See main article: 1884 United States presidential election.
| Election Name: | 1884 United States presidential election in Mississippi |
| Country: | United States |
| Flag Year: | 1877 |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1880 United States presidential election in Mississippi |
| Previous Year: | 1880 |
| Next Election: | 1888 United States presidential election in Mississippi |
| Next Year: | 1888 |
| Election Date: | November 4, 1884 |
| Turnout: | 10.67%[1] 0.32 pp |
| Image1: | StephenGroverCleveland.png |
| Nominee1: | Grover Cleveland |
| Party1: | Democratic Party (United States) |
| Home State1: | New York |
| Running Mate1: | Thomas A. Hendricks |
| Electoral Vote1: | 9 |
| Popular Vote1: | 77,653 |
| Percentage1: | 64.34% |
| Nominee2: | James G. Blaine |
| Party2: | Republican Party (United States) |
| Home State2: | Maine |
| Running Mate2: | John A. Logan |
| Electoral Vote2: | 0 |
| Popular Vote2: | 43,035 |
| Percentage2: | 35.66% |
| Map Size: | 205px |
| President | |
| Before Election: | Chester A. Arthur |
| Before Party: | Republican Party (United States) |
| After Election: | Grover Cleveland |
| After Party: | Democratic Party (United States) |
The 1884 United States presidential election in Mississippi took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose 9 representatives, or electors to the Electoral College, who voted for president and vice president.
Mississippi voted for the Democratic candidate, New York Governor Grover Cleveland over the Republican candidate, former Secretary of State James G. Blaine. Cleveland won Mississippi by a margin of 28.68%.
| United States presidential election in Mississippi, 1884[2] | ||||||
|---|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | ||
| Democratic | Grover Cleveland | 77,653 | 64.34% | 9 | ||
| Republican | James G. Blaine | 43,035 | 35.66% | 0 | ||
| Totals | 120,688 | 100.0% | 9 | |||