1996 United States presidential election in Wyoming information
Election in Wyoming
Main article: 1996 United States presidential election
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
Dole
40–50%
50–60%
60–70%
Clinton
40–50%
President before election
Bill Clinton
Democratic
Elected President
Bill Clinton
Democratic
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v
t
e
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[update], 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.
^"1996 Presidential General Election Results – Wyoming". Dave Leip’s U.S. Election Atlas.
^"1996 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
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