Featured image courtesy of Zimbio.com (Oct. 27, 2018 – Source: Harry How/Getty Images North America)
Everyone saw this coming from a mile away, but the Red Sox right fielder Mookie Betts has officially won the MVP award. Betts was one of three finalists along with the Cleveland Indians Jose Ramirez and the Los Angeles Angles Mike Trout. However, despite the worthy adversaries, Betts took home his first-ever MVP award.
Mookie Betts Wins MVP
The league definitely got this one right, as Betts was the best player in baseball throughout the 2018 season. On the season, Betts posted a .346/.438/.640 slash line with an accompanying 185 wRC+. In his 614 plate appearances, Betts hit 32 home runs and drove in 80 runs out of the leadoff spot. His .346 batting average led the league and his 185 wRC+ was second only to Mike Trout.
While Trout led the league in wRC+, Betts elite baserunning and defensive acumen, combined with his elite hitting, earned him a 10.4 fWAR on the season, the best in the league. While WAR isn’t a perfect stat by any means, Betts’ WAR was 0.6 wins higher than that of second-place Mike Trout.
Of course, Betts was also on the best team in baseball, which certainly helps his case. The Red Sox won 108 regular season games, with Betts being the best player on that stacked roster. While he disappeared a bit during playoff action, the postseason is not factored into the MVP voting process. However, Mookie still brought his patented elite defense and baserunning to the postseason and even hit a homer in the World Series clincher.
Betts is currently entering his second year of arbitration, so it is currently unknown exactly what he will make in 2019. Whatever it is, it will be a massive bargain, as the Red Sox have arguably the best young player in baseball under team control for the next two seasons. Betts hits free agency in 2021, at which point the Red Sox should pay him all the money.