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1912 United States presidential election in Wisconsin

  (Redirected from United States presidential election in Wisconsin, 1912)

The 1912 United States presidential election in Wisconsin was held on November 5, 1912. Wisconsin voters chose thirteen electors to the Electoral College, who voted for president and vice president.

United States presidential election, 1912

← 1908 November 5, 1912 1916 →
  Woodrow Wilson-H&E.jpg William Howard Taft - Harris and Ewing.jpg
Nominee Woodrow Wilson William Howard Taft
Party Democratic Republican
Home state New Jersey Ohio
Running mate Thomas R. Marshall Nicholas M. Butler
(replacing James S. Sherman)
Electoral vote 13 0
Popular vote 164,230 130,596
Percentage 41.06% 32.65%

  Theodore Roosevelt-Pach.jpg EugeneVictorDebs.png
Nominee Theodore Roosevelt Eugene V. Debs
Party Progressive Social Democratic
Home state New York Indiana
Running mate Hiram Johnson Emil Seidel
Electoral vote 0 0
Popular vote 62,448 33,476
Percentage 15.61% 8.37%

Wisconsin1912PresidentalElection.svg
County Results

President before election

William Howard Taft
Republican

Elected President

Woodrow Wilson
Democratic

Democratic Party candidate Woodrow Wilson won the state with 41% of the popular vote, winning Wisconsin's thirteen electoral votes.[1] As of the 2016 presidential election, this is the only election in which Walworth County voted for the Democratic candidate.

ResultsEdit

United States presidential election in Wisconsin, 1912
Party Candidate Votes Percentage Electoral votes
Democratic Woodrow Wilson 164,230 41.06% 13
Republican William Howard Taft 130,596 32.65% 0
Progressive Theodore Roosevelt 62,448 15.61% 0
Social Democratic Eugene V. Debs 33,476 8.37% 0
Prohibition Eugene Chafin 8,584 2.15% 0
Totals 399,334 100.0% 13

ReferencesEdit

  1. ^ "1912 Presidential General Election Results - Wisconsin". Retrieved August 18, 2016.