1916 Vermont gubernatorial election
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The 1916 Vermont gubernatorial election took place on November 7, 1916. Incumbent Republican Charles W. Gates, per the "Mountain Rule",[1] did not run for re-election to a second term as Governor of Vermont. Republican candidate Horace F. Graham defeated Democratic candidate William B. Mayo to succeed him.
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November 7, 1916
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Graham: 40â50% 50â60% 60â70% 70â80% 80â90% 90-100% Mayo: 40â50% 50â60% 60â70% Tie: 50% No Vote/Data: | |||||||||||||||||
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Republican primary
Results
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Republican | Horace F. Graham | 33,244 | 99.9 | ||
| Republican | Other | 37 | 0.1 | ||
| Total votes | '33,281' | '100' | |||
Democratic primary
Results
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Democratic | William B. Mayo | 3,562 | 99.9 | ||
| Democratic | Other | 1 | 0.0 | ||
| Total votes | '3,563' | '100' | |||
General election
Results
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Republican | Horace F. Graham | 43,265 | 71.1 | ||
| Democratic | William B. Mayo | 15,789 | 25.9 | ||
| Socialist | W. R. Rowland | 920 | 1.5 | ||
| Prohibition | Lester W. Hanson | 876 | 1.4 | ||
| N/A | Other | 4 | 0.0 | ||
| Total votes | '60,854' | '100' | |||
