<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>I think that the optimum proportions using the measure I described would be A(56), B(60), C(93), so adding up to 100 it would be A(26.79), B(28.71), C(44.50). The total of the squared representations would be 102.87, which is indeed the lowest of those I've measured.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>You could probably make a case for then making these the probability weights in a
single-winner lottery method as well. But again, this is purely considering proportionality and not positive support for candidates generally. For example:</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>2 to elect, approval voting</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size:
13px; font-style: normal; background-color: transparent;"><span>10: A, B</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>10: A, C</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>This measure would find BC to be the most proportional, even though A has universal support, because AB and AC are unbalanced.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande,
Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>Toby</span></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: 13px;"> <div style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 16px;"> <div dir="ltr"> <div class="hr" 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> Juho Laatu <juho4880@yahoo.co.uk>; EM <election-methods@lists.electorama.com> <br> <b><span style="font-weight: bold;">Sent:</span></b> Friday, 14 November 2014, 0:52<br> <b><span style="font-weight: bold;">Subject:</span></b> Re: [EM] Weighted PR question<br> </font> </div> <div class="y_msg_container"><br><div id="yiv0809991634"><div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; background-color: rgb(255, 255, 255);"><span></span><div><span>Looking at this, no intuitive answer sprung out at me.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color:
transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>Leaving aside considerations of positive support for the moment, I think the best measure of actual proportionality is to minimise the sum of the squares of the individual voters' "total representation". If v voters vote for a particular candidate, then if that candidate is elected then voters who voted for
that candidate get a representation of 1/v from that candidate and everyone else gets 0. Assuming I've made no mistakes:</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>For A(70), C(20), B(10), the total would be</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif;
font-size: 13px; font-style: normal; background-color: transparent;"><span>40*(70/70+10/50)^2 + 30*(70/70+20/60)^2 + 20*(20/60)^2 + 10*(10/50+20/60)^2 = 116.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>For A(70), C(30) it is 115.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida
Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>For B(50), C(50) it is 108.3.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>For Juho's A(35), B(25), C(40) it is 103.3.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif;
font-size: 13px; font-style: normal; background-color: transparent;"><span>This
would make Juho's ratios the most proportional of those suggested. But that doesn't rule out others being even better. </span><span>However, this proportionality criterion is not monotonic and can violate Pareto in some cases, so most proportional doesn't necessarily mean best even in an election that's supposed to be proportional. That being said, I'm not aware of a monotonic proportional approval system that I find to be reasonable enough for use, and this isn't a case where I currently see an intuitively obvious answer.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color:
transparent;"><span>With a
single winner lottery, I'd probably go for Juho's answer. I am aware of a website that uses random ballot approval voting for a choice that has to be made daily. Essentially, a random ballot is picked and a random candidate approved on that ballot is picked as the winner. It seems to work well and there's no incentive to vote dishonestly. And it's equivalent to what Juho has described.</span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span><br clear="none"></span></div><div style="color: rgb(0, 0, 0); font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 13px; font-style: normal; background-color: transparent;"><span>Toby</span></div><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: 13px;"> <div style="font-family: HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, Sans-Serif; font-size: 16px;"> <div dir="ltr"> <div class="qtdSeparateBR"><br><br></div><div class="yiv0809991634yqt0554963143" id="yiv0809991634yqtfd89406"><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;"></div> <font face="Arial" size="2"> <b><span style="font-weight: bold;">From:</span></b> Juho Laatu <juho4880@yahoo.co.uk><br clear="none"> <b><span style="font-weight: bold;">To:</span></b> EM <election-methods@lists.electorama.com> <br clear="none"> <b><span style="font-weight: bold;">Sent:</span></b> Thursday, 13 November 2014, 23:39<br
clear="none"> <b><span style="font-weight: bold;">Subject:</span></b> Re: [EM] Weighted PR question<br clear="none"> </font> </div></div><div class="yiv0809991634yqt0554963143" id="yiv0809991634yqtfd72308"> <div><br clear="none"><br clear="none">On 14 Nov 2014, at 00:26, Forest Simmons <<a href="mailto:fsimmons@pcc.edu" target="_blank" rel="nofollow" shape="rect" ymailto="mailto:fsimmons@pcc.edu">fsimmons@pcc.edu</a>> wrote:<br clear="none"><br clear="none">> Consider the following approval ballot set:<br clear="none">> <br clear="none">> 40 AB<br clear="none">> 30 AC<br clear="none">> 20 C<br clear="none">> 10 BC<br clear="none">> <br clear="none">> Of the following weighted representations, which would be better?<br clear="none">> <br clear="none">> (I) A(70), C(20), B(10)<br clear="none">> <br clear="none">> (II) A(70), C(30)<br clear="none">> <br clear="none">> (III) B(50), C(50)<br
clear="none">> <br clear="none">> or some other?<br clear="none"><br clear="none">How about<br clear="none">A 35 = 40 * 1/2 + 30 * 1/2<br clear="none">B 25 = 40 * 1/2 + 10 * 1/2<br clear="none">C 40 = 30 * 1/2 + 20 + 10 * 1/2<br clear="none"><br clear="none">> <br clear="none">> How about if the weights stood for probabilities in a single winner lottery. Which would be best?<br clear="none"><br clear="none">Same 35/25/40 approach could do also here. Or maybe 100% to A (the most approved candidate). Or maybe 70/50/60 (based on the number of approvals). I guess different single winner lotteries could have different targets.<br clear="none"><br clear="none">Juho<br clear="none"><br clear="none"><br clear="none">> <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<div
id="yiv0809991634yqtfd78441"><br clear="none"><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"></div><br clear="none"><br clear="none"></div> </div></div><div class="yiv0809991634yqt0554963143" id="yiv0809991634yqtfd00815"> </div></div><div class="yiv0809991634yqt0554963143" id="yiv0809991634yqtfd34303"> </div></blockquote><div class="yiv0809991634yqt0554963143" id="yiv0809991634yqtfd29566"><div></div>
</div></div></div></div></div><br><div class="yqt0554963143" id="yqtfd79047">----<br clear="none">Election-Methods mailing list - see <a href="http://electorama.com/em" target="_blank" shape="rect">http://electorama.com/em</a>for list info<br clear="none"></div><br><br></div> </div> </div> </blockquote><div></div> </div></body></html>