United States presidential election in Mississippi, 1948
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The 1948 United States presidential election in Mississippi took place on November 2, 1948 in Mississippi as part of the wider United States presidential election of 1948.
The Democratic Party candidate, South Carolina governor Strom Thurmond, overwhelmingly won Mississippi against fellow Democrat, incumbent President Harry S. Truman by a margin of 148,154 votes, or 77.08%. Although Truman was the national Democratic Party candidate, Thurmond managed to be placed on the ballot in Mississippi, South Carolina, Louisiana, and Alabama as the official Democratic candidate. Outside of these four states, Thurmond was forced to run under the label of the States' Rights Democratic Party. The Republican Party candidate, New York governor Thomas E. Dewey, had no impact on the race in Mississippi, only obtaining 2.62% of the popular vote, or 5,043 votes total.
Analysis
Mississippi in this time period was a one-party state dominated by the Democratic Party. The Republican Party was virtually nonexistent as a result of disenfranchisement among poor whites and African Americans, including voter intimidation against those who refused to vote Democrat. In the 1948 election, rifts in the national Democratic Party began to occur.
Southern Democrats walked out at the party's national convention in Philadelphia because of Truman's endorsement of civil rights for African Americans. This segregationist faction met on July 17, 1948, in Birmingham, Alabama, nominating South Carolina governor as its nominee for president. Mississippi governor Fielding L. Wright was nominated for Vice President.
Thurmond carried all of Mississippi's 82 counties, the vast majority with over 90% of the vote. The "weakest" region for Thurmond in the state came from the northeastern corner where he failed to break 60% in 4 counties. As of 2016, this is the last presidential election in which the Democratic candidate carried every county in Mississippi.[1]