The 1988 United States presidential election in Iowa took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president.
Quick Facts Nominee, Party ...
1988 United States presidential election in Iowa|
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County Results
Dukakis
40–50%
50–60%
60–70% |
Bush
40–50%
50–60%
60–70%
70–80%
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Iowa was won by Democratic Governor Michael Dukakis of Massachusetts with 54.71% of the popular vote over Republican Vice President George H.W. Bush's 44.50%, a victory margin of 10.22%, which made Iowa exactly 18% more Democratic than the nation-at-large. This made it one of 10 states (plus the District of Columbia) to vote for Dukakis, while Bush won a convincing electoral victory nationwide. Bush became the first Republican to ever win the presidency without carrying Iowa, a feat that would only be repeated once more, when George W. Bush 2000 lost the state but won the national election by extremely narrow margins. Bush Jr. would go on to carry the state in 2004, making the elder Bush the only Republican to have won the presidency without ever winning the state, as well as the only Republican to have ever won the national popular vote without winning the state. Dukakis was also the first Democratic presidential candidate to carry the state since Lyndon B. Johnson’s landslide victory in 1964. The state would not vote for the losing candidate again until 2000, the loser of the popular vote until 2016, and the loser of both the popular and electoral vote until 2020.