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1992 United States presidential election in Nevada

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1992 United States presidential election in Nevada

← 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 James Stockdale
Electoral vote 4 0 0
Popular vote 189,148 175,828 132,580
Percentage 37.36% 34.73% 26.19%

County Results

President before election

George H. W. Bush

Elected President

Bill Clinton

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.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]

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. 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 2016 presidential election, this is the last time a Democratic presidential candidate carried White Pine County in a presidential election,[3] and the last one where a third-party candidate carried any Nevada county. In this case, Ross Perot carried Storey County.[3]

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