1890 Idaho gubernatorial election
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 County results Shoup: 50–60% 60–70% Wilson: 50–60% | |||||||||||||||||
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| Elections in Idaho |
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The 1890 Idaho gubernatorial election was held on 1 October 1890, in order to elect the first governor of Idaho upon Idaho acquiring statehood in July 1890. Incumbent Republican governor of the Idaho Territory George L. Shoup defeated Democratic nominee Benjamin Wilson.[1]
General election
[edit]On election day, 1 October 1890, Republican nominee George L. Shoup won the election by a margin of 2,314 votes against his opponent Democratic nominee Benjamin Wilson, thereby retaining Republican control over the new office of governor. Shoup was sworn in as the first governor of the new state of Idaho on 8 December 1890.[2]
Results
[edit]| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | George L. Shoup (incumbent[a]) | 10,262 | 56.35% | |
| Democratic | Benjamin Wilson | 7,948 | 43.65% | |
| Total votes | 18,210 | 100.00% | ||
| Republican hold | ||||
Notes
[edit]- ^ Incumbent governor of Idaho Territory.
References
[edit]- ^ "George L. Shoup". National Governors Association. Retrieved May 12, 2023.
- ^ "ID Governor". ourcampaigns.com. September 26, 2005. Retrieved May 12, 2023.
Categories:
- 1890 Idaho elections
- Idaho gubernatorial elections
- October 1890 events
- 1890 United States gubernatorial elections
- United States gubernatorial elections in the 1890s
- 1890 in Idaho
- 1890 elections
- 1890 elections in North America
- 1890 elections in the United States
- 1890s in Idaho
- 1890s Idaho elections
- Government of Idaho