The 1992 United States presidential election in Wisconsin took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.

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1992 United States presidential election in Wisconsin

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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 11 0 0 Popular vote 1,041,066 930,855 544,479 Percentage 41.13% 36.78% 21.51%

County Results Precinct Results Clinton   30–40%   40–50%   50–60%   60–70%   70–80%   80–90%   90–100% Bush   30–40%   40–50%   50–60%   60–70%   70–80%   80–90%   90–100% Perot   30–40%   40–50%   50–60% Tie/No Data

President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

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Wisconsin was won by Governor Bill Clinton (D-Arkansas) with 41.13 percent of the popular vote over incumbent President George H. W. Bush (R-Texas) with 36.78 percent. Businessman Ross Perot (I-Texas) finished in third, with 21.51 percent of the popular vote.^[1] Clinton ultimately won the national vote, defeating incumbent President Bush.^[2]

As of the 2020 presidential election^[update], this is the last election in which Florence County voted for a Democratic presidential candidate, and the last time that Door County backed a losing candidate.^[3] Despite Clinton's decently solid victory in Wisconsin, it marks his narrowest one in a state that Michael Dukakis won in 1988.

• United States presidential elections in Wisconsin