1892 United States presidential election in Delaware
Election in Delaware
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 County Results Cleveland 40-50% 50-60%
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The 1892 United States presidential election in Delaware took place on November 8, 1892. All contemporary 44 states were part of the 1892 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.
Delaware was won by the Democratic nominees, former President Grover Cleveland of New York and his running mate Adlai Stevenson I of Illinois.
Despite reclaiming the presidency, Cleveland performed much worse in Delaware than he did in his 1888 re-election bid. The State shifted 10 points for Harrison, voting for Cleveland by only 1.4 percentage points. This would be one of four states where Harrison outperformed his initial run for the presidency in 1888, these other states were New Hampshire, Georgia and, South Carolina.
Results
| 1892 United States presidential election in Delaware[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Grover Cleveland | 18,581 | 49.90% | 3 | |
| Republican | Benjamin Harrison (incumbent) | 18,077 | 48.55% | 0 | |
| Prohibition | John Bidwell | 564 | 1.51% | 0 | |
| Write-ins | Scattered | 13 | 0.03% | 0 | |
| Totals | 37,235 | 100.00% | 3 | ||
| Voter turnout | — | ||||
See also
References
- ^ Dave Leip's U.S. Election Atlas; Presidential General Election Results – Delaware
Notes
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