t the Hobart
election instead of one only; and this is exactly what we advocate. It
is also admitted that when two candidates ran together at the first
Launceston election the more popular candidate was defeated; and again
the _Argus_ correspondent writes of the recent Hobart election:--"The
defeat of Mr. Nicholls was doubtless due to the fact of his supporters'
over-confidence--nothing else explains it. Many people gave him No. 2
votes who would have given him No. 1 votes had they not felt assured of
his success."

A second reason why the wrong candidates are liable to be elected is
that the process of elimination adopted by all the Hare methods has no
mathematical justification. The candidate who is first excluded has one
preference only taken account of, while others have many preferences
given effect to. We have shown that this glaring injustice was
recognized by Mr. Hare, and only adopted as a last resort. Professor
Nanson admits that "the process of elimination which has been adopted by
all the exponents of Hare's system is not satisfactory," and adds--"I do
not know a scientific solution of the difficulty." To bring home the
inequity of the process, consider a party which nominates six
candidates, A, B, C, D, E, and F, and whose numbers entitle it to three
seats, and suppose the electors to vote in the proportions and order
shown below on the first count.

FIRST SECOND THIRD FOURTH
COUNT. COUNT. COUNT. COUNT.
7-vote ADEFBC ADEBC AEBC ABC
6-vote EFDACB EDACB EACB ACB
5-vote CEBDFA CEBDA CEBA CBA
4-vote BDFACE BDACE BACE BAC
4-vote DCEFBA DCEBA CEBA CBA
3-vote FBAECD BAECD BAEC BAC

It will be noted that F, having fewest first votes, is eliminated from
the second count, D from the third count, and E from the fourth. A has
then 13 votes, B 7, and C 9. If the quota be 9 votes, A's surplus would
be passed on to B, and A, B, and C would be declared elected. But D, E,
and F are the candidates most in general favour, and ought to have been
elected. For if any one of the rejected candidates be compared with any
one of the successful candidates it will be found that in every case the
rejected candidate is higher in order of favour on a majority of the
papers. Again, if the Block Vote be applied, by counting three effective
votes, the result would be--A 10 votes, B 12, C 9, D 21, E 22, and F