1892 World Series

In the first split season in Major League Baseball history, the first-half champion Boston Beaneaters (102–48) played the second-half champion Cleveland Spiders (93–56) in a best-of-nine postseason series to determine the overall champion of the 1892 baseball season. After the first game ended in a 0–0 tie due to darkness, Boston won the next five games to win the championship.

1892 World Series
NL: Boston Beaneaters (5–0–1)
NL: Cleveland Spiders (0–5–1)

Background

Boston manager Frank Selee worried that late-October weather conditions would lead to postponements and low attendance. Cleveland's player-manager Patsy Tebeau suggested that “the [Boston] Beaneaters fear the humiliation of possible defeat.” Tebeau told Sporting Life that the cold weather was a “dodge … simply an excuse to avoid playing Cleveland.”[1]

Bookies had the Spiders as the favorite due to their pitching staff. Cy Young had gone 36–12 with a 1.93 earned-run average. Meanwhile, Boston superstar Mike "King" Kelly, described as “one of the biggest failures of the base ball season,” slumped with a batting average of only .189.[1][2]

Series summary

GameScoreDateBallpark, City
1Boston 0 – 0 Cleveland (11 innings)October 17League Park, Cleveland, Ohio
2Boston 4 – 3 ClevelandOctober 18League Park, Cleveland, Ohio
3Boston 3 – 2 ClevelandOctober 19League Park, Cleveland, Ohio
4Boston 4 – 0 ClevelandOctober 21South End Grounds, Boston, Massachusetts
5Boston 12 – 7 ClevelandOctober 22South End Grounds, Boston, Massachusetts
6Boston 8 – 3 ClevelandOctober 24South End Grounds, Boston, Massachusetts

Aftermath

Hall of Famer Hugh Duffy (Boston) batted .462 with 9 runs batted in and 6 extra-base hits including a home run.[3] The National League abolished the split-season format for 1893.[1]

References

See also

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