1992 United States presidential election in Oklahoma

1992 United States presidential election in Oklahoma

← 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 James Stockdale
Electoral vote 8 0 0
Popular vote 592,929 473,066 319,878
Percentage 42.65% 34.02% 23.01%

County Results

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

The 1992 United States presidential election in Oklahoma took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president.

Oklahoma was won by incumbent President George H. W. Bush (R-Texas) with 42.65 percent of the popular vote over Governor Bill Clinton (D-Arkansas) with 34.02 percent. Businessman Ross Perot (I-Texas) finished in third, with 23.01 percent of the popular vote.[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.[2] As of the 2020 presidential election, this is the last election in which Beckham County, Delaware County, Washita County, and Stephens County voted for a Democratic presidential candidate.[3] This was the first time that the Democrats won the White House without Jackson County and the first time since Oklahoma statehood that a losing Republican managed to carry any of the 10 counties bordering the Red River. This is the most recent time that a Democrat has carried a majority of Oklahoma's counties in a presidential election.

  1. ^ Cite error: The named reference results was invoked but never defined (see the help page).
  2. ^ "1992 Presidential General Election Data - National". Dave Leip’s U.S. Election Atlas. Retrieved June 8, 2012.
  3. ^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016

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