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| Elections in Ohio |
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The 1988 United States presidential election in Ohio took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. State voters chose 23 electors to the Electoral College, which selected the president and vice president.
Ohio 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.
Ohio weighed in for this election as 3% more Republican than the national average. This is the first election since 1924 in which Ohio did not vote the same way as Wisconsin, something that would reoccur in 2000, 2004, and 2020.
The presidential election of 1988 was a very partisan election for Ohio, with more than 99% of the electorate voting for either the Democratic or Republican parties.[1] Most counties in the state turned out more for Bush than Dukakis. Two notable exceptions to this trend were Cleveland's Cuyahoga County, and residents of several counties on the Eastern border with Pennsylvania, who voted largely for Dukakis. This was the last occasion until 2016 that Erie County, Montgomery County and Portage County voted for a Republican presidential candidate.[2]
Bush won the election in Ohio with 55.00% of the vote to Dukakis's 44.15%, for a margin of 10.85%, the last time a presidential candidate won the state by double digits until 2024.
- ^ "1988 Presidential General Election Results – Ohio". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved February 7, 2018.
- ^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016