1988 United States presidential election in Kentucky

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.

Bush carried Kentucky by 11.6 percentage points on election day, the state weighing in as 4 points more Republican than the national average. The presidential election of 1988 was a very partisan election for Kentucky, with more than 99 percent of the state's electorate voting for either the Democratic or Republican nominees, and only five parties appearing on the ballot. Most of the state's counties turned out for Bush, including highly populated Jefferson County, home to Louisville. Dukakis' strength was mostly isolated to rural counties in the Eastern Coalfield and in the Jackson Purchase.

, 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. Bush carried Kentucky by a solid 11-point margin, winning 84 out of the state's 120 counties. Kentucky in this era was a swing state, with Republicans making substantial gains in federal elections in the state over the course of the 1980s.

Counties that flipped from Republican to Democratic

 * Bath
 * Boyd
 * Carlisle
 * Carroll
 * Carter
 * Gallatin (previously tied)
 * Graves
 * Greenup
 * Henderson
 * Henry
 * Hickman
 * Marion
 * McCracken
 * McLean
 * Owen
 * Robertson
 * Trimble