United States presidential election in Wyoming, 1992

United States presidential election in Wyoming, 1992

← 1988 November 3, 1992 1996 →
Nominee George H.W. Bush Bill Clinton Ross Perot Party Republican Democratic Independent Home state Texas Arkansas Texas Running mate Dan Quayle Al Gore Electoral vote 3 0 0 Popular vote 79,347 68,160 51,263 Percentage 39.7% 34.1% 25.7%
County Results   Clinton—>70%   Clinton—60-70%   Clinton—50-60%   Clinton—40-50%   Bush—40-50%   Bush—50-60%   Bush—60-70%   Bush—>70%   Perot—40-50%
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

George H. W. Bush
Republican

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% of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.10%. Businessman Ross Perot (I-Texas) finished in third with 25.65% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush. This election was one of the closest results Wyoming ever had in U.S. history, since 1948, when it narrowly voted for Harry S. Truman over Thomas E. Dewey.