1988 United States presidential election in Nebraska

The 1988 United States presidential election in Nebraska 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 five electors to the Electoral College, which selected the president and vice president.

Nebraska 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. Bush won Nebraska by a powerful 21% margin. His 60.15% vote share made it his seventh-best state in the nation, and one of only seven where he exceeded 60% of the vote. Even in this drought-influenced election, Nebraska remained largely immune to any significant softening of the Republican vote, as occurred in nearby, generally solidly Republican states at the time such as South Dakota, Montana, Colorado, and Kansas.

The election was very partisan, with more than 99 percent of the electorate voting for either the Democratic or Republican parties, and only four candidates appearing on the ballot. As has been typical ever since the 1950s, nearly every county in Nebraska turned out a majority for the Republican Party. However, by carrying Saline County, which had long been the most Democratic county in the state, plus Dakota County and Thurston County in the northeast with their substantial Hispanic and Native American populations.

Nebraska weighed in for this election as 7% more Republican than the national average. As of the 2020 United States presidential election, this remains the last time that Nebraska has voted under a winner-take-all system in a presidential election; it would start allocating its electoral votes by congressional district in 1992.

Counties that flipped from Republican to Democratic

 * Dakota
 * Saline
 * Thurston