1988 United States presidential election in South Dakota

The 1988 United States presidential election in South Dakota took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Voters chose three electors to the Electoral College, which selected the president and vice president.

1988 United States presidential election in South Dakota

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Nominee George H. W. Bush Michael Dukakis Party Republican Democratic Home state Texas Massachusetts Running mate Dan Quayle Lloyd Bentsen Electoral vote 3 0 Popular vote 165,415 145,560 Percentage 52.85% 46.51%

County Results Bush   40–50%   50–60%   60–70%   70–80% Dukakis   40–50%   50–60%   60–70%   80–90%

President before election Ronald Reagan Republican Elected President George H. W. Bush Republican

South Dakota was won by incumbent United States Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.

South Dakota weighed in for this election as 1.5% more Democratic than the national average.^[1] This is the last of only four elections since statehood when South Dakota has voted more Democratic than the national average,^[a] an anomaly probably caused by the persistent crisis in the United States' farming sector during the decade.

Bush won the election in South Dakota with a 6-point margin.

The presidential election of 1988 was a very partisan election for South Dakota, with more than 99 percent of the electorate voting for either the Republican or the Democratic parties, and only five candidates appearing on the ballot.^[2] While most counties turned out in this election for Bush, the highly populated centers of Brown County and Sioux Falls's Minnehaha County, voted in majority for Dukakis.