United States presidential election in Maryland, 1996

United States presidential election in Maryland, 1996
Maryland
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%

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.[1] Clinton ultimately won the national vote, defeating both challengers and becoming re-elected to a second term as U.S. President.[2]

Results

United States presidential election in Maryland, 1996[1]
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

References

  1. 1 2 "1996 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 8 June 2012.
  2. "1996 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.
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