1912 United States presidential election in Rhode Island

1912 United States presidential election in Rhode Island

← 1908 November 5, 1912 1916 →

Nominee Woodrow Wilson William Howard Taft Theodore Roosevelt Party Democratic Republican Progressive Home state New Jersey Ohio New York Running mate Thomas R. Marshall Nicholas M. Butler Hiram Johnson Electoral vote 5 0 0 Popular vote 30,412 27,703 16,878 Percentage 39.04% 35.56% 21.67%

County Results Wilson   30-40% Taft   30-40%   40-50%

President before election William Howard Taft Republican Elected President Woodrow Wilson Democratic

Elections in Rhode Island
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Ballot measures 2020 Question 1
Providence Mayoral elections 1998 2002 2006 2010 2014 2018 2022
v t e

The 1912 United States presidential election in Rhode Island took place on November 5, 1912, as part of the 1912 United States presidential election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island was won by the Democratic Party nominees, New Jersey Governor Woodrow Wilson and Indiana Governor Thomas R. Marshall. Wilson and Marshall defeated incumbent President William Howard Taft, and his running mate Vice President James S. Sherman and Progressive Party candidates, former President Theodore Roosevelt and his running mate California Governor Hiram Johnson.

Wilson won Rhode Island by a narrow margin of 3.48 points, becoming the first Democratic presidential candidate since Franklin Pierce in 1852 to win the state. Wilson also became the first Democrat since Pierce to win Providence County or any county in the state. Another Democratic candidate wouldn't win Rhode Island again until Al Smith won it in 1928. This was the first of three times that the state voted differently than Minnesota, along with 1928 and 1984.