1996 United States presidential election in Montana

1996 United States presidential election in Montana

← 1992 November 5, 1996 (1996-11-05) 2000 →
 
Nominee Bob Dole Bill Clinton Ross Perot
Party Republican Democratic Reform
Home state Kansas Arkansas Texas
Running mate Jack Kemp Al Gore James Campbell
Electoral vote 3 0 0
Popular vote 179,652 167,922 55,229
Percentage 44.11% 41.23% 13.56%

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, as part of the 1996 United States presidential election. Voters chose three 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.[1] As of the 2020 presidential election, this is the last election in which Sheridan County, Dawson County, and Mineral County voted for the Democratic candidate.

With 13.56% of the popular vote, Montana would prove to be Ross Perot's second strongest state in the 1996 election after Maine.[2]

Montana was one of three states won by Clinton in 1992 that Bob Dole was able to flip, the others being Colorado and Georgia. This was also the first election since 1912 in which Montana voted differently than nearby California.

  1. ^ a b "Dave Leip's Atlas of U.S. Presidential Elections".
  2. ^ "1996 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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