1918 United States Senate election in Wyoming
The 1918 United States Senate election in Wyoming took place on November 5, 1918. Incumbent Republican Senator Francis E. Warren ran for re-election to his fifth consecutive term in the Senate, in his first popular election. He was opposed by former Governor John E. Osborne, the Democratic nominee, who had until recently served as the U.S. Assistant Secretary of State. Owing in part to Warren's long record of service, and also to the strong Republican performance throughout the country in the 1918 elections, Warren defeated Osborne by a wide margin.
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| Elections in Wyoming |
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Democratic primary
Candidates
- John E. Osborne, former Governor of Wyoming, former U.S. Assistant Secretary of State
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | John E. Osborne | 4,186 | 100.00% | |
| Total votes | 4,186 | 100.00% | ||
Republican primary
Candidates
- Francis E. Warren, incumbent U.S. Senator
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Francis E. Warren (inc.) | 12,709 | 100.00% | |
| Total votes | 12,709 | 100.00% | ||
General election
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Francis E. Warren (inc.) | 23,975 | 57.77% | |
| Democratic | John E. Osborne | 17,528 | 42.23% | |
| Total votes | 41,503 | 100.00% | ||
| Republican hold | ||||
References
- Chaplin, W. E. (1919). 1919 Official Directory of Wyoming and Election Returns for 1918. pp. 35–37.
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