1948 American League tie-breaker game

1948 American League
tie-breaker game

Fenway Park in Boston, game venue
1 2 3 4 5 6 7 8 9 R H E Cleveland Indians 1 0 0 4 1 0 0 1 1 8 13 1 Boston Red Sox 1 0 0 0 0 2 0 0 0 3 5 1
Date	October 4, 1948
Venue	Fenway Park
City	Boston, Massachusetts
Umpires	HP: Bill McGowan 1B: Bill Summers 2B: Eddie Rommel 3B: Charlie Berry
Attendance	33,957
Television	WBZ-TV (BOS)[1]
Radio	Mutual WHDH (BOS) WHK (CLE)
Radio announcers	Mutual: Mel Allen and Jim Britt[2][3]

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 before 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.