United States presidential election in Washington (state), 1996

Last updated on 12 October 2017

The 1996 United States presidential election in Washington took place on November 5, 1996 as part of the 1996 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for President and Vice President.

Washington was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 49.84% to 37.30% by a margin of 12.54%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third with 8.92% of the popular vote .[1]

Bill Clinton.jpg
Bill Clinton.jpg
Bob Dole, PCCWW photo portrait.JPG
Bob Dole, PCCWW photo portrait.JPG


United States presidential election in Washington, 1996
Party Candidate Running mate Votes Percentage Electoral votes
Democratic Bill Clinton (incumbent) Al Gore 1,123,323 49.84% 11
Republican Bob Dole Jack Kemp 840,712 37.30% 0
Reform Ross Perot Patrick Choate 201,003 8.92% 0
Green Party Ralph Nader Winona LaDuke 60,322 2.68% 0
Libertarian Harry Browne Jo Jorgensen 12,522 0.56% 0
Natural Law Dr. John Hagelin Dr. V. Tompkins 6,076 0.27% 0
U.S. Taxpayers' Howard Phillips Herbert Titus 4,578 0.20% 0
Independent Charles Collins Rosemary Giumarra 2,374 0.11% 0
Workers World Party Monica Moorehead Gloria La Riva 2,189 0.10% 0
Socialist Workers Party James Harris Laura Garza 738 0.03% 0
Totals 2,253,837 100.0% 11


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