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1944 United States presidential election in Rhode Island
Roosevelt 50-60% 60-70% 70-80%	Dewey 50-60% 60-70% 70-80%
President before election; Franklin D. Roosevelt; Democratic	Elected President ; Franklin D. Roosevelt; Democratic

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.

Rhode Island voted Democratic candidate, incumbent President Franklin D. Roosevelt of New York, over the Republican candidate, Governor Thomas E. Dewey of New York. Roosevelt ran with Senator Harry S. Truman of Missouri as his running mate, while Dewey ran with Governor John W. Bricker of Ohio as his running mate.

Roosevelt won the state by a margin of 17.33%. Rhode Island was one of six states that became more Democratic compared to 1940, alongside Michigan, Minnesota, New York, North Dakota, and Illinois.

Results

1944 United States presidential election in Rhode Island^[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	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%

By county

1944 United States presidential election in Rhode Island (by county) ^[2]
County	Franklin D. Roosevelt Democratic		Thomas Dewey Republican		Other candidates Various parties		Total
County	%	#	%	#	%	#	#
Bristol	56.0%	6,287	43.8%	4,919	0.1%	16	11,222
Kent	50.5%	14,059	49.3%	13,710	0.2%	57	27,826
Newport	54.6%	11,375	45.3%	9,435	0.1%	21	20,831
Providence	61.1%	137,216	38.8%	87,190	0.1%	321	224,727
Washington	43.8%	6,419	56.1%	8,233	0.1%	18	14,670