1860 United States presidential election in Minnesota

1860 United States presidential election in Minnesota

November 6, 1860 1864 →

Nominee Abraham Lincoln Stephen A. Douglas Party Republican Democratic Home state Illinois Illinois Running mate Hannibal Hamlin Herschel V. Johnson Electoral vote 4 0 Popular vote 22,069 11,920 Percentage 63.44% 34.27%

County Results Lincoln   40-50%   50-60%   60-70%   70-80%   80-90%   90-100% Douglas   50-60%   60-70% Unknown/No Vote

President before election James Buchanan Democratic Elected President Abraham Lincoln Republican

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The 1860 United States presidential election in Minnesota took place on November 6, 1860, as part of the 1860 United States presidential election. Minnesota voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.

Minnesota voted in its first ever United States presidential election, having been admitted as the 32nd state on May 11, 1858. The state was won by Illinois Representative Abraham Lincoln (Republican Party (United States)), running with Senator Hannibal Hamlin, with 63.44% of the popular vote, against Senator Stephen A. Douglas (D–Illinois), running with 41st Governor of Georgia Herschel V. Johnson, with 34.27% of the popular vote.

With 63.44% of the popular vote, Lincoln's victory within the state would be his second strongest victory in terms of percentage in the popular vote in the 1860 election after Vermont.^[1] Minnesota would never vote Democratic until Franklin D. Roosevelt won it in 1932, 72 years later.