Talk:Counting Single Transferable Votes

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Contents 1 Changes 2 Warren? 3 Senatorial/Gregory 4 Elimination 5 Meek's method?

[edit] Changes

I re-jigged the changes to establish the difference between Hare and Cincinnati - they both use random redistribution of votes, but Hare's are from a randomly determined surplus, whereas cincinnati is drawn at regular intervals from the whole set, to ensure a fair spread. Also emmphasised that the transfers from Hare can be from first preferences as well as from later transfers.

This is not the essential difference. The method of randomisation is a separate issue from the set from which the surplus is selected. The term "Cincinnati method" does not encompass every aspect of the process used in Cambridge. I have created a separate heading for randomisation. I also added a new grouping to emphasise the difference between surplus for previously-unelected candidate, and transfers to already-elected candidate. Joestynes 03:22, 27 May 2005 (UTC)

Right, that looks fair enough, though I still think that the example isn't entirely up to snuff, since the surplus is transfered with Hare whether it's a surplus from transfers or a simple first preference - i.e. the last 20 of 220 1st preferences would likewise be transfered...--Red Deathy 07:02, 2005 May 27 (UTC)

That's true (except saying the "last" 20 votes presupposes some ordering, whereas votes are usually randomised). I've just added the statement that, where the surplus arises from first preferences, the Hare System and the Cincinnati System are equivalent. The example is designed to highlight the difference between them, which occurs when the surplus arises from transfers. Joestynes 07:29, 27 May 2005 (UTC)

I reformatted the page so that it would be a lot easier to read. Basically, just changed indents and such so that each of the different methods could be distinguished as a separate topic and bumped the Example to its own major heading. --Leep4life 20:37, 14 December 2005 (UTC)

[edit] Warren?

There is another technique, which also requires computer counting, called "Warren" - I assume, like "Meek" it's named for it's inventor. Can anyone explain the differences between the Meek and Warren methods? I think it's described in the Tidemann paper that introduced CPO-STV, but I can't find a copy of that any more - the link in CPO-STV is broken--Po8crg 02:03, 8 February 2006 (UTC)

[edit] Senatorial/Gregory

Do these methods really transfer fractional surpluses from all votes, or just those which arrived in the counting round which put the candidate over quota? I.e. are they a fraction version of Hare or of Cincinnati. The article currently suggests of Cincinnati (while saying Clarke) but I have seen a fractional version of Hare used. And the link [1] has a commentary for Dromore in 21 May 1997 saying "Of the 602 votes transferred to Gribben, 327 transferred further" suggesting that this is more Hare-like than Cincinnati-like. Similarly, is Clarke Hare-like or Cincinnati-like? --Henrygb 00:55, 22 April 2006 (UTC)

[edit] Elimination

Of course it's an extremely unlikely scenario, but what happens if too many candidates have the same number of votes after, say, a few counts? What is the mechanism for deciding whom to eliminate? When using examples to explain the system it's naturally easier to take round numbers, and now my example has come to a standstill after two elections and one elimination. --Dub8lad1 20:42, 17 May 2007 (UTC)

It depends, all could be eliminated at once, or, if I recall correctly, they can toss a coin to see who goes out first.--Red Deathy 07:56, 18 May 2007 (UTC)

With any voting system, one can construct a hypothetical election where each candidate receives identical votes; some random lottery will be needed in such cases. That aside, there are some points in McDougall Trust articles [I've added a link to the index on the article page]:

From Random tie-breaking in STV:

Ties can arise in any STV election during exclusion. With some methods ties can arise at other stages as well; Jeffrey O’Neill [2] lists the cases. O’Neill also lists four tie-breaking methods. Two methods use the first or last difference in prior rounds to break a tie, and two methods use later preferences — Borda scores or most (fewest) last place preferences. Brian Wichmann [3] proposes to examine all possible outcomes.

From A new way to break STV ties in a special case:

• The ERS rules [6] and the Church of England rules use the first-difference method in an attempt to break a tie.
• The Meek algorithm [7] uses a deterministic algorithm based upon a random number generator to break a tie. No manual intervention is used. The New Zealand variant uses a similar method.
• When the Church of England rules are applied using a computer, then the software must break the ties without manual intervention in a manner which is not defined (by the rules).
• For Ireland, the manual rules are being computerized and have been used for three trial constituencies in 2002. Here, tie-breaking invokes a manual procedure, ie, the computer software does not break the tie.

  A curiosity is that in the Irish rules if when allocating surplus remainders there is a tie of the fractional part, the surplus vote is given to the candidate with the largest total number of papers from that surplus; if that is also tied then first difference is used.

• It seems that a Condorcet comparison has been used to resolve a strong tie between A and B (i.e. tie can’t be broken by first/last difference) in very small manual counts i.e. examine the papers to see how many times A is ahead of B compared to vice versa.

jnestorius^(talk) 22:19, 20 May 2007 (UTC)

[edit] Meek's method?

On the subject of this pages meek's method section. Does anybody follow it? The paper referenced at the bottom gives a perfectly understandable explanation for anybody with the skills to reproduce it here.

Where does the formula 1-(quota/candidatesVotes) come from for the weighting of an elected candidate. If you read the paper everybody starts out with a hopeful weighting. When he becomes elected we use the equation w1j=w0j*q/vj , where in this case w0j=1. We then iterate this formula repeatedly till q/vj=1.

see section 2.9 in the paper for details.

Zfishwiki (talk) 16:07, 7 May 2008 (UTC)

Okay, I've corrected the formula for the weightings, but there is still a lot of room for improvement in the section.

Zfishwiki (talk) 18:00, 9 May 2008 (UTC)