United States presidential election in Indiana, 1992

United States presidential election in Indiana, 1992

← 1988 November 3, 1992 1996 →
Nominee George H.W. Bush Bill Clinton Ross Perot Party Republican Democratic Independent Home state Texas Arkansas Texas Running mate Dan Quayle Al Gore Electoral vote 12 0 0 Popular vote 989,375 848,420 455,934 Percentage 42.9% 36.8% 19.8%
County Results   Clinton—>70%   Clinton—60-70%   Clinton—50-60%   Clinton—40-50%   Bush—40-50%   Bush—50-60%   Bush—60-70%   Bush—>70%   Perot—40-50%
President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

George H. W. Bush
Republican

Bill Clinton
Democratic

The 1992 United States presidential election in Indiana took place on November 3, 1992, throughout all 50 states and D.C., which was part of the 1992 United States presidential election. Voters chose 12 representatives, or electors to the Electoral College, who voted for President and Vice President.

Indiana was won by President George H.W. Bush (R-TX). The presidential contest in Indiana was not a surprise, with Bush winning 42.91% to 36.79% over Governor Bill Clinton (D). Still, the margin of victory was lesser than in earlier elections; Clinton won in every state bordering Indiana, and except for the Midwest Bush finished first in no state more northern than Indiana. Billionaire Businessman Ross Perot (I-TX) finished in third with a significant 19.77% of the popular vote in Indiana. Indiana would continue to vote Republican in presidential elections until 2008, in which Barack Obama won by a close margin, becoming the first Democrat to carry the state since 1964.