<html><body><div style="color:#000; background-color:#fff; font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif;font-size:10pt"><div><span></span></div><div>For completeness, on the score voting version of this, I don't think the way I suggested is actually the best way to go about it. I suggested that if, for example, someone gives scores of 9/10 and 7/10 to A and B respectively, then this would count as 0.9 * 0.7 people approving both, 0.9 * 0.3 approving just A, 0.1 * 0.7 approving just B, and 0.1 * 0.3 approving neither. This seemed to me at the time to be the most balanced way of doing it. However, this causes the maximum score to have relevance causing a possible failure of independence of multiplication factors.</div><div><br></div><div>For example, someone gives scores of 10/10 and 5/10 to A and B respectively. This would result in 0.5 people approving both and 0.5 approving just A.</div><div><br></div><div>But
consider scores of 10/20 and 5/20 to A and B respectively. This would mean that 0.125 would approve both, 0.375 would approve just A, 0.125 would approve just B, and 0.375 would approve neither. This gives completely different ratios, and so could potentially lead to different result if everyone gave the same scores but out of a different maximum.</div><div><br></div><div>So it seems that a better way would be to clump them together as much as possible and do away with the mixing and matching. So for example, scores of 10/10, 8/10 and 6/10 to A, B and C would mean 0.6 of a voter would approve ABC, 0.2 would approve AB and 0.2 would approve A. For the same scores out of 20, it would be the same but scaled down - 0.3, 0.1 and 0.1. This method also means far less "splitting" of the voter. Instead of splitting into up to 2^c parts where c candidates are awarded a non-zero score, voters would only ever be split into a maximum of c
parts.</div><div><br></div><div>While I initially thought this way if doing it was asymmetrical and lopsided, I now see it as simpler and less arbitrary, and it's trivial to see that it passes independence of multiplication factors. There is still the question - is it The Right Way, but I now think it is. I initially saw it as asymmetrical because I pictured each voter like a square with approvals going down from left to right. Using the example above (scores of 10, 8, 6), it would be 3333332211 where the left 6 tenths approve all 3 of ABC, the next 2 tenths approve of AB and the right two tenths approve of just A. But it could equally be seen like this: 1233333321, which is nice and symmetrical, like a pyramid. Because of this, I now view it in a more positive light and it assuages my doubts on whether it's The Right Way. But that's just an insight into my weird mind I suppose.</div><div><br></div><div><div>One final
thing. On this "paradox":</div><div><br></div><div><div><div>2 to elect</div><div><br></div><div>10 voters: A, B, C</div><div>10 voters: A, B, D</div><div><br></div></div><div>where AB is no better than CD, it would only ever really come up where there is an exact tie. In all other cases, such as:</div><div><br></div><div><div>2 to elect</div><div><br></div><div>99 voters: A, B, C</div><div>99 voters: A, B, D</div></div></div><div>1 voter: C</div><div>1 voter: D</div><div><br></div><div>then CD is justified because it is objectively more proportional. If you pick AB because of something like greater overall satisfaction, you have to come up with some potentially arbitrary trade-off between satisfaction and proportionality, as well as a definition of satisfaction. The sequential version would still pick the most popular overall candidate first anyway, and subsequent candidates would be picked to get the best proportionality, so I
would see it as "Reweighted Range Voting done right". On the non-sequential version, there could be some sort of tie-break system based on something like total votes where there is an exact tie, which could then elect AB over CD in the first example. But that would be an add-on rather than an intrinsic part of the system.</div></div><div><br><blockquote style="padding-left: 5px; margin-top: 5px; margin-left: 5px; border-left-color: rgb(16, 16, 255); border-left-width: 2px; border-left-style: solid;"> <div style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 10pt;"> <div style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 12pt;"> <div dir="ltr"> <div style="margin: 5px 0px; padding: 0px; border: 1px solid rgb(204, 204, 204); border-image: none; height: 0px; line-height: 0; font-size: 0px;" contenteditable="false" readonly="true"></div>
<font face="Arial" size="2"> <b><span style="font-weight: bold;">From:</span></b> Toby Pereira <tdp201b@yahoo.co.uk><br> <b><span style="font-weight: bold;">To:</span></b> Toby Pereira <tdp201b@yahoo.co.uk>; "election-methods@lists.electorama.com" <election-methods@lists.electorama.com> <br> <b><span style="font-weight: bold;">Sent:</span></b> Monday, 2 June 2014, 18:05<br> <b><span style="font-weight: bold;">Subject:</span></b> Re: [EM] Proportional Approval System + Score conversion<br> </font> </div> <div><br><div id="yiv2089800436"><div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 10pt; background-color: rgb(255, 255, 255);"><div><span></span></div><div dir="ltr"><div>I was thinking about a score voting version of this, and I think I've got it to work. I was struggling because the system only works with exact approvals rather than
