1944 United States presidential election in Rhode Island
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 County Results
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| Elections in Rhode Island |
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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 4 electors to the Electoral College, which selected the president and vice president.
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% | ||||
References
- ^ "1944 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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