<div dir="ltr"><div><div><div><div><div><div>I have tweaked things slightly in order to make things slightly simpler and to make sure that when Range style ballots are used the method reduces to standard Range in the one winner case.<br><br></div>The general method takes any proportional lottery based on score/range (including approval) style ballots and converts it into a PR election method.<br><br></div>Recall that a lottery method is a system of assigning probabilities to candidates. A lottery method is "proportional" if it assigns probabilities in proportion to the respective faction sizes when all faction members vote with single-minded and exclusive loyalty in favor of their favorite.<br><br></div>Let's suppose that we are in the context of an election where w>0 candidates are to be elected, and that there is at least one subset of candidates W of size w, such that when our lottery method is applied to W, all members of W are assigned positive probabilities. <br><br></div>If there is only one such subset, then that is the subset selected by our method.<br><br></div>Otherwise, for each subset W of candidates of the requisite size w, we do the following steps:<br></div><br>(1) Partition the ballots into sets S and S' that do and don't, respectively, give at least one candidate of W a positive rating.<br><br><div><div><div><div><div><div>(2) Let p be the probability that a randomly drawn ballot would be a member of the set S. Let q = 1 - p. [Note this is a change; the definitions of p and q have been switched, for esthetic reasons.]<br><br></div><div>(3) Let p1, p2, ... be the respective probabilities assigned to the members of W by our lottery method, when that method is restricted to the ballot set S.<br><br></div><div>(4) Let a1, a2, ... be the respective averages of the candidate ballot scores over the respective ballots that rate the respective candidates positively. [In case of approval, all of these averages will be ones]<br><br></div><div>Having completed these four steps for each candidate set W of size w, elect the set W that maximizes the value of the expression<br><br></div><div>min(a1*p1, a2*p2, ...)*(p),<br><br></div><div>Now you I will show you the reason for the tweaks: suppose that w=1. Then the value of p1 is 1, and the value of a1 is the average rating of W's only candidate over the ballots in the set S, The average rating of W's member over S' is zero, so the average score over the union of S and S' is the weighted average a1*p1*p + 0*q, which simplifies to a1*p1*p . Therefore in the case w=1, the standard range winner wins!<br><br></div><div>We have mentioned several of the various random ballot lotteries. More variations are possible. Ordinal ballots can be used via the "Implicit approval cutoff," or with the help of an explicit one. If the lottery is random nallot the only step needing an approval measure is step (4) above. There are other possible ways to adapt to ordinal ballots (ranked preference ballots).<br><br></div><div>There are other proportional lotteries besides the random ballot ones. One is the so called Ultimate Lottery, a restricted version of which is called the Nash Lottery. The Nash Lottery Method picks the lottery that maximizes the product of the ballot expectations based on the lottery. It turns out that the crucial factor that makes this lottery proportional is the "homogeneity" of the ballot expectations in the probabilities. So if we widen the admissible ballots to include any homogeneous functions of the probabilities (along with the natural requirement that such functions not be decreasing in any of their arguments), the we get the Ultimate Lottery Method.<br><br></div><div>Jobst Heitzig has come up with many lotteries with special attention to those with low entropy, which makes those lotteries useful in single winner elections. Why would we want to use a lottery in single winner elections? It turns out that the element of chance, when skillfully incorporated, can (more or less) remove incentives for insincere voting. The best known example of this is the "benchmark" standard random ballot lottery. In that method there is no incentive for insincere voting, but the resulting lotteries tend to be high entropy lotteries, hence not good for single winner elections.<br><br></div><div>One of Jobst's best methods has two stages. <br><br>The first stage generates a set of approval ballots from "thresh-hold" information supplied by the voters on their ballots. I won't go into the details of this stage, but the voters give tentative approvals that allow the method to automatically build approval ballots that would back-fire on defectors.<br><br></div><div>The second stage calculates the total approvals for the respective candidates and assigns each ballot B to the candidate with the greatest total approval of any candidate approved on ballot B. The lottery probabilities are proportional to the number of ballots assigned to each candidate.<br><br></div><div>For our multi-winner purposes we can replace the first stage with direct use of approval ballots or by conversion of range ballots into approval ballots via Toby's idea or by other ideas, including the "sincere approval strategy" technique, for example.<br><br></div><div>Where else can we find suitable proportional lotteries?