Talk:Limited voting

2006 comments

I don't really understand the article about limited voting. Does this system mean that the number of POSITIONS for which a voter can vote is fewer than the number of vacancies (while presumably more than 1)? If so, I think the text should be edited to reflect this.

I would also like to know where this method has been tried, or is in place.

This system has been used in the UK in the 19th Century. the voter has fewer votes than there are seats to be filled - that's the point.

thus if there three seats available a voter would have two or one votes. --Red Deathy 14:53, 22 June 2006 (UTC)[reply]

Copyright

Copyright in the UK is dated from the death of the Author, 70 yrs IIRC - some of Bernard Shaw's works published in the 1890's are still in cop[yright because he didn't die until 1950. Industry practise is generally to assume items over 100 years old are out of copyright - I'd assume Charles Seymour did die more than 70 years ago, but could you check. Regardless, I think citation of passages is allowable, so I'll not delete the passage unless some wiser head informs me.--Red Deathy 07:10, 2 October 2006 (UTC)[reply]

Just checked Library of Congress Authorities, Charles Seymour, author of Electoral reform in England and Wales died in 1963, so if that book was published in the UK it is still in copyright - be careful people - I'll try and read up on whether citation is OK.--Red Deathy 07:14, 2 October 2006 (UTC)[reply]

Thank you for your research invalidating my assumptions on English copyright law and how long ago the author died. The title page of the original book is reproduced in the reprint. It reads 'New Haven: Yale University Press/ London: Humphrey Milford/ Oxford University Press/ MDCCCCXV' so I think it was published in the UK as well as the United States. The copyright notice says 'Copyright, 1915 by Yale University Press/ First printed from type October, 1915, 600 copies'. I think the short passage I have quoted qualifies as fair use, but I am no expert on copyright. --Gary J 02:25, 23 November 2006 (UTC)[reply]

The example

Not to split hairs, but I make it that the liberals achieve well over a droop quota of votes for each candidate, i.e. the example given the liberals are merely managing to realise their vasdt proponerance of votes (i.e. avoiding the pitfall illustrated in my own example) - could an example be found, or made up, showing how tight organisation could be super-proportional and allow a party without substantial support to wion all three? I suppose if we could find an example of one party where a lot of their votes go to one candidate, or a three/three (or whatever) battle?--Red Deathy 08:14, 2 October 2006 (UTC)[reply]

I take your point. I have substituted the 1880 Birmingham election for that in 1868. This was a closer election. An example of things going wrong was Leeds 1874 (also a three member seat). The Liberals polled more votes than the Conservatives, but only elected one member. L1 15,390; C1 14,864; C2 13,194; L2 11,850; L3 5,954. There does not seem to be a British example of both parties trying to win all the seats in the same limited vote election.

Applying the simplest concept of a droop quota to this kind of complex non-proportional election seems to be difficult. I take the equivalent calculation to be the minimum number of votes which must elect a candidate, although in practice some (or indeed all as in the 1880 Birmingham election) candidates might be elected with fewer votes depending upon the number of candidates and the relative distribution of support. An example, taking the simplest possible limited vote situation - two places to fill and three candidates, with voters limited to one vote; is that the minimum vote on which a candidate must be returned is one vote over one-third of the total. However the minimum vote on which a candidate might be elected is just one vote more than the runner up. The most extreme situation would be, if there were 100 voters and 99 voted for candidate A, 1 for candidate B and none for candidate C. This approach to identifying the minimum level of support which must elect a candidate does however break-down when contemplating an election with three places to be filled, in which the elector can cast one or two votes, as no individual candidate can receive more votes than the number of electors participating but if the elector uses his second vote then this may count against the other choice. It is beyond my mathematical ability to formulate an equation to precisely identify the must elect point for all examples of that situation.

Looking at the 1880 Birmingham election, the minimum number of votes which must elect a candidate is one-quarter plus 1 (disregarding any fractional remainder). There were 94,635 votes cast, so the simplistic must elect figure is 23,659 (as if three candidates receive that number of votes only 23,658 would be left for other candidates). However in practice none of the five candidates reached that "quota" figure. It is not possible, as it would be in a proportional election, to simply sum the total votes for a party and then divide that figure by the must elect quota figure to calculate how many seats the party "should" win because of the effect of the second votes. By dividing the Liberal votes by three and multiplying by two we get an approximation of the number of Liberal supporters (43,061). The leading Conservative candidate got 15,735 votes. The relative shares are Liberal 73.24%, Conservative 26.76%. I think we can therefore say, that if instead of a limited vote election it had been a proportional one, that vote distribution must have elected a Conservative MP to the third seat. --Gary J 10:49, 23 November 2006 (UTC)[reply]

The position was worse than that: you should have divided the sum of the Liberal votes by two (as each voter had two votes) to get about 32296 and the sum of the Conservative votes by two to get about 15021 making the ratio about 68.3% to 31.7%. --Rumping (talk) 11:23, 21 March 2013 (UTC)[reply]