1992_United_States_presidential_election_in_Nevada
1992 United States presidential election in Nevada
Election in Nevada
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.
.jpg)
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.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]
Clinton's win in the state, although narrow, marked the first time since 1964 that a Democrat would carry Nevada and the beginning of its transition from a fairly Republican state into a swing state, like it had been in most elections prior to 1968. The state would go on to back the national winner in every election onward until 2016. Although Clinton carried only two of the state's seventeen 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.
Clinton's percentage of the vote was the lowest to win a state since Woodrow Wilson carried only 32.08 percent of the vote in Idaho in 1912. As of the 2020 presidential election[update], this is the last time a Democratic presidential candidate carried White Pine County in a presidential election,[3] and the last one where a candidate outside the two major parties carried any Nevada county. In this case, Ross Perot carried Storey County.[3]