1884 United States presidential election in West Virginia
The 1884 United States presidential election in West Virginia took place on November 4, 1884, as part of the 1884 United States presidential election. West Virginia voters chose six representatives, or electors, to the Electoral College, who voted for president and vice president.[2]
 | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| Turnout | 21.37% of the total population  3.16 pp[1] | |||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| ||||||||||||||||||||||||||
 County Results
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| Elections in West Virginia |
|---|
 |
West Virginia 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 50.94% of the popular vote, against Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 47.75% of the vote.[2]
The Prohibition Party chose John St. John, the former governor of Kansas, and former Maryland State Senator William Daniel as their presidential and vice-presidential candidates and received 0.71% of the vote. The Greenback and Anti-Monopoly Parties both chose major general and former governor of Massachusetts Benjamin Butler and Absolom M. West, an unseated Mississippi representative, received 0.60% of the vote.
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | Grover Cleveland | 67,311 | 50.94% | |
| Republican | James G. Blaine | 63,096 | 47.75% | |
| Prohibition | John St. John | 939 | 0.71% | |
| Greenback | Benjamin Butler | 799 | 0.60% | |
| Total votes | 132,145 | 100.00% | ||
References
