The 1988 United States presidential election in Rhode Island took place on November 8, 1988, as part of the 1988 United States presidential election, which was held throughout all 50 states and D.C. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

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1988 United States presidential election in Rhode Island

← 1984 November 8, 1988 1992 →
Turnout	70.2%[1] 7.8 pp
Nominee Michael Dukakis George H. W. Bush Party Democratic Republican Home state Massachusetts Texas Running mate Lloyd Bentsen Dan Quayle Electoral vote 4 0 Popular vote 225,123 177,761 Percentage 55.64% 43.93%
County Results Municipality Results Dukakis   40–50%   50–60%   60–70% Bush   50–60%   60–70%
President before election Ronald Reagan Republican Elected President George H. W. Bush Republican

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Rhode Island voted for the Democratic nominee, Massachusetts Governor Michael Dukakis, over Republican Vice President George H. W. Bush. Dukakis took 55.64% of the vote to Bush's 43.93%, a margin of 11.71%. This made it one of 10 states (plus the District of Columbia) to vote for Dukakis, while Bush won a convincing electoral victory nationwide.

A liberal New England state, Rhode Island gave Dukakis his strongest state victory in the nation, with only the District of Columbia voting more Democratic. It was one of just two states (along with Iowa) to vote Democratic by a double-digit margin, and one of only two states (along with Hawaii) to have all of its counties go to Dukakis. Despite this, it was still a relatively strong Republican performance compared to how the state has trended since. The state has voted Democratic in every presidential election that followed. As of the 2020 presidential election, this is the last time a Republican candidate won over 40% of the vote in Rhode Island.

• United States presidential elections in Rhode Island
• Presidency of George H. W. Bush