<html><body><div style="color:#000; background-color:#fff; font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif;font-size:13px"><div><span></span></div><div id="yui_3_16_0_1_1412332069246_52951" dir="ltr">Kathy</div><div id="yui_3_16_0_1_1412332069246_52952" dir="ltr"><br></div><div id="yui_3_16_0_1_1412332069246_52959" dir="ltr">The problem I have with it is that the result changes when irrelevant alternatives are removed. Assign the obvious seats as before:</div><div id="yui_3_16_0_1_1412332069246_53007" dir="ltr"><br></div><div id="yui_3_16_0_1_1412332069246_53008" dir="ltr">5 voters (A): 2 seats</div><div id="yui_3_16_0_1_1412332069246_53009" dir="ltr">3 voters (B): 1 seat</div><div id="yui_3_16_0_1_1412332069246_53010" dir="ltr"><br></div><div id="yui_3_16_0_1_1412332069246_53011" dir="ltr">And according to your method C is first in line for the next seat followed by B and then A. But what if there was no C? It should still be B ahead of A, right? But the ideal share of seats becomes:</div><div id="yui_3_16_0_1_1412332069246_52997" dir="ltr"><br></div><div id="yui_3_16_0_1_1412332069246_52967" dir="ltr">5 voters (A): 2.5</div><div id="yui_3_16_0_1_1412332069246_53003" dir="ltr">3 voters (B): 1.5</div><div id="yui_3_16_0_1_1412332069246_53030" dir="ltr"><br></div><div dir="ltr">And the seats owed to the groups now are:</div><div id="yui_3_16_0_1_1412332069246_53049" dir="ltr"><br></div><div dir="ltr">5 voters (A): 0.5</div><div dir="ltr">3 voters (B): 0.5</div><div id="yui_3_16_0_1_1412332069246_53031" dir="ltr"><br></div><div id="yui_3_16_0_1_1412332069246_53029" dir="ltr">This happens with any pair. A v B, A v C or B v C all produce ties for the final seat when the third faction is ignored. Sainte-Laguë (which I would argue works proportionally for party voting) also gives a three-way tie. I specifically came up with this scenario a while ago as a test for methods that seem to work for two factions, and it was important that my own method passed it before I deemed it acceptable.</div><div id="yui_3_16_0_1_1412332069246_53068" dir="ltr"><br></div><div id="yui_3_16_0_1_1412332069246_53069" dir="ltr">Toby</div><div id="yui_3_16_0_1_1412332069246_52958"><br> </div><blockquote id="yui_3_16_0_1_1412332069246_52946" 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 id="yui_3_16_0_1_1412332069246_52945" style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px;"> <div id="yui_3_16_0_1_1412332069246_52944" style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 12px;"> <div id="yui_3_16_0_1_1412332069246_52943" dir="ltr"> <hr size="1" id="yui_3_16_0_1_1412332069246_52950"> <font id="yui_3_16_0_1_1412332069246_52942" face="Arial" size="2"> <b><span style="font-weight: bold;">From:</span></b> Kathy Dopp <kathy.dopp@gmail.com><br> <b><span style="font-weight: bold;">To:</span></b> Toby Pereira <tdp201b@yahoo.co.uk> <br><b><span style="font-weight: bold;">Cc:</span></b> Kristofer Munsterhjelm <km_elmet@t-online.de>; EM <election-methods@lists.electorama.com> <br> <b><span style="font-weight: bold;">Sent:</span></b> Friday, 3 October 2014, 21:03<br> <b><span style="font-weight: bold;">Subject:</span></b> Re: [EM] General PR question (from Andy Jennings in 2011)<br> </font> </div> <div class="y_msg_container" id="yui_3_16_0_1_1412332069246_52947"><br>Toby,<br clear="none"><br clear="none">I argue that the three possible sets of winning candidates are not<br clear="none">equally proportionately fair; and, further, my method finds the<br clear="none">closest to proportional of the three possible sets of winning<br clear="none">candidates you gave -- and here is why:<br clear="none"><br clear="none">First, to reiterate the example you gave:<br clear="none"><br clear="none">5: A1, A2, A3, A4<br clear="none">3: B1, B2, B3, B4<br clear="none">1: C1, C2, C3, C4<br clear="none"><br clear="none">Simply calculate what proportions out of the total voters, each group<br clear="none">is times the number of total seats:<br clear="none"><br clear="none">i.e. the share of the voters should, ideally, be as close as possible<br clear="none">to the share of seats they receive:<br clear="none"><br clear="none">5 voters: 2.222222222... seats<br clear="none">3 voters: 1.333333333... seats<br clear="none">1 voter: 0.44444444... seats<br clear="none"><br clear="none">First assign the obvious seats:<br clear="none"><br clear="none">5 voters -- 2 seats<br clear="none">3 voters -- 1 seat<br clear="none"><br clear="none">Now there is one seat left to assign and the following remainders of<br clear="none">seats owed proportionately to each group:<br clear="none"><br clear="none">5 voters: 0.222222222....<br clear="none">3 voters: 0.333333333...<br clear="none">1 voter: 0.444444444....<br clear="none"><br clear="none">The winner who is still owed the largest proportion of seats is the 1<br clear="none">voter. Thus, the MOST proportionately fair set of winning candidates,<br clear="none">as minimizing my sum says, is:<br clear="none"><br clear="none">5 voters: 2 seats<br clear="none">3 voters: 1 seat<br clear="none">1 voter: 1 seat<br clear="none"><br clear="none">That is the best that could be done, proportionately for 4 seats for<br clear="none">those voters.<br clear="none"><br clear="none">The numerical results of my Proportionality measurement sum for the<br clear="none">three sets of winners you suggested is:<br clear="none"><br clear="none">A1, A2, A3, B1: 0.148148<br clear="none">A1, A2, B1, B2: 0.098765<br clear="none">A1, A2, B1, C1: 0.074074<br clear="none"><br clear="none">Thus, the method I suggest identifies a set of winning candidates that<br clear="none">is arithmetically closest to fairly proportional. Of course for those<br clear="none">hypothetical voter groups any set of winners having two candidates<br clear="none">from the group with 5 voters and any one each from the groups with 3<br clear="none">voters and 1 voter will have equal proportionality measurement with<br clear="none">the third winning set you suggested above.<br clear="none"><br clear="none">I believe there is no other measure that can measure closeness to<br clear="none">proportional fairness of a winning set better for approval ballots,<br clear="none">although the measure could obviously be changed by multiplying all<br clear="none">terms by a constant since that would not change the ordering of any<br clear="none">winning set relative to another.<br clear="none"><br clear="none">Kathy<br clear="none"><br clear="none"><br></div> </div> </div> </blockquote> </div></body></html>