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1904 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

United States presidential election in Rhode Island, 1904

← 1900 November 8, 1904 1908 →
President Theodore Roosevelt, 1904.jpg
Nominee Theodore Roosevelt Alton B. Parker
Party Republican Democratic
Home state New York New York
Running mate Charles W. Fairbanks Henry G. Davis
Electoral vote 4 0
Popular vote 41,605 24,839
Percentage 60.60% 36.18%

President before election

Theodore Roosevelt

Elected President

Theodore Roosevelt

The 1904 United States presidential election in Rhode Island took place on November 8, 1904. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island overwhelmingly voted for the Republican nominee, President Theodore Roosevelt, over the Democratic nominee, former Chief Judge of New York Court of Appeals Alton B. Parker. Roosevelt won Rhode Island by a margin of 24.42%.


United States presidential election in Rhode Island, 1904[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Theodore Roosevelt of New York Charles Warren Fairbanks of Indiana 41,605 60.60% 4 100.00%
Democratic Alton Brooks Parker of New York Henry Gassaway Davis of West Virginia 24,839 36.18% 0 0.00%
Socialist Eugene Victor Debs of Indiana Benjamin Hanford of New York 956 1.39% 0 0.00%
Prohibition Silas Comfort Swallow of Pennsylvania George Washington Carroll of Texas 768 1.12% 0 0.00%
Socialist Labor Charles Hunter Corregan of New York William Wesley Cox of Illinois 488 0.71% 0 0.00%
Total 68,656 100.00% 4 100.00%


  1. ^ "1904 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.

This page was last edited on 4 May 2019, at 13:25
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