United States presidential election in Nevada, 1948

United States presidential election in Nevada, 1948
November 2, 1948[1]
All 3 Nevada votes to the Electoral College

 
Nominee Harry S. Truman Thomas Dewey
Party Democratic Republican
Home state Missouri New York
Running mate Alben W. Barkley Earl Warren
Electoral vote 3 0
Popular vote 31,291 29,357
Percentage 50.4% 47.3%
President before election

Harry S. Truman
Democratic

 Elected President

Harry S. Truman
Democratic

The 1948 United States presidential election in Nevada took place on November 2, 1948, as part of the 1948 United States presidential election. Nevada voters chose three[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Nevada was won by incumbent President Harry S. Truman (D–Missouri), running with Senator Alben W. Barkley, with 50.37% of the popular vote, against Governor Thomas Dewey (R–New York), running with Governor Earl Warren, with 47.26% of the popular vote.[3][4]

Results

United States presidential election in Nevada, 1948
Party Candidate Votes %
Democratic Harry S. Truman (inc.) 31,291 50.37%
Republican Thomas Dewey 29,357 47.26%
Progressive Henry A. Wallace 1,469 2.36%
Total votes 62,117 100%

References

  1. ↑ "United States Presidential election of 1948 - Encyclopædia Britannica". Retrieved October 26, 2017.
  2. ↑ "1948 Election for the Forty-First Term (1949-53)". Retrieved October 26, 2017.
  3. ↑ "1948 Presidential General Election Results - Nevada". Retrieved October 26, 2017.
  4. ↑ "The American Presidency Project - Election of 1948". Retrieved October 26, 2017.
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