1948 American League tie-breaker game

The 1948 American League tie-breaker game was a one-game extension to Major League Baseball's (MLB) 1948 regular season, played between the Cleveland Indians and Boston Red Sox to determine the winner of the American League (AL) pennant. The game was played on October 4, 1948, at Fenway Park in Boston, Massachusetts. It was necessary after both teams finished the season with identical win–loss records of 96–58. This was the first-ever one-game playoff in the AL, and the only one prior to 1969 when the leagues were split into divisions.

The Indians defeated the Red Sox, 8–3, as the Indians scored four runs in the fourth inning and limited the Red Sox to five hits. The Indians advanced to the 1948 World Series, where they defeated the Boston Braves, four games to two, giving them their second and most recent World Series championship. In baseball statistics, the tie-breaker counted as the 155th regular season game by both teams, with all events in the game added to regular season statistics.

The 1948 Major League Baseball season was predicted to be a close race between the Yankees and Red Sox. In a United Press poll conducted just before the season started, the majority of sportswriters chose the Yankees, who had won last year's World Series, to face the Braves or St. Louis Cardinals that year, while others chose the Red Sox; only one sportswriter chose the Indians to reach the World Series. Most of the American League managers had the Yankees finishing first, followed by the Red Sox, Indians, and Detroit Tigers. Tension and confidence was evident between the teams, as two months into the season, after defeating the Red Sox 7–0, Yankees manager Bucky Harris declared that the Yankees would win the pennant, though they were currently at second place at the time.

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