1916 United States presidential election in Rhode Island

1916 United States presidential election in Rhode Island

← 1912 November 7, 1916 1920 →
 
Nominee Charles Evans Hughes Woodrow Wilson
Party Republican Democratic
Home state New York New Jersey
Running mate Charles W. Fairbanks Thomas R. Marshall
Electoral vote 5 0
Popular vote 44,858 40,394
Percentage 51.08% 46.00%

County Results
Hughes
  40-50%
  50-60%


President before election

Woodrow Wilson
Democratic

Elected President

Woodrow Wilson
Democratic

The 1916 United States presidential election in Rhode Island took place on November 7, 1916, as part of the 1916 United States presidential election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island was won by the Republican nominee, U.S. Supreme Court Justice Charles Evans Hughes of New York, and his running mate Senator Charles W. Fairbanks of Indiana. They defeated Democratic nominees, incumbent Democratic President Woodrow Wilson and Vice President Thomas R. Marshall. Hughes won the state by a narrow margin of 5.08 points.

This was the first election since 1892 where a Democratic candidate earned more than forty percent in at least one Rhode Island county. As of the 2020 United States presidential election, it is also the most recent election where the Democratic candidate lost Rhode Island but won the presidency.


© MMXXIII Rich X Search. We shall prevail. All rights reserved. Rich X Search