1940 United States presidential election in Oklahoma

The 1940 United States presidential election in Oklahoma took place on November 5, 1940, as part of the 1940 United States presidential election. Voters chose eleven[2] representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1940 United States presidential election
1940 United States presidential election in Oklahoma
November 5, 1940[1]
All 11 Oklahoma votes to the Electoral College
 
Nominee Franklin D. Roosevelt Wendell Willkie
Party Democratic Republican
Home state New York Indiana
Running mate Henry A. Wallace Charles L. McNary
Electoral vote 11 0
Popular vote 474,313 348,872
Percentage 57.4% 42.2%
President before election

Franklin D. Roosevelt
Democratic

 Elected President

Franklin D. Roosevelt
Democratic

Elections in Oklahoma
Government

Oklahoma was won by incumbent President Franklin D. Roosevelt (D–New York), running with Secretary Henry A. Wallace, with 57.41% of the popular vote, against Wendell Willkie (R–Indiana), running with Minority Leader Charles L. McNary, with 42.23% of the popular vote.[3][4]

Results
1940 United States presidential election in Oklahoma
Party Candidate Votes %
Democratic Franklin D. Roosevelt (inc.) 474,313 57.41%
Republican Wendell Willkie 348,872 42.23%
Prohibition Roger Babson 3,027 0.37%
Total votes 826,212 100%

References
  1. "United States Presidential election of 1940 - Encyclopædia Britannica". Retrieved September 11, 2018.
  2. "1940 Election for the Thirty-ninth Term (1941-45)". Retrieved September 11, 2018.
  3. "1940 Presidential General Election Results - Oklahoma". Retrieved September 11, 2018.
  4. "The American Presidency Project - Election of 1940". Retrieved September 11, 2018.


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