1966 Iowa gubernatorial election

The 1966 Iowa gubernatorial election was held on November 8, 1966. Incumbent Democrat Harold Hughes defeated Republican nominee William G. Murray with 55.34% of the vote.

1966 Iowa gubernatorial election
November 8, 1966
 
Nominee Harold Hughes William G. Murray
Party Democratic Republican
Popular vote 494,259 394,518
Percentage 55.34% 44.17%
County results
Hughes:      50–60%      60–70%      70–80%
Murray:      40-50%      50-60%      60-70%
Governor before election

Harold Hughes
Democratic

 Elected Governor

Harold Hughes
Democratic

Elections in Iowa

Primary elections

Primary elections were held on September 6, 1966.[1]

Democratic primary

Candidates

Results
Democratic primary results[1]
Party Candidate Votes %
Democratic Harold Hughes 80,198 100.00
Total votes 80,198 100.00

Republican primary

Candidates

Results
Republican primary results[1]
Party Candidate Votes %
Republican William G. Murray 87,371 50.5
Republican Robert K. Beck 85,733 49.5
Total votes 173,109 100.00

General election

Candidates

Major party candidates

  • Harold Hughes, Democratic
  • William G. Murray, Republican

Other candidates

  • David B. Quiner, Independent
  • Charles Sloca, Independent

Results
1966 Iowa gubernatorial election[2]
Party Candidate Votes % ±
Democratic Harold Hughes 494,259 55.34%
Republican William G. Murray 394,518 44.17%
Independent David B. Quiner 3,680 0.41%
Independent Charles Sloca 715 0.08%
Majority 99,741
Turnout 893,175
Democratic hold Swing
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References
  1. "Summary of Official Canvass of Votes Cast in Iowa Primary Election" (PDF). Secretary of State of Iowa. 1966. Retrieved March 20, 2020.
  2. "Summary of Official Canvass of Votes Cast in Iowa General Election" (PDF). Secretary of State of Iowa. 1966. Retrieved March 20, 2020.
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