A Zogby/Reuters Poll released July 16 shows these results when voters are presented with a list of 4 presidential candidates: Obama 46%, McCain 36%, Barr 3%, Nader 3%, other or undecided 12%. See here for more details. When the voter is only presented with the names of Obama and McCain, 3% still volunteer someone else. Thanks to ThirdPartyWatch for this news.
If Bob Barr polled 3% in each state, the Libertarian Party would become a qualified party, for the first time ever, in Arkansas, Connecticut (presidential status only), Iowa, and Kentucky. At that point, the only states in which the party would never have been a qualified party would be Minnesota, New Jersey, New York, Pennsylvania, Rhode Island, Tennessee, and Virginia. Minnesota requires a vote of 5% for any statewide race; New Jersey requires 10% for all votes cast for lower house of the legislature; New York requires 50,000 votes for Governor; Pennsylvania requires registration of 15%; Rhode Island requires 5% for president or governor; Virginia requires 10% for any statewide race. The Tennessee requirement is currently under attack in federal court.
The Libertarian Party has also never been a qualified party in the District of Columbia, but it has some reason to believe it can attain that status this year, since it has a strong candidate for Delegate to the U.S House. He needs 7,500 votes to give the party “qualified” status.
Finally, although the Libertarian Party has been a qualified party for statewide office in both Illinois and Georgia, it has never been a ballot-qualified party in those two states for district and county office. Those two states, along with Connecticut, are the only states in which it is possible for a party to be “qualified” for all statewide office, but not all office.