United States presidential election in Montana, 1992

United States presidential election in Montana, 1992

← 1988 November 3, 1992 1996 →
Nominee Bill Clinton George H. W. Bush Ross Perot Party Democratic Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle Electoral vote 3 0 0 Popular vote 154,507 144,207 107,225 Percentage 37.6% 35.1% 26.1%
County Results   Clinton—60-70%   Clinton—50-60%   Clinton—40-50%   Clinton—30-40%   Bush—30-40%   Bush—40-50%   Bush—50-60%
Results by county showing number of votes by size and candidates by color.
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 Montana was won on November 3, 1992 by Governor Bill Clinton (D-Arkansas) with a 37.63% of the popular vote plurality over incumbent President George H. W. Bush's (R-Texas) 35.12%, and businessman Ross Perot's (I-Texas) 26.12%. All of Montana's three electoral votes were assigned to Clinton, contributing to his 370 to 168 electoral vote win of the 1992 presidential election.

Official results by county from the Montana Secretary of State.

Montana was won by Governor Bill Clinton (D-Arkansas) with 37.63% of the popular vote over incumbent President George H. W. Bush (R-Texas) with 35.12%. Businessman Ross Perot (I-Texas) finished in third with 26.12% of the popular vote. It was the first time the state voted for a Democrat since it was won by President Lyndon B. Johnson in 1964. It is also the last time that the state voted Democratic as of 2016. Considered a safe Republican state, Clinton's win was due to independent Ross Perot splitting the Republican vote and allowing the Democrats to win the state for the first time since 1964.