United States presidential election in Montana, 1948
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| Elections in Montana | |||||||||
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The 1948 United States presidential election in Montana took place on November 2, 1948, throughout the continuous 48 states. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.
Montana voted for the Democratic nominee, President Harry Truman, over the Republican nominee, New York Governor Thomas E. Dewey. Truman won Montana by a large margin of 9.94%.
Results
| United States presidential [1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Harry Truman (incumbent) | 119,071 | 53.09% | 4 | |
| Republican | Thomas E. Dewey | 96,770 | 43.15% | 0 | |
| Progressive | Henry Wallace | 7,313 | 3.26% | 0 | |
| Socialist | Norman Thomas | 695 | 0.31% | 0 | |
| Prohibition | Claude Watson | 429 | 0.19% | 0 | |