1992 United States presidential election in Idaho

The 1992 United States presidential election in Idaho took place on November 3, 1992, as part of the 1992 United States presidential election. State voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

1992 United States presidential election in Idaho

November 3, 1992
 
Nominee George H. W. Bush Bill Clinton Ross Perot
Party Republican Democratic Independent
Home state Texas Arkansas Texas
Running mate Dan Quayle Al Gore James Stockdale
Electoral vote 4 0 0
Popular vote 202,645 137,013 130,395
Percentage 42.03% 28.42% 27.05%

County Results

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

Idaho was won by incumbent President George H. W. Bush (R-Texas) with 42.03% of the popular vote over Governor Bill Clinton (D-Arkansas) with 28.42%. Businessman Ross Perot (I-Texas) finished in a close third, with 27.05% of the popular vote. Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot. As of the 2020 presidential election, this is the last election in which Bonner County, Clearwater County, Lewis County, and Benewah County voted for a Democratic presidential candidate.

With 27.05 percent of the popular vote, Idaho would prove to be Perot's fourth strongest state after Maine, Alaska and Utah.

In Franklin County, Clinton was a mere 22 votes away from placing 4th behind Independent candidate, Bo Gritz who himself scored 12.40%, his strongest county in the entire state.

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