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1996 United States presidential election in Hawaii

  (Redirected from United States presidential election in Hawaii, 1996)

The 1996 United States presidential election in Hawaii took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for president and vice president.

1996 United States presidential election in Hawaii

← 1992 November 5, 1996 2000 →
  Bill Clinton.jpg Bob Dole, PCCWW photo portrait.JPG RossPerotColor.jpg
Nominee Bill Clinton Bob Dole Ross Perot
Party Democratic Republican Reform
Home state Arkansas Kansas Texas
Running mate Al Gore Jack Kemp Patrick Choate
Electoral vote 4 0 0
Popular vote 205,012 113,943 27,358
Percentage 56.93% 31.64% 7.60%

County Results

President before election

Bill Clinton

Elected President

Bill Clinton

Hawaii was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 56.93% to 31.64% by a margin of 25.29%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third, with 7.6% of the popular vote.[1]


1996 United States presidential election in Hawaii
Party Candidate Running mate Votes Percentage Electoral votes
Democratic Bill Clinton (incumbent) Al Gore 205,012 56.93% 4
Republican Bob Dole Jack Kemp 113,943 31.64% 0
Reform Ross Perot Patrick Choate 27,358 7.60% 0
Green Ralph Nader Winona LaDuke 10,386 2.88% 0
Libertarian Harry Browne Jo Jorgensen 2,493 0.69% 0
Natural Law Dr. John Hagelin Dr. V. Tompkins 570 0.16% 0
U.S. Taxpayers' Party Howard Phillips Herbert Titus 358 0.10% 0
Totals 360,120 100.0% 4


  1. ^ Leip, David. "Dave Leip's Atlas of U.S. Presidential Elections". Retrieved 2018-06-14.

See alsoEdit