1884 United States presidential election in Florida

The 1884 United States presidential election in Florida took place on November 4, 1884, as part of the 1884 United States presidential election. Florida voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.[2]

United States presidential election in Florida, 1884

November 4, 1884
Turnout	22.26% of the total population 3.11 pp[1]
Nominee Grover Cleveland James G. Blaine Party Democratic Republican Home state New York Maine Running mate Thomas A. Hendricks John A. Logan Electoral vote 4 0 Popular vote 31,769 28,031 Percentage 52.96% 46.73%
County Results Cleveland   50-60%   60-70%   70-80%   80-90%   90-100% Blaine   50-60%   60-70%   70-80%
President before election Chester A. Arthur Republican Elected President Grover Cleveland Democratic

Florida was won by Grover Cleveland, the 28th governor of New York, (D–New York), running with the former governor of Indiana Thomas A. Hendricks, with 52.96% of the popular vote, against Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 46.73% of the vote.[2]

Results

United States presidential election in Florida, 1884[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Grover Cleveland of New York	Thomas A. Hendricks of Indiana	31,769	52.96%	4	100.00%
	Republican	James G. Blaine of Maine	John A. Logan of Illinois	28,031	46.73%	0	0.00%
	Write-in	Write-in of	Write-in of	118	0.20%	0	0.00%
	Prohibition	John St. John of Kansas	William Daniel of Maryland	72	0.12%	0	0.00%
Total				59,990	100.00%	4	100.00%

References

1. "1884 Presidential Election Results Florida Total Population Turnout".
2. "1884 Presidential Election Results Florida".