United States presidential election in Vermont, 1992

United States presidential election in Vermont, 1992

← 1988 November 3, 1992 1996 →
Nominee Bill Clinton George H.W. Bush Ross Perot Party Democratic Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle Electoral vote 3 0 0 Popular vote 133,592 88,122 65,991 Percentage 46.11% 30.42% 22.78%
County Results   Clinton—50-60%   Clinton—40-50%   Clinton—<40%
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 Vermont 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.

Vermont was won by Democratic Governor Bill Clinton of Arkansas, taking 46.11% of the popular vote over incumbent Republican President George H.W. Bush of Texas, who took 30.42%, a victory margin of 15.70%. Independent Texas billionaire Ross Perot finished in third with 22.78% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush.

This election would prove to be a re-aligning one for the state of Vermont. Clinton's decisive double-digit victory in the state, including a Democratic sweep of all fourteen counties, represented a dramatic turnaround for a state that up to that point had been reliably Republican for over 130 years. This was only the second time in history that a Democratic presidential nominee had won Vermont's electoral votes, the first and only prior time being when Lyndon Johnson had carried the state in the 1964 Democratic landslide. Except for 1964, Vermont had voted Republican in every presidential election from 1856 to 1988. Beginning in 1992, however, Vermont instantly became a reliably Democratic state in presidential elections, voting Democratic in every election that has followed.