<html><head></head><body><div class="ydpfc924663yahoo-style-wrap" style="font-family:Helvetica Neue, Helvetica, Arial, sans-serif;font-size:16px;"><div dir="ltr">Hi Robert,<br clear="none"><br clear="none">Le dimanche 2 juin 2019 à 02:10:13 UTC−5, robert bristow-johnson <<a shape="rect" href="mailto:rbj@audioimagination.com" rel="nofollow" target="_blank">rbj@audioimagination.com</a>> a écrit : <br clear="none">><br clear="none">>> Hi Robert,<br clear="none">>Hi,<br clear="none">>> I know that you feel it's adequate that there is a probabilistic relationship between (x-y) and (x+y). I think<br clear="none">>> the relationship to expect in reality is unclear. Consider that in real world experience major contests that<br clear="none">>> involve the strongest candidates and most of the voters are often close races, in which case (x-y) doesn't<br clear="none">>> predict (x+y) well at all.<br clear="none">><br clear="none">>it seems to me that one would expect x-y to scale with x+y. if the decisiveness (which is the name that i give <br clear="none">>the percent margin, (x-y)/(x+y) ) is about the nature of the individuals voting, not about their quantity) is <br clear="none">>unchanged with size, then x-y is expected to get bigger as x+y does. if the decisiveness is the equal between <br clear="none">>two paired runoffs in RP or Schulze, which decision is more important to satisfy? the one involving more voters <br clear="none">>weighing in or the pairing with fewer? <br clear="none"><br clear="none">And: <br clear="none"><br clear="none">> > Personally I would say that (x-y) just reveals the lowest possible value for (x+y).<br clear="none">>No, i think that, if statistics are generated from actual elections you'll find (assuming x>y) that x-y will correlate<br clear="none">>with x+y. The expectation value of x-y will increase with x+y . I think viewing it as a binary probability distribution<br clear="none">>where the probability that a random voter votes for candidate X is x/(x+y) and the probability this randomly<br clear="none">>chosen voter votes for candidate Y is y/(x+y). In this group I am leaving out the voters that preferred neither X nor Y. <br clear="none"><br clear="none">Most candidates can't actually win, in a reasonable method, so their contests shouldn't be affecting the result (so as</div><div dir="ltr">to minimize spoilers). So even if you are correct that (x-y) scales with (x+y) for most candidates, the situation where</div><div dir="ltr">we need it to work is where (x+y) is high; whether it's true with unviable candidates shouldn't matter.<br clear="none"><br clear="none">In the real world with two candidates there is an incentive to compete over the median voter, because the candidate</div><div dir="ltr">who can do that has the majority (or close). If the two candidates are nominated and campaign effectively then (x-y)</div><div dir="ltr">should be small.<br clear="none"><br clear="none"> <div><div>Possibly RCV in some form could change this dynamic, although that doesn't seem like it would be a good thing</div><div>(i.e. higher margins between the most competitive candidates).</div></div><br clear="none">> there are people disappointed with the result of any election. the idea of majority rule is to reduce<br clear="none">>the number of disappointed persons with franchise. assuming all voters have equal franchise, the net<br clear="none">>number of people being disappointed are the Winning votes minus Losing Votes. If we want to minimize that,<br clear="none">>we emphasize the results of elections with larger margins over those of smaller margins.<br clear="none"><br clear="none"> <div><div>I like the idea of measuring disappointment, but I think it has to be done per voter and based on what results the</div><div>method can actually offer. When X beats Y with some margin or vote count, we don't know much about what would</div><div>disappoint X>Y voters or what their goals are. If Y is neither a voter's favorite nor least favorite, whether they want</div><div>Y to be defeated depends on what else is on the table. If the method can only realistically elect Y or Z, then</div><div>disappointment from the resolution of X vs. Y will depend a lot on the opinions on Y vs. Z.</div><div><br></div></div></div></div><div class="ydpfab5ebcdyqt6885168168" id="ydpfab5ebcdyqtfd83001">Kevin</div><div class="ydpfab5ebcdyqt6885168168" id="ydpfab5ebcdyqtfd83001"><br></div></body></html>