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] As of the 2024 presidential election [update] , this is the last election in which Kansas was decided by single digits and the last time that Ellis County , Cherokee County , Leavenworth County , Labette County , and Miami County voted for a Democratic presidential candidate,[ 3] as well as the last election in which Wabaunsee County , Anderson County , Jefferson County , and Morris County (the four counties Perot either won or tied in) did not support the Republican candidate.[ 3] This is the closest the Democrats have come since 1964 to winning Kansas.
With 26.99% of the popular vote, Kansas would prove to be Perot's fifth-strongest state after Maine , Alaska , Utah and Idaho . Perot took third in the state overall but defeated Clinton or Bush for second place in 63 of Kansas's 105 counties.[ 1]