2004 United States presidential election in Rhode Island

2004 United States presidential election in Rhode Island

← 2000 November 2, 2004 2008 →
Turnout62.1%[1] Increase 0.7 pp
 
Nominee John Kerry George W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate John Edwards Dick Cheney
Electoral vote 4 0
Popular vote 259,760 169,046
Percentage 59.42% 38.67%


President before election

George W. Bush
Republican

Elected President

George W. Bush
Republican

The 2004 United States presidential election in Rhode Island took place on November 2, 2004, and was part of the 2004 United States presidential election. Voters chose four 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 20.75% 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 he had in four years earlier, he was overwhelmingly defeated in a traditional Democratic stronghold, winning only 38% of the vote to 59% for Kerry.

As of 2020, this remains the only time in history that a Republican president has been re-elected without carrying Rhode Island.

  1. ^ This figure is calculated by dividing the total number of votes cast in 2004 (440,372) by an estimate of the number of registered voters in Rhode Island in 2004 (709,050). See "Presidential Turnout History" (PDF). Rhode Island Board of Elections. Retrieved February 6, 2018.[permanent dead link]