United States presidential election in Rhode Island, 2004
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County Results
Kerry—60-70%
Kerry—50-60%
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The 2004 United States presidential election in Rhode Island took place on November 2, 2004 throughout all 50 states and D.C., which was part of the 2004 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.
Rhode Island was won by Democratic nominee John Kerry by a 20.8% margin of victory. Prior to the election, all 12 news organizations considered this a state Kerry would win, or otherwise considered as a safe blue state. Even though President George W. Bush fared better than in 2000, he was overwhelmingly defeated in a traditional Democratic stronghold, winning only 39% of the vote to 59% for Kerry.
There were 12 news organizations who made state by state predictions of the election. Here are their last predictions before election day.
Kerry won every single pre-election poll, each with a double digit margin and with at least 49% of the vote. The final 3 poll average showed Kerry leading 55% to 38%.
Bush raised $282,237. Kerry raised $977,390.
Neither campaign advertised or visited here in the fall campaign.
Federally, Rhode Island is one of the most reliably Democratic states during presidential elections, regularly giving the Democratic nominees one of their best showings. In 1980, Rhode Island was one of only six states to vote against Ronald Reagan. Reagan did carry Rhode Island in his 49-state victory in 1984, but the state was the second weakest of the states Reagan won. Rhode Island was the Democrats' leading state in 1988 and 2000, and second-best in 1996 and 2004.
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