1912 United States presidential election in Idaho

1912 United States presidential election in Idaho

← 1908 November 5, 1912 1916 →
 
Nominee Woodrow Wilson William Howard Taft
Party Democratic Republican
Home state New Jersey Ohio
Running mate Thomas R. Marshall Nicholas M. Butler
Electoral vote 4 0
Popular vote 33,921 32,810
Percentage 32.08% 31.02%

 
Nominee Theodore Roosevelt Eugene V. Debs
Party Progressive Socialist
Home state New York Indiana
Running mate Hiram Johnson Emil Seidel
Electoral vote 0 0
Popular vote 25,527 11,960
Percentage 24.14% 11.31%

County Results

President before election

William Howard Taft
Republican

Elected President

Woodrow Wilson
Democratic

The 1912 United States presidential election in Idaho took place on November 5, 1912, as part of the 1912 United States presidential election. State voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.

Idaho was won by New Jersey Governor Woodrow Wilson (DVirginia), running with Indiana Governor Thomas R. Marshall, with 32.08 percent of the popular vote, against the 27th president of the United States William Howard Taft (ROhio), running with Columbia University President Nicholas Murray Butler, with 31.02 percent of the popular vote, the 26th president of the United States Theodore Roosevelt (PNew York), running with governor of California Hiram Johnson, with 24.14 percent of the popular vote and the five-time candidate of the Socialist Party of America for President of the United States Eugene V. Debs (SIndiana), running with the first Socialist mayor of a major city in the United States Emil Seidel, with 11.31 percent of the popular vote.[1]

Wilson's 32.08% of the popular vote is the smallest vote share by which any candidate has carried a state in any United States presidential election; slightly smaller than John Quincy Adams' 32.46% in Illinois in 1824, and Abraham Lincoln's 32.32% in California in 1860.

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