United States presidential election in Montana, 1996

From Wikipedia, the free encyclopedia
Jump to: navigation, search
United States presidential election in Montana, 1996

← 1992 November 5, 1996 (1996-11-05) 2000 →
  Bob Dole, PCCWW photo portrait.JPG 44 Bill Clinton 3x4.jpg RossPerotColor.jpg
Nominee Bob Dole Bill Clinton Ross Perot
Party Republican Democratic Reform
Home state Kansas Arkansas Texas
Running mate Jack Kemp Al Gore Patrick Choate
Electoral vote 3 0 0
Popular vote 179,652 167,922 55,229
Percentage 44.1% 41.2% 13.6%

MT1996president.png
County results

President before election

Bill Clinton
Democratic

Elected President

Bill Clinton
Democratic

The 1996 United States presidential election in Montana took place on November 5, 1996. Voters chose 3 representatives, or electors to the Electoral College, who voted for president and vice president.

Montana voted for Senate Majority Leader Bob Dole over President Bill Clinton by a slim margin of 2.88%.[1] Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third, with 13.56% of the popular vote in Montana.[2] As of the 2016 presidential election, this is the last election in which Sheridan County, Dawson County, and Mineral County voted for the Democratic candidate.

Results[edit]

United States presidential election in Montana, 1996[3]
Party Candidate Running mate Votes Percentage Electoral votes
Republican Bob Dole Jack Kemp 179,652 44.11% 3
Democratic Bill Clinton Al Gore 167,922 41.23% 0
Reform Ross Perot Patrick Choate 55,229 13.56% 0
Libertarian Harry Browne Jo Jorgensen 2,526 0.62% 0
Natural Law (write in) John Hagelin Mike Tompkins 1,754 0.43% 0
Constitutional (write in) Howard Phillips Herbert Titus 152 0.04% 0
(write in) Charles E. Collins Rosemary Giumarra 20 0.00% 0
Independent (write in) Write-ins 5 0.00 0
Socialist (write in) Mary Cal Hollis Eric Chester 1 0.00 0

References[edit]