Cleveland Indians clinch 4th seed to face Yankees in wild card round

The Cleveland Indians have officially clinched the 4th seed in the American League and will face the New York Yankees the best of three wild card round starting Tuesday.

The 2020 regular season is officially in the books. With an 8-6 victory over the Pittsburgh Pirates Sunday, the Cleveland Indians finished the shortened season 35-25 and are locked into 4th seed of the AL Playoffs.

This means they will face the New York Yankees in the three-game wild card round starting Tuesday at Progressive Field.

Sunday’s regular season finale saw the Indians rally back from a 6-2 deficit as Franmil Reyes hit a three-run home run to cut the lead to 6-5 and Carlos Santana hit a two-run double to give the Tribe a lead they’d never surrender.

The Indians took two out of three from the Pirates as they won eight of their final 10 games of the regular season after rebounding from an eight-game losing streak in early September.

Cleveland entered the day one game behind the Chicago White Sox for second place in the AL Central but owned the tiebreaker in the event the Indians won and the White Sox loss. That’s exactly what happened as Cleveland took care of Pittsburgh while the White Sox lost 10-8 to the NL Central champion Cubs Sunday.

The Indians will now shift their focus to the Yankees, who they last faced in the postseason in the 2017 ALDS in a series that saw New York storm back from a 2-0 deficit to win the best-of-five series.

Cleveland is expected to send ace and likely Cy Young Award winner Shane Bieber to the mound in Game 1 of their postseason series with Yankees ace Gerrit Cole likely to get the ball for Bombers.

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The best-of-three series will begin at Progressive Field Tuesday with no-days off in between games, meaning if the weather cooperates, the series will end by Thursday.