United States presidential election in Maryland, 1996

Main article: United States presidential election, 1996

United States presidential election in Maryland, 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 Popular vote 966,207 681,530 115,812 Percentage 54.25% 38.27% 6.50%

410px County Results   Clinton—80-90%   Clinton—70-80%   Clinton—60-70%   Clinton—50-60%   Clinton—40-50%   Dole—40-50%   Dole—50-60%   Dole—60-70%   Dole—70-80%   Dole—80-90%

President before election Bill Clinton Democratic Elected President Bill Clinton Democratic

The 1996 United States presidential election in Maryland 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.

Maryland was won by incumbent President Bill Clinton (D-Arkansas) with 54.25% of the popular vote over Senator Bob Dole (R-Kansas) with 38.27%. Businessman Ross Perot (Reform-Texas) finished in third with 6.50% of the popular vote. Clinton ultimately won the national vote, defeating both challengers and becoming re-elected to a second term as U.S. President.

Results

United States presidential election in Maryland, 1996
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Bill Clinton (incumbent)	966,207	54.25%	10
	Republican	Bob Dole	681,530	38.27%	0
	Reform	Ross Perot	115,812	6.50%	0
	Libertarian	Harry Browne	8,765	0.49%	0
	N/A	Others	8,556	0.48%	0
Totals			1,780,870	100.0%	10