Barr On Ballot Before 94.6% of Voters
October 23rd, 2008Since it is now almost certain that Barr will not be on the ballot in Connecticut, it can be presumed that his name will be on ballots containing 94.6% of the voters. The calculation uses 2004 turnout data. Of course, the distribution of where the votes come from across the U.S. will not be precisely what it was in 2004, and the exact figure can’t be known until all the votes are counted in November 2008.
The last time the Libertarian Party missed putting its presidential candidate on the ballot in more than two states was 1988. In that year, Ron Paul, the Libertarian nominee, was on ballots containing 92.3% of the total vote cast. Paul missed being on in North Carolina, West Virginia, Indiana and Missouri.
October 23rd, 2008 at 12:55 pm
Richard, can you please provide the number of potential EV’s for each candidate?
Thanks,
PS
Technically, each candidate is eligible in these States without pre-election write-in registration: Alabama, Delaware, Iowa, New Hampshire, New Jersey, Oregon, Pennsylvania, Rhode Island, Vermont, and Wyoming
October 23rd, 2008 at 1:11 pm
This is my latest article entitled “Self Destruction thy name is Bob Barr”
http://www.nolanchart.com/article5275.html
October 23rd, 2008 at 1:14 pm
The printed October 1 Ballot Access News has the number of electoral votes in the petitioning chart. But the number of electoral votes is misleading, since the number of electoral votes isn’t strictly proportional to population or number of votes cast. For example, Wyoming has 5 times as many electoral votes per person as California does.
October 23rd, 2008 at 1:21 pm
I understand that EV’s aren’t “strictly proportional”; I just haven’t seen anything showing how many EV’s each candidate is eligible for.
October 23rd, 2008 at 3:22 pm
You can see the electoral votes at http://www.thegreenpapers.com/G08/President-Details.phtml