1988 United States presidential election in Minnesota

1988 United States presidential election in Minnesota

← 1984 November 8, 1988 1992 →
Turnout	68.84%[1]
Nominee Michael Dukakis George H. W. Bush Party Democratic (DFL) Republican Home state Massachusetts Texas Running mate Lloyd Bentsen Dan Quayle Electoral vote 10 0 Popular vote 1,109,471 962,337 Percentage 52.91% 45.90%
County Results Dukakis   40-50%   50-60%   60-70% Bush   40-50%   50-60%   60-70%
President before election Ronald Reagan Republican Elected President George H. W. Bush Republican

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v t e

The 1988 United States presidential election in Minnesota took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for president and vice president.

Minnesota was won by Democrat Michael Dukakis, Governor of Massachusetts, with 52.91% of the popular vote over Republican Vice President George H. W. Bush's 45.90%, a victory margin of 7.01%.^[2] This made Minnesota roughly 14.8% more Democratic than the nation-at-large.

Four years earlier Minnesota had been the only state in the entire country to vote for Democrat Walter Mondale over Republican Ronald Reagan, and this Democratic strength in the state endured in 1988, as Minnesota chose Michael Dukakis by a comfortable margin despite George H.W. Bush winning a convincing victory nationwide. Minnesota has the longest streak of voting Democratic of any state, having not voted Republican since 1972.

As of the 2020 U.S. presidential election, this is the last time Minnesota voted to the right of neighboring Iowa, as well as the last time in which Clay County voted for a losing presidential candidate.