BOSTON (CBS) — World Series champ, first-time father, three-time Gold Glover and Most Valuable Player of the American League. What a month for Mookie Betts.
The Red Sox’ leadoff man was announced Thursday night as the AL’s MVP, edging out Mike Trout of the Angels and Jose Ramirez of the Indians for the honor.
The 26-year-old Betts posted a .346 batting average, a .640 slugging percentage and a 10.9 WAR in 2018, leading all of MLB in all three categories. His 1.078 OPS was second only to Trout (1.088), and Betts was one of just two players to hit at least 30 home runs and steal at least 30 bases this year. (Ramirez was the other.)
Betts set career highs in home runs (32), doubles (47), stolen bases (30), walks (81), and runs (129), and he did all of that while playing exceptional defense, as evidenced by the Gold Glove voting, which awarded him his third consecutive award for his work in right field. Betts also won a second Silver Slugger Award this year.
Betts’ offensive numbers dropped considerably in the postseason (.210/.300/.323), but he was still at the center of some key moments throughout the Red Sox’ run to the World Series.
Of the three MVP finalists, Betts was the leader in batting average, slugging percentage and doubles. Trout was the leader in OBP and OPS, while Ramirez led in home runs and RBIs.
Betts was the runner-up for the AL MVP in 2016, when Trout received 356 points in voting and Betts received 311. Betts finished sixth in MVP voting in 2017.