fractional scores. But instead of fractional scores, a score of e.g. 9 out of 10 can be translated into 0.9 people approving the candidate and 0.1 people not approving them. If I score a 9 and a 7 to two candidates, then this would be 0.9 * 0.7 approving both, 0.9 * 0.3 approving just the former, 0.1 * 0.7 approving the latter and 0.1 * 0.3 approving neither.</div><div><br clear="none"></div><div>If I give a candidate 1 out of 10, and no-one else gives them anything, then if that candidate gets elected then the 0.1 voter "possesses" 1/number of voters, so 10 of the candidate. This might seem
a weird anomaly, but it doesn't actually break the system. It's just what happens so that every candidate provides the same total amount of representation for the voters.</div><div><br clear="none"></div><div>This system should always elect the score winner where there is a single winner. For example with scores out of 100:</div><div><br clear="none"></div><div>Elect 1</div><div><br clear="none"></div><div>1 voter: A=100, B=49</div><div>1 voter: A=0, B=49</div><div><br clear="none"></div><div>Elected candidates/voters = 1/2 = 0.5, so that's the target score that we work out the deviation from.</div><div><br clear="none"></div><div>Normal score voting would obviously elect A. If A is elected, then 1 voter has a score of 1 and the other has a score of 0. So the total squared deviation from 0.5 (candidates/voters) is 0.5^2 + 0.5^2 = 0.5</div><div><br clear="none"></div><div>If B is elected, 0.98 voters have a score of 1/0.98 = 1.02. 1.02 voters have a
score of 0. The squared deviation from 0.5 is 0.98*0.52^2 + 1.02*0.5^2 = 0.52.
This is a higher deviation so A is elected.</div></div><div><br clear="none"><blockquote style="padding-left: 5px; margin-top: 5px; margin-left: 5px; border-left-color: rgb(16, 16, 255); border-left-width: 2px; border-left-style: solid;"> <div style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 10pt;"> <div style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 12pt;"> <div id="yiv2089800436yqt20990"><div dir="ltr"> <div style="margin: 5px 0px; padding: 0px; border: 1px solid rgb(204, 204, 204); border-image: none; line-height: 0; font-size: 0px; min-height: 0px;"></div> <font face="Arial" size="2"> <b><span style="font-weight: bold;">From:</span></b> Toby Pereira <tdp201b@yahoo.co.uk><br clear="none"> <b><span style="font-weight: bold;">To:</span></b> "election-methods@lists.electorama.com" <election-methods@lists.electorama.com> <br
clear="none"> <b><span style="font-weight: bold;">Sent:</span></b> Saturday, 31 May 2014, 17:32<br clear="none"> <b><span style="font-weight: bold;">Subject:</span></b> [EM] Proportional Approval System<br clear="none"> </font> </div> <div><br clear="none"><div id="yiv2089800436"><div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 10pt; background-color: rgb(255, 255, 255);"><div dir="ltr"><div>I've come up with a system of proportional representation using approval voting that I think avoids many of the problems of other systems.</div><div><br clear="none"></div><div>For each voter, their score is the number of elected candidates that they "possess". If v voters have voted for a particular candidate, then each voter possesses 1/v of that candidate. If there are c candidates and v voters, then as long as each elected candidate has received at least one approval, then
the average possession score for a voter is c/v. Full proportionality is
achieved if every voter has a score of c/v. The measure of a set of candidates is the average squared deviation from c/v for the voters' scores (lower deviation being better). For example:</div><div><br clear="none"></div><div>4 to
elect</div><div><br clear="none"></div><div>3 voters: A, B, C, D</div><div>1 voter: E, F, G, H</div><div><br clear="none"></div><div>The obvious proportional result is A, B, C, E. And you can see that the 3 ABCD voters would each possess 1/3 of each of A, B, C, so would have a score of 1. The single EFGH voter would fully possess E, so would also have a score of 1. c/v in this case is 1 so this is trivially a proportional result.</div><div><br clear="none"></div><div>But this also works where there are commonly rated candidates. For example:</div><div><br clear="none"></div><div>6 to elect</div><div><div><br clear="none"></div><div>20 voters: A, B, C, D, E, F</div><div>10 voters: A, B, C, G, H, I</div><div><br clear="none"></div></div><div>Reweighted Range Voting <a href="http://www.rangevoting.org/RRV.html" target="_blank" rel="nofollow" shape="rect">http://www.rangevoting.org/RRV.html</a> would elect A, B, C, D, E, F on the basis that
20 voters get 3 candidates and 10 voters get 3, making it "proportional". But I would
argue it's best to ignore
