Talk:Highest averages method

message: Im having trouble understanding the exact rationale behind these methods. Why do seemingly provisionaly choosen divisors result in a proportional result? Also, why are such complicated ways of allocation used at all; it would seem that simply dividing the number of votes each party won with the total votes and multiplying this with the number of available seets, rounded down, and then largest remaining fractions up to the remaining number of seets chosen (if you write this out, it is mathematically equivalent to Hare quota) would by definition result in the most proportional results, and its the most intuitive - mathematically most evident - way of doing this. --195.29.116.30 04:54, 19 July 2006 (UTC)
thread_title: why?

message: The table calculates modified S-L with 1.4, 3.4, 5.4,.... This results in a top row with a divisor of 1.4, which is incomparable with all the other tables. Why not use the mathematically equivalent series 1, 2 3/7 (2.429..), 3 6/7,...? This would lead to more comparable numbers (although it would take a footnote to explain it.) 187.143.7.74 (talk) 14:49, 12 February 2010 (UTC)
replies: text: No, it doesn't. Modified S-L (and the table) uses 1.4, 3, 5, .... --Roentgenium111 (talk) 18:05, 24 January 2013 (UTC)

text: The table has been modified to allow for direct comparisons across rows, by using 0.5, 1.5... for Sainte-Lague (which is the "more correct" fencepost sequence; using 1, 3, ... is just a convenience to avoid dealing with fractions). Closed Limelike Curves (talk) 21:02, 2 February 2024 (UTC)
thread_title:  Modified S-L 

message: I have radically changed the part which previously said: "D'Hondt favors the merging of parties, while Sainte-Laguë favours neither merging nor splitting parties which expect to gain more than 1 or 2 seats. (It does favor splitting of very small parties – expecting to gain only 1-2 seats – into still smaller ones). Modified Saint-Laguë prevents this splitting advantage for small parties, while remaining impartial towards party size for all larger parties" on the grounds that it is not true.  Just to illustrate from the current example, if the Yellow party split its 47,000 votes into 6 new parties with about 7,800 votes each, it would win 6 seats in both unmodified Sainte-Laguë and modified Sainte-Laguë (with unmodified Sainte-Laguë it might even successfully do a split into 7 parties with 6,700 votes each so going from 4 to 7 seats).  The reality is that Sainte-Laguë often rewards splits, even for large parties, rather like the Hare quotas in Hong Kong. --Rumping (talk) 16:45, 22 May 2015 (UTC)
thread_title:  Sainte-Laguë and splits 

message: For missing references and other useful material (including mathematical properties of the apportionments achieved and other algorithms always yielding the same apportionment) see http://de.wikipedia.org/wiki/Sitzzuteilungsverfahren . -- Wegner8 07:13, 19 October 2017 (UTC)
thread_title:  Fill gaps from the German Wikipedia 

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thread_title:  External links modified 

message: Hi
not sure on this but I'm not sure the huntington hill example is correct.
because the divisor is n(n+1) the first number in the series is infinity - hence all who meet the threshold get at least one seat. so the first round should both eliminate two of the parties, and assign four seats to the other parties leaving six to be allocated. these will go four to yellow, one to white, one to red, giving overall allocation of 5,2,2,1. In this example it is the same seat allocation as d'hondt and modified S-L.
--Dirtyrottenscoundrel (talk) 09:32, 23 October 2020 (UTC)
replies: text: The example was incorrect and has been corrected. Closed Limelike Curves (talk) 20:58, 2 February 2024 (UTC)
thread_title:  is huntington-hill example correct 

message: @ et al: The new column in the table "Allows zeros" might need some re-thinking.  Any of the methods that gives  can be said to not allow zeros in that every alternative will get one representative before any alternative gets a second alternative.  However, if giving one representative to every alternative makes too many representatives then a floor number of votes is established and only those alternatives with that many votes will get a representative.  In this case, some alternatives will get zero representatives even though "Allows zeros" is listed as "no". —Quantling (talk | contribs) 21:26, 26 January 2024 (UTC)
replies: text: Seeing no objections, I am going to remove the column "Allows zeros" from the table, so that it does not mislead the reader into believing that some systems disallow zeros in all cases.  —Quantling (talk | contribs) 14:50, 31 January 2024 (UTC)

text: .  Your recent edit to restore the "Allows zeros" column to the table didn't quite work / appears incomplete.  I look forward to seeing your ideas for this.  Thank you —Quantling (talk | contribs) 14:31, 2 February 2024 (UTC)
replies: text: Corrected, thanks. Closed Limelike Curves (talk) 20:54, 2 February 2024 (UTC)
thread_title:  Allows zeros