1944 United States presidential election in Rhode Island

The 1944 United States presidential election in Rhode Island took place on November 7, 1944, as part of the 1944 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.

Main article: 1944 United States presidential election
1944 United States presidential election in Rhode Island
November 7, 1944
 
Nominee Franklin D. Roosevelt Thomas E. Dewey
Party Democratic Republican
Home state New York New York
Running mate Harry S. Truman John W. Bricker
Electoral vote 4 0
Popular vote 175,356 123,487
Percentage 58.59% 41.26%
County Results
President before election

Franklin D. Roosevelt
Democratic

 Elected President

Franklin D. Roosevelt
Democratic

Elections in Rhode Island
Federal government
U.S. President
U.S. Senate
U.S. House
State government
General elections
Gubernatorial elections
State Senate elections
State House of Representatives elections
Providence
Mayoral elections

Rhode Island was won by Democratic candidate, incumbent President Franklin D. Roosevelt won the state over the Republican candidate New York governor Thomas E. Dewey.

Roosevelt won the state by a margin of 17.33%.

Results
1944 United States presidential election in Rhode Island[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Franklin Delano Roosevelt of New York Harry S. Truman of Missouri 175,356 58.59% 4 100.00%
Republican Thomas Edmund Dewey of New York John William Bricker of Ohio 123,487 41.26% 0 0.00%
Prohibition Claude A. Watson of California Andrew Nathan Johnson of Kentucky 433 0.14% 0 0.00%
Total 299,276 100.00% 4 100.00%

See also

References
  1. "1944 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.


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