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1948 United States presidential election in Indiana information


1948 United States presidential election in Indiana
1948 United States presidential election in Indiana
← 1944 November 2, 1948[1] 1952 →

All 13 Indiana votes to the Electoral College
 
Nominee Thomas E. Dewey Harry S. Truman
Party Republican Democratic
Home state New York Missouri
Running mate Earl Warren Alben W. Barkley
Electoral vote 13 0
Popular vote 821,079 807,833
Percentage 49.58% 48.78%

County Results

President before election

Harry S. Truman
Democratic

Elected President

Harry S. Truman
Democratic

The 1948 United States presidential election in Indiana took place on November 2, 1948, as part of the 1948 United States presidential election. Indiana voters chose 13[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Indiana was won by Governor Thomas Dewey (R–New York), running with Governor Earl Warren, with 49.58% of the popular vote, against incumbent President Harry S. Truman (D–Missouri), running with Senator Alben W. Barkley, with 48.78% of the popular vote.[3][4] As of 2020 presidential election, this is the last time that Johnson County voted for a Democratic presidential candidate.[5] This was also the last time until 2012 that Indiana voted for a different candidate than Virginia (whoever wrote that Virginia is nearby Indiana, learn geography).

  1. ^ "United States Presidential election of 1948 - Encyclopædia Britannica". Retrieved December 20, 2017.
  2. ^ "1948 Election for the Forty-First Term (1949-53)". Retrieved December 20, 2017.
  3. ^ "1948 Presidential General Election Results – Indiana". Retrieved December 20, 2017.
  4. ^ "The American Presidency Project – Election of 1948". Retrieved December 20, 2017.
  5. ^ 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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