United States presidential election in Kansas, 1992

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United States presidential election in Kansas, 1992
← 1988 November 3, 1992 1996 →
  43 George H.W. Bush 3x4.jpg 44 Bill Clinton 3x4.jpg RossPerotColor.jpg
Nominee George H.W. Bush Bill Clinton Ross Perot
Party Republican Democratic Independent
Home state Texas Arkansas Texas
Running mate Dan Quayle Al Gore James Stockdale
Electoral vote 6 0 0
Popular vote 446,951 390,434 312,358
Percentage 38.9% 33.7% 27.0%

County Results
  Tie (Perot-Bush)—<40%

President before election

George H. W. Bush

Elected President

Bill Clinton

The 1992 United States presidential election in Kansas took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose six representatives, or electors to the Electoral College, who voted for President and Vice President.

Kansas was won by incumbent President George H.W. Bush (R-Texas) with 38.88% of the popular vote over Governor Bill Clinton (D-Arkansas) with 33.74%. Businessman Ross Perot (I-Texas) finished in third with 26.99% of the popular vote. Kansas was surprisingly close, likely because Ross Perot split the vote. Had Clinton won in Kansas, it would have been a major upset victory.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]


United States presidential election in Kansas, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush (incumbent) 449,951 38.88% 6
Democratic Bill Clinton 390,434 33.74% 0
Independent Ross Perot 312,358 26.99% 0
Libertarian Andre Marrou 4,314 0.37% 0
N/A Other 199 0.02% 0
Totals 1,157,256 100.0% 6


  1. ^ a b "1992 Presidential General Election Results - Kansas". U.S. Election Atlas. Retrieved 8 June 2012. 
  2. ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.