1948 Major League Baseball postseason

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DatesOctober 4-11, 1948[1]
Teams3
ChampionsCleveland Indians
(2nd title)
Runners-upBoston Braves
1948 Major League Baseball postseason
Tournament details
DatesOctober 4-11, 1948[1]
Teams3
Final positions
ChampionsCleveland Indians
(2nd title)
Runners-upBoston Braves
 1946
1951 

The 1948 Major League Baseball season resulted in a tie for the American League pennant between the Cleveland Indians and Boston Red Sox. Both finished the regular season with identical 96-58 records.

The tie thus required a one-game playoff to be held between the two teams. This game was played on October 4, 1948 at Fenway Park in Boston. The Indians defeated the Red Sox and advanced to the 1948 World Series, where they defeated the National League champion Boston Braves in six games.

American League playoff World Series
AL Cleveland 4
AL Cleveland 1 NL Boston Braves 2
AL Boston Red Sox 0

American League tiebreaker playoff

Cleveland Indians vs. Boston Red Sox

Monday, October 4, 1948 1:15 pm (ET) at Fenway Park in Boston, Massachusetts
Team123456789RHE
Cleveland1004100118131
Boston100002000351
WP: Gene Bearden (20–7)   LP: Denny Galehouse (8–8)
Home runs:
CLE: Lou Boudreau 2 (18), Ken Keltner (31)
BRS: Bobby Doerr (27)
Attendance: 33,957
Notes: Game duration 2 hours and 24 minutes
Boxscore

The Indians defeated the Red Sox to advance to the World Series for the first time since 1920 (in the process denying what would have been the only all-Boston World Series between the Red Sox and Braves).

Gene Bearden pitched a complete game and Lou Boudreau hit two homers as the Indians blew out the Red Sox to clinch the pennant.

This was the last time the Indians and Red Sox played for the pennant until the ALCS in 2007, where the Red Sox returned the favor and defeated the Indians in seven games en route to a World Series title after trailing 3–1 in the series.

1948 World Series

References

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