At this point, enough votes are in to project (wth a pretty high level of certainty) that the order of finish in the Presidential election will be as follows…
1 – Barack Obama
2 – John McCain
3 – Ralph Nader (ind)
4 – Bob Barr (lib)
5 – Cynthia McKinney (grn)
6 – Chuck Baldwin (con)
7 – Alan Keyes (aip)
Those are the seven candidates who have or appear likely to surpass 100,000 votes each.
Ron Paul will likely come in 8th, followed by a wide field of Socialists and other smaller parties.
5 responses so far ↓
1 gls // Nov 5, 2008 at 1:55 am
http://news.aol.com/elections/2008/president#presResults
is showing Baldwin ahead of Mckinney by over 25K votes with 77% reporting
2 Austin Cassidy // Nov 5, 2008 at 2:02 am
Baldwin isn’t on the ballot in California though.
3 Austin Cassidy // Nov 5, 2008 at 2:05 am
Looks like McKinney could do 55,000 or a little better in CA. Baldwin might get 5,000-10,000 write-in votes. Maybe.
4 rdupuy // Nov 5, 2008 at 2:39 am
Sounds like a large number for McKinney in CA. She may do only 30,000.
However I admit I, at first, also thought McKinney might top Baldwin due to Ca.
5 paulie cannoli // Nov 5, 2008 at 4:50 am
With 90% counted
Ralph Nader, 611,125 3rd
Bob Barr, 460,619 4th
Chuck Baldwin, 167,795 5th
Cynthia McKinney, 128,153 6th
Alan Keyes, AIP 26,001
Ron Paul, CST 17,812
Paul and Baldwin are best positioned to benefit from write-in votes, especially in California.
There is thus a decent chance Ron Paul will have more than Keyes in the end.
Doesn’t look like much chance of McKinney passing Baldwin at this point.
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