1992 United States presidential election in Wyoming

1992 United States presidential election in Wyoming
Bush 30–40% 40–50% 50–60%	Clinton 30–40% 40–50%
President before election; George H. W. Bush; Republican	Elected President ; Bill Clinton; Democratic

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.

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.^[1] 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.^[2] 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^[update], this is the last election in which Natrona County and Carbon County voted for the Democratic candidate.^[3] 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.^[4]

^ Cite error: The named reference results was invoked but never defined (see the help page).
^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved June 11, 2012.
^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016
^ Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 342-343 ISBN 0786422173