United States presidential election in Nevada, 1992

United States presidential election in Nevada, 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 4 0 0 Popular vote 189,148 175,828 132,580 Percentage 37.4% 34.7% 26.2%
County Results   Clinton—40-50%   Clinton—<40%   Bush—<40%   Bush—40-50%   Perot—<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 Nevada took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Nevada was won by Governor Bill Clinton (D-Arkansas) with 37.36% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 34.73%. Businessman Ross Perot (I-Texas) finished in third with 26.19% of the popular vote. Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.

Clinton's win, although narrow, would mark the beginning of Nevada's transition from a safe Republican state into a swing state; in all three presidential elections held in the 1980s, Republicans had dominated Nevada by double digit margins and swept every county in the state. Although Clinton carried only 2 of the state's 17 counties, his win in Clark County, where Las Vegas is located, and the most populous county in the state, would prove the key to this and future Democratic victories in Nevada.