United States presidential election in New York, 1988

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United States presidential election in New York, 1988
New York
1984 ←
November 8, 1988
→ 1992

  Dukakis1988rally cropped.jpg 43 George H.W. Bush 3x4.jpg
Nominee Michael Dukakis George H.W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 36 0
Popular vote 3,347,882 3,081,871
Percentage 51.62% 47.52%

New york presidential results 1988.svg

County Results
  Dukakis—70-80%
  Dukakis—60-70%
  Dukakis—50-60%
  Dukakis—<50%
  Bush—<50%
  Bush—50-60%
  Bush—60-70%

President before election

Ronald Reagan
Republican

Elected President

George H.W. Bush
Republican


The 1988 United States presidential election in New York took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose thirty-six representatives, or electors to the Electoral College, who voted for President and Vice President.

New York was won by Governor Michael Dukakis (D-Massachusetts) with 51.62% of the popular vote over Vice President George H.W. Bush (R-Texas) with 47.52%.[1]

Results [edit]

United States presidential election in New York, 1988
Party Candidate Votes Percentage Electoral votes
Democratic Michael Dukakis 3,347,882 51.62% 36
Republican George H.W. Bush 3,081,871 47.52% 0
Right to Life William Marra 20,497 0.32% 0
New Alliance Lenora Fulani 15,845 0.24% 0
Libertarian Ron Paul 12,109 0.19% 0
Workers World Larry Holmes 4,179 0.06% 0
Socialist Workers James Warren 3,287 0.05% 0
Write-in Edward Winn 10 0.00% 0
Write-in Willa Kenoyer 3 0.00% 0
Totals 6,485,683 100.0% 36

References [edit]

  1. ^ "1988 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 13 October 2012.