1948 United States presidential election in Wisconsin

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1948 United States presidential election in Wisconsin

← 1944 November 2, 1948 1952 →
  Harry S. Truman.jpg ThomasDewey.png
Nominee Harry S. Truman Thomas E. Dewey
Party Democratic Republican
Home state Missouri New York
Running mate Alben W. Barkley Earl Warren
Electoral vote 12 0
Popular vote 647,310 590,959
Percentage 50.7% 46.3%

Wisconsin Presidential Election Results 1948.svg
County Results

President before election

Harry S. Truman
Democratic

Elected President

Harry S. Truman
Democratic

The 1948 United States presidential election in Wisconsin was held on November 2, 1948 as part of the 1948 United States presidential election. State voters chose twelve electors to the Electoral College, who voted for president and vice president.

Democratic Party candidate Harry S. Truman won Wisconsin with 51% of the popular vote, winning the state's twelve electoral votes.[1]

Results[edit]

1948 United States presidential election in Wisconsin
Party Candidate Votes Percentage Electoral votes
Democratic Harry S. Truman 647,310 50.70% 12
Republican Thomas E. Dewey 590,959 46.28% 0
Progressive Henry A. Wallace 25,282 1.98% 0
Socialist Norman Thomas 12,547 0.98% 0
Totals 1,276,098 100.0% 12

References[edit]

  1. ^ "1948 Presidential General Election Results - Wisconsin". Retrieved August 19, 2016.