CLEVELAND -- The Cleveland Indians will have home-field advantage against the Boston Red Sox in the American League Division Series.
The Indians secured the second-best record in the American League and earned home-field advantage for the Division series with a 3-2 win over the Kansas City Royals at Kauffman Stadium and Boston’s loss to the Toronto Blue Jays at Fenway Park Sunday.
Winners of the American League Central Division for the first time since 2007, the Indians (94-67) will host Games 1 and 2 of the Division Series against the A.L. East champion Red Sox (93-69) on Thursday and Friday nights, October 6 and 7, at Progressive Field.
A potential Game 5 of the first round would be played in Cleveland on Wednesday, October 12.
Game 3 at Fenway Park will be played on Sunday, October 9. A potential Game 4 matchup in Boston would take place on Monday, October 12.
The Indians finished the regular-season with a 2-4 record against the Red Sox. In those six meetings, Boston outscored Cleveland, 31-18, including a 15-11 edge at Progressive Field and 16-7 advantage at Fenway Park.
The Indians have announced that Trevor Bauer will start Game 1 against Boston, and potential A.L. Cy Young Award candidate Corey Kluber will take the mound in Game 2 despite experiencing a strained right quadriceps muscle in the final week of the regular season.
In their history, the Indians have met the Red Sox in postseason play on four separate occasions, and they hold a 2-2 record in those series. However, they are on a two-series losing streak against Boston, as the Red Sox bested the Indians in the 1999 American League Division Series and 2007 A.L. Championship Series.
The Indians are 3-3 in six A.L. Division Series appearances, with their last coming against the New York Yankees in 2007, when they defeated the Bronx Bombers and came within one inning of a World Series trip.