1868 United States presidential election in New York

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United States presidential election in New York, 1868

← 1864 November 3, 1868 1872 →
Turnout91.7%[1] Increase 2.4 pp
  Horatio Seymour - Brady-Handysmall.jpg Ulysses S Grant by Brady c1870-restored (cropped).jpg
Nominee Horatio Seymour Ulysses S. Grant
Party Democratic Republican
Home state New York Illinois
Running mate Francis Preston Blair, Jr. Schuyler Colfax
Electoral vote 33 0
Popular vote 429,883 419,888
Percentage 50.59% 49.41%

President before election

Andrew Johnson

Elected President

Ulysses S. Grant

The 1868 United States presidential election in New York took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose 33 representatives, or electors to the Electoral College, who voted for president and vice president.

New York voted for the Democratic nominee, former Governor of New York Horatio Seymour, over the Republican nominee, General Ulysses S. Grant. Seymour won his home state by a very narrow margin of 1.18%, making him the first Democratic candidate since Franklin Pierce in 1852 to win the state. Seymour also became the first losing Democratic presidential candidate to win New York.


United States presidential election in New York, 1868[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Horatio Seymour of New York Francis Preston Blair, Jr. of Missouri 429,883 50.59% 33 100.00%
Republican Ulysses S. Grant of Illinois Schuyler Colfax of Indiana 419,888 49.41% 0 0.00%
Total 849,771 100.00% 33 100.00%


  1. ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
  2. ^ "1868 Presidential General Election Results - New York".