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1888 United States presidential election in Minnesota

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1888 United States presidential election in Minnesota

← 1884 November 6, 1888 1892 →
 
Nominee Benjamin Harrison Grover Cleveland Clinton Fisk
Party Republican Democratic Prohibition
Home state Indiana New York New Jersey
Running mate Levi P. Morton Allen G. Thurman John Anderson Brooks
Electoral vote 7 0 0
Popular vote 142,192 104,385 15,511
Percentage 54.12% 39.65% 5.82%

President before election

Grover Cleveland
Democratic

Elected President

Benjamin Harrison
Republican

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

Minnesota voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 14.47%.

With 5.82% of the popular vote, Minnesota would be the Prohibition Party candidate Clinton Fisk’s strongest victory in terms of percentage in the popular vote.[1]

Results

1888 United States presidential election in Minnesota[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Benjamin Harrison of Indiana Levi Parsons Morton of New York 142,492 54.12% 7 100.00%
Democratic Grover Cleveland of New York Allen Granberry Thurman of Ohio 104,385 39.65% 0 0.00%
Prohibition Clinton Bowen Fisk of New Jersey John Anderson Brooks of Missouri 15,311 5.82% 0 0.00%
Labor Alson Streeter of Illinois Charles E. Cunningham of Arkansas 1,097 0.42% 0 0.00%
Total 263,285 100.00% 7 100.00%

Notes

References

  1. ^ "1888 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1888 Presidential General Election Results - Minnesota". U.S. Election Atlas. Retrieved 23 December 2013.