1996 United States presidential election in Washington (state)

1996 United States presidential election in Washington (state)
Turnout	74.52% 8.08%
Clinton 40-50% 50-60%	Dole 40-50% 50-60%
President before election; Bill Clinton; Democratic	Elected President ; Bill Clinton; Democratic

The 1996 United States presidential election in Washington took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.

The State of Washington was won by President Bill Clinton (D–AR) over Senator Bob Dole (R–KS), with Clinton winning 49.84% to 37.30% for a margin of 12.54%. Billionaire businessman Ross Perot (Reform–TX) finished in third, with 8.92% of the popular vote.^[2] As of the 2020 presidential election^[update], this is the last election in which Spokane County, Kittitas County, Pend Oreille County, Ferry County, and Asotin County voted for a Democratic presidential candidate.^[3]

^ Secretary of State: Kim Wyman. "Voter Turnout by Election". www.sos.wa.gov. Retrieved May 25, 2020.
^ Dave Leip’s U.S. Election Atlas; 1996 Presidential General Election Results – Washington
^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016