United States presidential election in Oregon, 1992

United States presidential election in Oregon, 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 7 0 0 Popular vote 621,314 475,757 354,091 Percentage 42.5% 32.5% 24.2%
County Results   Clinton—50-60%   Clinton—40-50%   Clinton—<40%   Bush—<40%   Bush—40-50%   Bush—50-60%
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 Oregon took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Oregon was won by Governor Bill Clinton (D-Arkansas) with 42.48% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 32.53%. Businessman Ross Perot (I-Texas) finished in third with 24.21% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush.

As of 2017, this is the last occasion when the following counties have given a plurality to the Democratic Party nominee: Curry, Deschutes and Jefferson. It was also the first election since 1888 when Crook County had not backed the winning candidate.