1912 United States presidential election in Wisconsin

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

1912 United States presidential election in Wisconsin

← 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%

County Results

President before election

William Howard Taft

Elected President

Woodrow Wilson

Democratic Party candidate Woodrow Wilson won Wisconsin with 41% of the popular vote, winning the state's thirteen electoral votes.[1]

With his win, Wilson became the first Democratic presidential candidate to win Wisconsin since Grover Cleveland in 1892. Another Democrat would not carry the state again until Franklin D. Roosevelt in 1932.

As of the 2016 presidential election, this is the only election in which Walworth County voted for the Democratic candidate.


1912 United States presidential election in Wisconsin
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


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