1992 United States presidential election in Michigan

1992 United States presidential election in Michigan

← 1988 November 3, 1992 1996 →
Turnout	62.5% [1]
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 18 0 0 Popular vote 1,871,182 1,554,940 824,813 Percentage 43.77% 36.38% 19.30%
County Results Congressional District Results Clinton   30-40%   40-50%   50-60%   60-70%   70–80%   80–90% Bush   30-40%   40-50%   50-60%
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

Elections in Michigan
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The 1992 United States presidential election in Michigan took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose 18 representatives, or electors to the Electoral College, who voted for president and vice president.

Michigan was won by Governor Bill Clinton (D-Arkansas) with 43.77% of the popular vote over incumbent President George H. W. Bush (R-Texas) with 36.38%. Businessman Ross Perot (I-Texas) finished in third, with 19.30% of the popular vote.^[2] Clinton ultimately won the national vote, defeating incumbent President Bush.^[3] This was the first time that a Democratic presidential candidate carried the state since Hubert Humphrey in 1968. It would not vote Republican again until 2016.

As of the 2020 presidential election^[update], this is the last time that Oakland County voted for a Republican presidential candidate and the last time that Charlevoix County voted for a Democratic presidential candidate.^[4] Michigan voted most similarly to the national results of the election.

This is the most recent election where Michigan voted more Republican than nearby Pennsylvania. Afterward, Michigan would become the most Democratic of the three Rust Belt swing states (including Wisconsin and Pennsylvania), a status it has held since.