1912 United States presidential election in Wyoming

1912 United States presidential election in Wyoming

← 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 Murray Butler
Electoral vote 3 0
Popular vote 15,310 14,560
Percentage 36.20% 34.42%

 
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 9,232 2,760
Percentage 21.83% 6.53%

County Results

President before election

William Howard Taft
Republican

Elected President

Woodrow Wilson
Democratic

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

Wyoming was won by Princeton University President Woodrow Wilson (DVirginia), running with governor of Indiana Thomas R. Marshall, with 36.20 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 34.42 percent of the popular vote, the 26th president of the United States Theodore Roosevelt (PNew York), running with governor of California Hiram Johnson, with 21.83 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 6.53 percent of the popular vote.[1]

Despite having a member of the Progressive Party (Joseph M. Carey) as the governor of the state, Theodore Roosevelt came in third and Wyoming was one of the areas Roosevelt received lower levels of support.

  1. ^ "1912 Presidential Election Results Wyoming".