United States presidential election in Minnesota, 1996

United States presidential election in Minnesota, 1996
Minnesota
November 5, 1996
 
Nominee Bill Clinton Bob Dole Ross Perot
Party Democratic Republican Reform
Home state Arkansas Kansas Texas
Running mate Al Gore Jack Kemp Pat Choate
Electoral vote 10 0 0
States carried 0
Popular vote 1,120,438 766,476 257,704
Percentage 51.10% 34.96% 11.75%

County Results
  Clinton—60-70%
  Clinton—50-60%
  Clinton—<50%
  Dole—<50%
President before election

Bill Clinton
Democratic

 Elected President

Bill Clinton
Democratic

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

A Democratic-leaning state, Minnesota was comfortably won by incumbent Democratic President Bill Clinton. Clinton took 51.10% of the popular vote over Republican challenger Bob Dole, who took 34.96%, a victory margin of 16.14%. Reform Party candidate Ross Perot finished in third with 11.75% of the popular vote.[1]


Results

United States presidential election in Minnesota, 1996
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton (incumbent) 1,120,438 51.10% 10
Republican Bob Dole 766,476 34.96% 0
Reform Ross Perot 257,704 11.75% 0
Green Ralph Nader 24,908 1.14% 0
Libertarian Harry Browne 8,271 0.38% 0
Grassroots Dennis Peron 4,898 0.22% 0
U.S. Taxpayer Howard Phillips 3,416 0.16% 0
Write-ins Write-ins 2,903 0.13% 0
Natural Law John Hagelin 1,808 0.08% 0
Ind. Grassroots John Birrenback 787 0.04% 0
Socialist Workers James Harris 684 0.03% 0
Socialist Equality Jerome White 347 0.02% 0
Totals 2,192,640 100.0% 10

References

  1. ↑ "1996 Presidential General Election Results - Minnesota". U.S. Election Atlas. Retrieved 30 January 2013.