The Boston Red Sox are still celebrating their 2018 World Series victory this past month over the Los Angeles Dodgers. On Friday, the celebration entailed signing the Most Valuable Player of that series, first baseman Steve Pearce, to a one-year contract for the 2019 season:
The #RedSox today signed first baseman Steve Pearce to a one-year contract through the 2019 season. pic.twitter.com/q82gMEozvm
— Boston Red Sox (@RedSox) November 16, 2018
Pearce will reportedly make more than $6 million as part of his deal:
Source: Steve Pearce and the Boston Red sox have agreed on a one-year, $6.25M deal.
— Jeff Passan (@JeffPassan) November 16, 2018
Pearce, 36 in April, came over in a midseason trade from the Toronto Blue Jays. Though he was viewed as a bench or platoon piece, he played a bigger role down the stretch and into the postseason due to Mitch Moreland's health. Pearce ended up appearing in 50 regular-season games for the Red Sox, batting .279/.394/.507 with 16 extra-base hits, including seven homers.
Had Pearce done nothing in October, his acquisition would've been worthwhile. Instead, he hit .289/.426/.658 in 47 plate appearances, with four home runs. Three of those home runs came in Games 4 and 5 of the World Series, as he drove in seven of Boston's 14 total runs.
Pearce again figures to split time with Moreland at the cold corner.