1948 United States presidential election in Wisconsin
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| Elections in Wisconsin |
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The 1948 United States presidential election in Wisconsin was held on November 2, 1948 as part of the 1948 United States presidential election. State voters chose twelve electors to the Electoral College, who voted for president and vice president.
Democratic Party candidate Harry S. Truman won Wisconsin with 51% of the popular vote, winning the state's twelve electoral votes.[1]
Results
| 1948 United States presidential election in Wisconsin | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Harry S. Truman | 647,310 | 50.70% | 12 | |
| Republican | Thomas E. Dewey | 590,959 | 46.28% | 0 | |
| Progressive | Henry A. Wallace | 25,282 | 1.98% | 0 | |
| Socialist | Norman Thomas | 12,547 | 0.98% | 0 | |
| Totals | 1,276,098 | 100.0% | 12 | ||
References
- ^ "1948 Presidential General Election Results - Wisconsin". Retrieved August 19, 2016.
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