1948 United States presidential election in Nebraska

1948 United States presidential election in Nebraska

← 1944 November 2, 1948[1] 1952 →
All 6 Nebraska 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 6 0 Popular vote 264,774 224,165 Percentage 54.15% 45.85%
County Results Dewey   50-60%   60-70%   70-80% Truman   50-60%   60-70%
President before election Harry S. Truman Democratic Elected President Harry S. Truman Democratic

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

Nebraska was won by Governor Thomas E. Dewey (R–New York), running with Governor Earl Warren, with 54.15% of the popular vote, against incumbent President Harry S. Truman (D–Missouri), running with Senator Alben W. Barkley, with 45.85% of the popular vote.^[3][4] As of the 2020 presidential election, this is the last occasion Wheeler County has voted for a Democratic presidential candidate.^[5]

Dewey won Nebraska by an 8.3% margin, a significantly reduced margin from his 1944 run, when he won by 17.2%.

Though he failed to win the state, Truman's performance in Nebraska was the best for a Democrat since Franklin Roosevelt's victory in 1936, and would remain the best until Lyndon B. Johnson's narrow victory in 1964. This was the only election between 1900 and 1992 in which Nebraska voted differently than neighboring Colorado.

With 54.15% of the popular vote, Nebraska would prove to be Dewey's third strongest state in the nation after Vermont and Maine.^[6]