1900 United States presidential election in Wyoming

1900 United States presidential election in Wyoming

← 1896 November 6, 1900 1904 →
 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai Stevenson I
Electoral vote 3 0
Popular vote 14,482 10,164
Percentage 58.71% 41.20%

County Results
McKinley
  40-50%
  50-60%
  60-70%


President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in Wyoming took place on November 6, 1900, as part of the 1900 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 the President William McKinley (ROhio), running with the 33rd Governor of New York, Theodore Roosevelt, with 58.66 percent of the popular vote, against representative William Jennings Bryan (DNebraska), running with the 23rd Vice President Adlai Stevenson I, with 41.17 percent of the popular vote.[1] McKinley won the state by a margin of 17.49%

McKinley had previously lost Wyoming to Bryan four years earlier while Bryan would later go on to lose the state again to William Howard Taft in 1908.

  1. ^ "1900 Presidential Election Results Wyoming".