candidates approved by all when considering who else to elect. In this case the 20 ABCDEF voters would all have a score of 3 * 1/30 + 3 * 1/20 = 1/4. The 10 ABCGHI voters would score 3 * 1/30 = 1/10.</div><div><br clear="none"></div><div>Whereas with A, B, C, D, E, G, the 20 ABCDEF voters would score 3 * 1/30 + 2 * 1/20 = 1/5. The 10 ABCGHI voters would score 3 * 1/30 + 1 * 1/10 = 1/5. The equal scores imply a proportional result. To check, c/v = 6/30 = 1/5 so it all adds up.</div><div><br clear="none"></div><div>This also works for other examples that I've tried. Where there isn't an exact proportional result available, the squared deviation seems to work as the best measure. For example:</div><div><br clear="none"></div><div>4 to elect</div><div><br clear="none"></div><div>5 voters: A, B, C, D</div><div>3 voters: E, F, G, H</div><div>1 voter: I, J, K, L</div><div><br clear="none"></div><div>I won't do the maths here, but this would give an exact 3-way
tie between ABCE, ABEF and ABEI. This is the same as the result
that Sainte-Laguë would give (which I would argue is objectively more proportional than D'Hondt).</div><div><br clear="none"></div><div>It's not obvious how to score a set of candidates where at least one candidate has no approvals, but this should never cause a problem. As long as there are at least as many approved candidates as seats, there is no need to consider unapproved candidates. If there aren't, you would always elect all the approved candidates anyway.</div><div><br clear="none"></div><div>For a similar reason, it's not obvious what to do with score voting. For example, if a single voter has given a candidate a score of 1 out of 10, then does this voter possess 1/10 of this candidate or the whole candidate? And if 10 voters have given a score of 1/10 then do they each possess 1/100 or 1/10? If it's 1/100, then candidates won't be "fully possessed" and we end up with a no obvious way of working out the best set of candidates because the
average c/v would not be the
same for each set of candidates and we'd be looking at more than just the squared deviation. We'd have to consider both average and deviation, which would make it messy. But otherwise we end up with a voter fully possessing a candidate that they've given a low score to, and this would count against that voter. So at the moment, this is just a system for approval voting.</div><div><br clear="none"></div><div>I think this is superior to the approval version of Reweighted Range Voting because it is independent of commonly approved candidates, and so it also avoids the proportionality problems that this can cause, which I have discussed elsewhere.</div><div><br clear="none"></div><div>It is also superior to STV methods because in these methods a voter is considered to just have a single vote and if that vote is fully assigned to a candidate then it doesn't matter what else happens. For example:</div><div><br clear="none"></div><div>4 to
elect</div><div><br clear="none"></div><div>10 voters: A, B,
C, D</div><div>10 voters: A, B, C, D, E, F</div><div><br clear="none"></div><div>The best result here seems to be A, B, C, D, but STV is indifferent between this and, say, A, B, E, F. The 10 ABCD voters can have their votes assigned to A and B (5 to each) and the 10 ABCDEF voters can have their votes assigned to E and F (5 to each). This would heavily favour the ABCDEF voters. In practice, most STV methods wouldn't do this because of the sequential way they operate. But STV methods that consider whole sets of candidates non-sequentially would be indifferent to these results. So avoiding this result would be more accidental than ideological.</div><div><br clear="none"></div><div>However, my method is also indifferent between results that you might expect a method not to be. For example:</div><div><br clear="none"></div><div>2 to elect</div><div><br clear="none"></div><div>10 voters: A, B, C</div><div>10 voters: A, B, D</div><div><br
clear="none"></div><div>You might think that A, B is the obvious
result. But my system would say this is equal to C, D. So in this sense it could be said to fail monotonicity. However, I do think it makes sense in the way that proportional systems work. If A and B are elected, then they are both "shared" among all 20 voters. Where if it's C and D then although each voter only has one candidate, they share the candidate with half as many people, and their possession score would be the same under either result.</div><div><br clear="none"></div><div>If this all stands up, I think this is better than the other systems of proportional representation based on approval voting that I've seen.</div></div><div><br clear="none"></div></div></div></div><br clear="none">----<br clear="none">Election-Methods mailing list - see <a href="http://electorama.com/em" target="_blank" rel="nofollow" shape="rect">http://electorama.com/em</a>for list info<br clear="none"><br clear="none"><br clear="none"></div></div> </div> </div>
</blockquote><div></div> </div></div></div></div><br><br></div> </div> </div> </blockquote><div></div> </div></body></html>