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

← 1964 November 8, 1966 1968 →
 
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

Primary elections edit

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

Democratic primary edit

Candidates edit

Results edit

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

Republican primary edit

Candidates edit

Results edit

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 edit

Candidates edit

Major party candidates

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

Other candidates

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

Results edit

1966 Iowa gubernatorial election[2]
Party Candidate Votes % ±%
Democratic Harold Hughes (incumbent) 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

References edit

  1. ^ a b c "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.