1988 United States presidential election in Kentucky|
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Bush
40–50%
50–60%
60–70%
70–80%
80–90%
90–100% |
Dukakis
40–50%
50–60%
60–70%
70–80%
80–90%
90–100% |
Other
Tie
No Votes
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The 1988 United States presidential election in Kentucky took place on November 8, 1988. All 50 states and the District of Columbia were part of the 1988 United States presidential election. Kentucky voters chose nine electors to the Electoral College, which selected the president and vice president. Kentucky was won by incumbent Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle for vice president, and Dukakis ran with Texas Senator Lloyd Bentsen.
In the 1980s, Kentucky was a swing state, having voted for the winner of every presidential election from 1964 on. However, Mitch McConnell's narrow win in the 1984 Senate election represented a significant shift of the state towards the Republican Party at the federal level.
Bush carried Kentucky by 11.6 percentage points on election day, the state weighing in as 4 points more Republican than the national average. Bush carried a majority of Kentucky's counties, including the most highly populated counties: Jefferson County, home to Louisville; Fayette County, home to Lexington; Daviess County, home to Owensboro; and Kenton and Campbell Counties, in the Cincinnati area. Dukakis' strength was mostly isolated to rural counties in the Eastern Coalfield and the Jackson Purchase.
As of the 2024 presidential election, this remains the last time that Jefferson County has voted for a Republican presidential candidate, as well as the last time that Kentucky has voted more Democratic than neighboring Tennessee.
