1968 United States presidential election in Wyoming

1968 United States presidential election in Wyoming

← 1964 November 5, 1968 1972 →
 
Nominee Richard Nixon Hubert Humphrey George Wallace
Party Republican Democratic Independent
Home state New York[a] Minnesota Alabama
Running mate Spiro Agnew Edmund Muskie Marvin Griffin
Electoral vote 3 0 0
Popular vote 70,927 45,173 11,105
Percentage 55.76% 35.51% 8.73%

County Results

President before election

Lyndon B. Johnson
Democratic

Elected President

Richard Nixon
Republican

The 1968 United States presidential election in Wyoming took place on November 5, 1968. All 50 states and the District of Columbia were part of the 1968 United States presidential election. State voters chose three electors to the Electoral College, who voted for president and vice president.

Wyoming was won by the Republican nominee, former vice president Richard Nixon, running with Governor of Maryland Spiro Agnew. Together they defeated the Democratic nominee, incumbent Vice President Hubert Humphrey, and his running mate, Senator Edmund Muskie from Maine.

Nixon carried Wyoming with 55.76% of the vote to Humphrey's 35.51%, a victory margin of 20.25%. This marked a 33.45% swing to the right from 1964, when the state had voted for Democrat Lyndon B. Johnson by 13.2%. Also on the ballot was former and future Alabama Governor George Wallace, running as an Independent in Wyoming. Although Wallace carried five states in the South, he had only modest appeal in Wyoming. His performance was the best by any third-party candidate in Wyoming since Robert La Follette won nearly a third of the vote in 1924, but nonetheless did not equal his vote share in the fellow Western states of Alaska, Idaho and Nevada.

With 55.76 percent of the popular vote, Wyoming would prove to be Nixon's fifth strongest state in the 1968 election after Nebraska, Idaho, Utah and North Dakota.[1]


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  1. ^ "1968 Presidential Election Statistics". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.