1903 World Series – Boston Wins over the Pirates

By | Oct 1, 2017

In 1903 Baseball’s American League was just two years old, but they were popular enough that it was decided that the champion of each league would play a series of nine games for the national championship. The series began on October 1st with the last game held on October 13th.

The Pittsburgh Pirates led by shortstop Honus Wagner won their 3rd straight National League pennant. They ended the year with a .650 winning percentage and a 6.5 game lead over the New York Giants.

The American League team was the team from Boston, they wouldn’t adopt the official nickname Red Sox until 1908. They ended the year 14.5 games ahead of the Philadelphia Athletics with 91 wins, the same number of wins as the Pittsburgh Pirates.

The series was planned for 9 games, but only 8 were played since Boston won 5 of the 8. Pittsburgh won 3 of the first 4, 7-3 in game 1, 4-2 in game 3 and 5-4 in game 4. Before Boston won the next 4, 11-2, 6-3, 7-3 and 3-0. Boston also won the 2nd game of the series on a 3 hit shut out.

Wagner would only bat .222 in the series. The entire Pittsburgh team batted .237 for the series. They had batted .286 for the season.

Cy Young pitching for the Boston home team lost the first game to Pittsburgh’s Deacon Phillippe. In the first inning he gave up 4 of the 7 runs that the pirates scored, although there was also three Boston errors leading to 3 of the 4 runs being unearned. Young would go on to win two of the five games won by Boston. He pitched in 4 games to an ERA of 1.85 Phillippe would win all three games won by Pittsburgh, but he also lost 2 with a 3.07 ERA.


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