United States presidential election in New York, 1988

United States presidential election in New York, 1988

← 1984 November 8, 1988 1992 →
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%
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

Ronald Reagan
Republican

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 Democratic Governor Michael Dukakis of Massachusetts with 51.62% of the popular vote over Republican Vice President George H.W. Bush of Texas, who took 47.52%.

1988 would mark the end of an era in New York's political history. Since the 1940s, New York had been a Democratic-leaning swing state, usually voting Democratic in close elections, but often by small margins. Republicans would dominate much of upstate New York and populated suburban counties like Nassau County, Suffolk County, and Westchester County. However they would be narrowly outvoted statewide by the fiercely Democratic and massively populated New York City area, along with some upstate cities like Buffalo, Albany, and the college town of Ithaca. This pattern would endure in 1988 for the final time, allowing Bush to keep the race fairly close, only losing the state to Dukakis by 4 points.