1890 Rhode Island gubernatorial election

The 1890 Rhode Island gubernatorial election was held on April 2, 1890. Democratic nominee John W. Davis defeated incumbent Republican Herbert W. Ladd with 48.76% of the vote.

1890 Rhode Island gubernatorial election

April 2, 1890

Nominee John W. Davis Herbert W. Ladd Party Democratic Republican Popular vote 20,548 18,988 Percentage 48.76% 45.06%

Governor before election Herbert W. Ladd Republican Elected Governor John W. Davis Democratic

General election

Candidates

Major party candidates

• John W. Davis, Democratic
• Herbert W. Ladd, Republican

Other candidates

• John H. Larry, Prohibition
• Arnold B. Chace, Union[1][2][3]

Results

1890 Rhode Island gubernatorial election[4]

Party		Candidate	Votes	%	±%
	Democratic	John W. Davis	20,548	48.76%
	Republican	Herbert W. Ladd	18,988	45.06%
	Prohibition	John H. Larry	1,820	4.32%
	Independent	Arnold B. Chace	752	1.78%
Majority			1,560
Turnout
	Democratic gain from Republican		Swing

References

1. "A Quaker for Governor". The sun. New York, N.Y. March 2, 1890. p. 18. Retrieved 9 January 2021.
2. "For the Suppression of Rum". Pittsburg dispatch. Pittsburg, Pa. February 26, 1890. p. 4. Retrieved 9 January 2021.
3. "Personal and Political". The herald. Milbank, S.D. March 4, 1890. p. 2. Retrieved 9 January 2021.
4. Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN 9780871879967. Retrieved July 27, 2020.