1888 United States presidential election in Delaware explained

See main article: 1888 United States presidential election.

Election Name:1888 United States presidential election in Delaware
Country:Delaware
Flag Year:1888
Type:presidential
Ongoing:no
Previous Election:1884 United States presidential election in Delaware
Previous Year:1884
Next Election:1892 United States presidential election in Delaware
Next Year:1892
Election Date:November 6, 1888
Image1:StephenGroverCleveland.png
Nominee1:Grover Cleveland
Party1:Democratic Party (United States)
Home State1:New York
Running Mate1:Allen G. Thurman
Electoral Vote1:3
Popular Vote1:16,414
Percentage1:55.15%
Nominee2:Benjamin Harrison
Party2:Republican Party (United States)
Home State2:Indiana
Running Mate2:Levi P. Morton
Electoral Vote2:0
Popular Vote2:12,950
Percentage2:43.51%
Map Size:210px
President
Before Election:Grover Cleveland
Before Party:Democratic Party (United States)
After Election:Benjamin Harrison
After Party:Republican Party (United States)

The 1888 United States presidential election in Delaware took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

Delaware voted for the Democratic nominee, incumbent President Grover Cleveland, over the Republican nominee, Benjamin Harrison. Cleveland won the state by a margin of 11.64%.

This was the last time until 2000 that a Republican was elected president without winning Delaware.

See also