1964 United States presidential election in Rhode Island
The 1964 United States presidential election in Rhode Island took place on November 3, 1964, as part of the 1964 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.
Rhode Island voted overwhelmingly for the Democratic nominee, incumbent President Lyndon B. Johnson of Texas, over the Republican nominee, Senator Barry Goldwater of Arizona. Johnson ran with Senator Hubert H. Humphrey of Minnesota, while Goldwater’s running mate was Congressman William E. Miller of New York.
Johnson carried Rhode Island in a landslide, taking 80.87% of the vote to Goldwater’s 19.13%, a Democratic victory margin of 61.74%. This made Rhode Island Lyndon Johnson’s strongest state in the nation: even in the midst of a massive nationwide Democratic landslide, Rhode Island weighed in as 39 percent more Democratic than the national average during the 1964 election.
The staunch conservative Goldwater was widely seen in the Northeastern United States as a right-wing extremist; he had voted against the Civil Rights Act of 1964, and the Johnson campaign portrayed him as a warmonger who as president would provoke a nuclear war. While John F. Kennedy had won 63.63 percent in Rhode Island in 1960 mostly by sweeping the ethnic Catholic vote, for 1964, this traditional Democratic coalition was joined by mass defections of moderate Yankee Republicans who had voted for Eisenhower and Nixon but could not support the extremist Goldwater. His landslide was so large, he won a record 315,463 votes, a record that still hasn't been beat. Consequently, the incumbent Johnson was able to take more than eighty percent of the vote in liberal Rhode Island.
Johnson swept all five counties in Rhode Island with over seventy percent of the vote. In Providence County, the most populated county, home to the state’s capital and largest city, Providence, Johnson took 83.5% of the vote. This was the strongest showing ever for a Democratic presidential candidate in Providence.
Johnson’s 80.87 percent remains the highest vote share percentage any presidential candidate of either party has ever received in Rhode Island, and his 61.74 percent victory margin remains the widest margin by which any candidate of either party has ever won the state. In fact, since 1968 no state of the Union has ever given any candidate so large a proportion of the vote.
|Presidential candidate||Party||Home state||Popular vote||Electoral
|Count||Percentage||Vice-presidential candidate||Home state||Electoral vote|
|Lyndon B. Johnson||Democratic||Texas||315,463||80.87%||4||Hubert Humphrey||Minnesota||4|
|Barry Goldwater||Republican||Arizona||74,615||19.13%||0||William E. Miller||New York||0|
|Needed to win||270||270|
- David Leip’s Atlas of U.S. Presidential Elections; 1964 Presidential General Election Data Graphs – Rhode Island
- Counting the Votes; Rhode Island
- Donaldson, Gary; Liberalism’s Last Hurrah: The Presidential Campaign of 1964; p. 190 ISBN 1510702369
- Edwards, Lee and Schlafly, Phyllis; Goldwater: The Man Who Made a Revolution; pp. 286-290 ISBN 162157458X