United States presidential election in Oregon, 1988

United States presidential election in Oregon, 1988

November 8, 1988

Nominee Michael Dukakis George H.W. Bush Party Democratic Republican Home state Massachusetts Texas Running mate Lloyd Bentsen Dan Quayle Electoral vote 7 0 Popular vote 616,206 560,126 Percentage 51.28% 46.61%

County Results   Dukakis—60-70%   Dukakis—50-60%   Dukakis—<50%   Bush—<50%   Bush—50-60%   Bush—60-70%

President before election Ronald Reagan Republican Elected President George H. W. Bush Republican

The 1988 United States presidential election in Oregon took place on November 8, 1988 as part of the 1988 United States presidential election. Voters chose seven electors of the Electoral College, who voted for president and vice president.

Oregon was won by the Democratic nominee, Massachusetts governor Michael Dukakis, over the Republican nominee, Vice President George H. W. Bush. Oregon was one of just 10 states won by Dukakis in an election won by Vice President Bush. It also marked the first victory by a Democratic presidential candidate in Oregon since 1964; Democrats have won every presidential election in Oregon since, albeit it nearly went red in 2000.^[1]

Results

United States presidential election in Oregon, 1988[2]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Michael Dukakis	616,206	51.28%	7
	Republican	George H. W. Bush	560,126	46.61%	0
	Libertarian	Ron Paul	14,811	1.23%	0
	New Alliance Party	Lenora Fulani	6,487	0.54%	0
	No party	Write-in	3,974	0.33%	0
	No party	David Duke (write-in)	90	0.01%	0

References

1. "Oregon Presidential Election Voting History". 270ToWin.com. Retrieved June 12, 2012.
2. "1988 Presidential General Election Results - Oregon". Dave Leip's U.S. Election Atlas. Retrieved June 12, 2012.