1886 World Series

The 1886 World Series was won by the St. Louis Browns of the American Association over the Chicago White Stockings of the National League, four games to two. It was played from October 18–23 in Chicago and St. Louis.

Background

In 1886, the St. Louis Browns won the American Association championship with a record of 93–46, while the Chicago White Stockings won the National League championship with a record of 90–34. The two teams agreed to meet each other in a best-of-seven pre-modern-era World Series, with the winner taking all the prize money.[1][2] It was the second straight year that the Browns and White Stockings met in the World Series.[3] The six games of the series were played on six consecutive days.[4]

Game summaries

Tip O'Neill

Overview

The Browns' O'Neill led all players with a .400 batting average, eight hits, and two home runs in the series. Welch had the second-highest batting average, at .350. Caruthers, who started three games for St. Louis, went 2–1 with a 2.42 earned run average. Clarkson started four games for Chicago and went 2–2 with a 2.03 ERA.[2]

For winning the series, St. Louis earned $13,920 in prize money. This was the American Association's only undisputed championship over the National League.[4]

See also

References

  1. 1 2 3 "The Chronology – 1886". baseballlibrary.com. Retrieved February 3, 2014.
  2. 1 2 "1886 World Series". baseball-reference.com. Retrieved February 3, 2014.
  3. 1 2 3 4 5 Snyder, John (2013). Cardinals Journal. Clerisy Press. pp. 34–35.
  4. 1 2 3 4 5 Nemec, David (2004). The Beer and Whisky League. Globe Pequot. pp. 6, 118–120.
  5. 1 2 James, Bill (2010). The New Bill James Historical Baseball Abstract. Simon & Schuster. p. 47.
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