# 1918 United States Senate election in Wyoming

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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.

## Democratic primary
### Candidates

* John E. Osborne, former Governor of Wyoming, former U.S. Assistant Secretary of State

### Results

Democratic primary :
| Party | Party | Candidate | Votes | % |
| | Democratic | John E. Osborne | 4,186 | 100.00% |
| Total votes | Total votes | Total votes | 4,186 | 100.00% |

## Republican primary
### Candidates

* Francis E. Warren, incumbent U.S. Senator

### Results

Republican primary :
| Party | Party | Candidate | Votes | % |
| | Republican | Francis E. Warren (inc.) | 12,709 | 100.00% |
| Total votes | Total votes | Total votes | 12,709 | 100.00% |

## General election
### Results

1918 United States Senate election in Wyoming :
| Party | Party | Candidate | Votes | % |
| | Republican | Francis E. Warren (inc.) | 23,975 | 57.77% |
| | Democratic | John E. Osborne | 17,528 | 42.23% |
| Total votes | Total votes | Total votes | 41,503 | 100.00% |
| | Republican hold | Republican hold | Republican hold | Republican hold |