1992 United States presidential election in Wyoming

The 1992 United States presidential election in Wyoming took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

1992 United States presidential election in Wyoming

November 3, 1992
 
Nominee George H. W. Bush Bill Clinton Ross Perot
Party Republican Democratic Independent
Home state Texas Arkansas Texas
Running mate Dan Quayle Al Gore James Stockdale
Electoral vote 3 0 0
Popular vote 79,347 68,160 51,263
Percentage 39.70% 34.10% 25.65%

County Results

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

Wyoming was won by incumbent President George H. W. Bush (R-Texas) with 39.70 percent of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.10 percent. Businessman Ross Perot (I-Texas) finished in third, with 25.65 percent of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush. This election was the closest result in the state since 1948, when it narrowly voted for Harry S. Truman over Thomas E. Dewey. In fact, as of 2020 this is the last time in which the state was decided by single digits. As of the 2020 presidential election, this is the last election in which Natrona County and Carbon County voted for the Democratic candidate. It was also the first time any Wyoming county had voted for a Democrat since Jimmy Carter won Sweetwater County in 1976, and the first time Teton County voted for a Democratic presidential candidate since it voted for Franklin D. Roosevelt over Wendell Willkie in 1940.

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