1988 United States presidential election in New York

1988 United States presidential election in New York

← 1984 November 8, 1988 1992 →
 
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Alliance Liberal Conservative
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

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

International policy with the buckling Soviet Union was a critical component of the political landscape in the late 1980s. Vice President Bush can be seen here standing with United States President Ronald Reagan and Soviet General Secretary Mikhail Gorbachev, on the New York waterfront, 1988.

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 36 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%, a victory margin of 4.10%.[1] This result made New York roughly 12% more Democratic than the nation-at-large. Dukakis’ statewide victory is largely attributable to winning four of five boroughs of New York City overall with 66.2% of the vote. However, even though losing the city in a landslide, Bush's 32.8% share of the vote was a relatively respectable showing for a Republican in NYC, particularly in retrospect. In the 8 elections that followed 1988, Republican presidential candidates have received only 17% to 24% of the vote in New York City. This would be the last time until 2016 that the state would vote differently than neighboring Pennsylvania.

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%. Bush became the first Republican ever to win the White House without carrying Broome County and the first to win without Montgomery County since Rutherford B. Hayes in 1876.

This was the last election in which a Republican presidential nominee won heavily populated Nassau and Westchester Counties, as well as Monroe, Onondaga, and Ulster Counties,[2] and also the last election in which New York was decided by a single-digit margin. Beginning in 1992, the Democrats would make substantial inroads in the suburbs around New York City as well as parts of upstate, making New York a solid blue state that has gone Democratic by double-digit margins in every election since, consequently, this is the last time a Democrat lost the Upstate region. Rensselaer, Franklin, and St. Lawerence counties would not vote Republican again until 2016.

  1. ^ "1988 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved October 13, 2012.
  2. ^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016

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