Billie Weiss/Boston Red Sox/Getty Images

Two years after finishing second in the American League MVP voting, Boston Red Sox star Mookie Betts took home the hardware for his spectacular performance during the 2018 season. 

Per the Boston Globe, Betts won the voting over Mike Trout and Jose Ramirez.

MLB Network shared the full voting results:

Betts started the season coming off a down year—by his standards—in 2017. The 26-year-old hit .264/.344/.459 with 24 home runs, 102 RBI and 26 stolen bases. 

In the September issue of Baseball Digest (h/t NBC Sports), Betts told Evan Drellich he came into 2018 feeling nervous because of changes he made to his swing during the offseason: 

“This whole season, hitting has been kind of that anxiety—you know, fear. We all saw how I started in spring training, and I was trying to learn something new. I was like, ‘Phew, I don’t know, this may be my end.’ … It was bad. I was actually really scared. Luckily, I was able to use my abilities that God gave me at picking up on things and kind of adapting. You give me a little something, I’ll work it and kind of make it my own within that structure. Seemed to work so far.”

The changes paid off for Betts and the Red Sox. He led Major League Baseball with a .346 batting average, .640 slugging percentage and finished second with a .438 on-base percentage. 

In addition to his offensive prowess, Betts continued to be one of the best defensive players in MLB. His 20 defensive runs saved were tied with Milwaukee Brewers center fielder Lorenzo Cain for second-most among all outfielders. 

With 10.4 wins above replacement, per FanGraphs, Betts led all big leaguers in total value this season. He was instrumental in the Red Sox putting together a historic season that included a franchise-record 108 wins, a third straight American League East title and a World Series championship.

Betts is the first Boston player to be named AL MVP since Dustin Pedroia in 2008 and 10th overall since 1931. 

Source link