Even though Trout has scored 54 runs so far (two more than Betts), he has mostly hit second in the lineup. However, Betts is a leadoff hitter whose primary job is to score runs. Also, Betts has another MVP candidate – J.D. Martinez – hitting behind him in the Red Sox lineup, which enhances his run total.
The Angels’ chances of making the playoffs have been vastly improved by Trout’s success. Without him, the team is batting just .242 this season. Even without Betts, the Red Sox would still be batting 14 points higher than the Angels without Trout, and they would still have a decent shot at the playoffs.
Trout has hit more home runs than Betts, driven in more runs, and scored just as many runs. He is more durable than Betts, as he has not missed any games, and he has not made a single error this season.
Betts is in his fifth year with the Boston Red Sox. He has been named to the All-Star team twice, and he finished second in MVP voting in 2016. He is on pace for career highs in batting average, home runs, and steals.
Like Trout, Betts excels at the role he was hired to play. He leads the league in batting average, and he is tied for seventh in steals. He is also tied with Trout for the league lead in runs scored. But that’s not all he does for the Red Sox. Betts has also hit 17 homers and knocked in 37 runs, totals that are great for a player hitting in any spot in the lineup.
Due to injuries, including a hamstring problem and abdominal strain, Betts has played 48 games this season to Trout’s 64. In games that Betts has played, the Red Sox are 33-15 (.688). Without him, they are 11-7 (.611). In other words, the team’s chances of winning drop nearly eight percent without Betts in the lineup. In a 162 game season, that’s significant. Trout may be more durable, but Betts has put up similar numbers in 18 fewer games than Trout. And he doesn’t just have a higher batting average; his average is 53 points higher, a huge margin.
In a tightly contested race, you can find both Trout and Betts in the top five of many statistical categories. But Betts has put up similar or better numbers than Trout in most areas, all while playing in fewer games. His batting average is considerably higher than Trout’s. His absence has a dramatic impact on the Red Sox’s ability to win, even though they are a very strong team overall. Finally, Betts is the best player on one of the best teams in the league. If he can stay healthy and continue to perform as he has, Mookie Betts should be the 2018 AL MVP.
(stats as of June 12)
Photographs (from top): Jim McIsaac/Getty Images; Billie Weiss/Boston Red Sox/Getty Images; John Capella/Sports Imagery/ Getty Images