1892 United States presidential election in Montana

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United States presidential election in Montana, 1892

November 8, 1892 (1892-11-08) 1896 →
 
Nominee Benjamin Harrison Grover Cleveland James Weaver
Party Republican Democratic Populist
Home state Indiana New York Iowa
Running mate Whitelaw Reid Adlai Stevenson James Field
Electoral vote 3 0 0
Popular vote 18,871 17,690 7,338
Percentage 42.44% 39.79% 16.50%

County results

President before election

Benjamin Harrison
Republican

Elected President

Grover Cleveland
Democratic

The 1892 United States presidential election in Montana took place on November 8, 1892, as part of the 1892 United States presidential election. Voters chose 3 representatives, or electors to the Electoral College, who voted for President and Vice President.

Montana voted for the Republican nominee, incumbent President Benjamin Harrison, over the Democratic nominee, former President Grover Cleveland, who was running for a second, non-consecutive term and over the People's Party (Populists) nominee James B. Weaver. Harrison won Montana by a margin of 2.65%.

This was the first time that Montana voted in a national electtion.

Results

United States presidential election in Montana, 1892[1]
Party Candidate Votes Percentage Electoral votes
Republican Benjamin Harrison (incumbent) 18,871 42.44% 3
Democratic Grover Cleveland 17,690 39.79% 0
People's James Weaver 7,338 16.50% 0
Prohibition John Bidwell 562 1.26% 0
Totals 44,461 99.99% 3

References

  1. ^ "1892 Presidential General Election Results - Montana". U.S. Election Atlas. Retrieved 1 August 2012.