1996 United States presidential election in Wyoming

1996 United States presidential election in Wyoming

← 1992 November 5, 1996 2000 →
 
Nominee Bob Dole Bill Clinton Ross Perot
Party Republican Democratic Reform
Home state Kansas Arkansas Texas
Running mate Jack Kemp Al Gore Patrick Choate
Electoral vote 3 0 0
Popular vote 105,388 77,934 25,928
Percentage 49.81% 36.84% 12.25%

County Results

President before election

Bill Clinton
Democratic

Elected President

Bill Clinton
Democratic

The 1996 United States presidential election in Wyoming took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

Wyoming was won by Senator Bob Dole (R-KS), with Dole winning 49.81 percent to 36.84 percent over President Bill Clinton (D) by a margin of 12.97 points. Billionaire businessman Ross Perot (Independent-TX) finished in third, with 12.25 percent of the popular vote.[1]

With 12.25 percent of the popular vote, Wyoming would prove to be Perot's fourth-strongest state in the 1996 election after Maine, Montana and Idaho.[2] This is the last time a Democrat won any counties other than Albany or Teton counties. As of the 2020 presidential election, this remains the last time that Sweetwater County has voted for a Democratic presidential candidate, as well as the last time that a Republican failed to win a majority of the state's vote, or won by a margin of less than 30 points. This was the first election ever where a president was elected twice without ever carrying the state.

  1. ^ "1996 Presidential General Election Results – Wyoming". Dave Leip’s U.S. Election Atlas.
  2. ^ "1996 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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