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1896 United States presidential election in Wyoming information


1896 United States presidential election in Wyoming
1896 United States presidential election in Wyoming
← 1892 November 3, 1896 1900 →
 
Nominee William Jennings Bryan William McKinley
Party Democratic Republican
Alliance Populist
Home state Nebraska Ohio
Running mate Arthur Sewall (Democratic)[a]
Thomas E. Watson (Populist)[b]
Garret Hobart
Electoral vote 3 0
Popular vote 10,861 10,072
Percentage 51.49% 47.75%

County Results

President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in Wyoming took place on November 3, 1896, as part of the 1896 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 representative William Jennings Bryan (D–Nebraska), running with shipbuilder, railroad president and director, bank president Arthur Sewall, with 51.49 percent of the popular vote, against the 39th Governor of Ohio William McKinley (R–Ohio), running with New Jersey State Senator, Garret Hobart, with 47.75 percent of the popular vote. Bryan won the state by a narrow margin of 3.74%.

This is also the only election where the Republican candidate won the election without Wyoming.

Bryan's support for many Populist goals resulted in him being nominated by both the Democratic Party and the People's Party (Populists), though with different running mates. One electoral vote from Wyoming was cast for the Populist Bryan-Watson ticket with Thomas E. Watson as vice president and two votes were cast for the Bryan-Sewall ticket.

Bryan would later lose Wyoming to William McKinley four year later and would later lose the state again to William Howard Taft in 1908.
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