United States presidential election in Minnesota, 1996

United States presidential election in Minnesota, 1996

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 51.1% 35.0% 11.8%

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.