<br><br></div><div>Andy Jennings found that he could take any sequential PR method and convert it into a proportional lottery by cloning all of the candidates as many times as needed, and allowing a huge number of winners. Then the candidate lottery probabilities are proportional to the number of clones they have in the winning circle.<br><br></div><div>In fact it turned out that Andy;s first application of that technique generated the Nash Lottery by use of sequential PAV adapted to range ballots.<br><br></div><div>More recently, Andy and I have seen that there are many ways to convert practlcally any single winner method into a sequential PR method.<br><br></div><div>Putting all of this together we have the following diagram:<br><br></div><div>single winner -> sequential PR -> Lottery -> multi-winner PR<br><br></div><div>So in a standard way we can convert any single winner method into a multi-winner PR method that retains some of the flavor of the single winner method. Why not just stop at the sequential PR stage? That's always a possibility, but sequential methods tend to generate an hierarchy among the winners in order of election. It is sometimes better philosophically to have a method that compares (many if not all) possible winning sets (including the winning sets generated by various sequential PR methods) against each other. In this regard, Jameson has just invented (and posted the EM list) a method that compares methods on the basis of sum of squared envy.<br></div><div><br></div><br></div><div>That's it for now ...<br></div><div><div><div><div><div><div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Dec 3, 2015 at 2:55 PM, Forest Simmons <span dir="ltr"><<a href="mailto:fsimmons@pcc.edu" target="_blank">fsimmons@pcc.edu</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div>Now I just have time to point out that in place of the random ballot lottery we could use any proportional lottery, such as PAV or The Ultimate Lottery.<br><br></div>More on next time ...<br></div><div><div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Dec 3, 2015 at 2:42 PM, Forest Simmons <span dir="ltr"><<a href="mailto:fsimmons@pcc.edu" target="_blank">fsimmons@pcc.edu</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div><div><div><div><div><div><div><div><div><div><div>Continuing as promised...<br><br></div>Now let's allow range ballots with greater resolution than mere approval ballots.<br><br></div>For each w-tuple of candidates we ...<br><br></div>(1) first divide the ballots into two sets, the set S of ballots that have positive support for at least one candidate of our w-tuple, and the complementary set S' of ballots that rate every member of our w-tuple at zero.<br><br></div>(2) Then we imagine an experiment of drawing a ballot B at random from the entire set of ballots. Let p0 be the probability that B is a member of S', and let p1, p2, ... pw be the respective probabilities for the choices of the respective members of our w_tuple given that B is a member of S. [Multiplication by (1-p0) would give the respective unconditioned probabilities.]<br><br></div>(3) Let a1, a2, ... be the respective candidate score averages for ballots in S.<br><br></div>(4) elect the w-tuple with the greatest value of the expression<br><br></div>min(a1*p1, a2*p2, ...) - p0.<br><br></div>There are at least three good ways of defining the details of the random ballot experiment.<br><br></div>(1) The value p3, say, is the probability that candidate 3 would be the highest rated candidate (from our w-tuple) on a randomly drawn ballot from the set S. If several candidates are tied in this respect, divide up the probability (from Ballot B) equally among them..<br><br></div>(2) The positive scores on every ballot B are normalized with respect to the members of our w-tuple. These normalized scores are averaged over S to get the respective "random ballot" probabilities.<br><br></div>(3) Use a Toby Pereira transformation to convert each range style ballot into one hundred approval ballots. Then use a random approval ballot lottery on that set. I'll explain this in my next post ...<br></div><div><div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Dec 3, 2015 at 1:44 PM, Forest Simmons <span dir="ltr"><<a href="mailto:fsimmons@pcc.edu" target="_blank">fsimmons@pcc.edu</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div><div><div><div>Continuing as promised ..<br><br></div>Suppose that there are to be w winners, and the ballots are still approval style:<br><br></div>For each w_tuple of candidates, let p1, p2, ... be the respective probabilities of selection of the respective candidate by random ballot (restricted to the w-tuple), and let p0 be the probability that a random ballot would not approve any of the candidates in the w-tuple.<br><br></div>Elect the w-tuple with the largest value of min(p1, p2, ...) - p0.<br><br></div>To Be Continued ...<div><div><br><div><div><div><div><div><div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Dec 3, 2015 at 1:30 PM, Forest Simmons <span dir="ltr"><<a href="mailto:fsimmons@pcc.edu" target="_blank">fsimmons@pcc.edu</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><br><div class="gmail_extra"><br><div class="gmail_quote">This query has lead me to some interesting ideas:<br><br></div><div class="gmail_quote">Approval Ballots. Two to elect:<br><br></div><div class="gmail_quote">For each pair of candidates {X1, X2}, let p1 and p2 be the respective probabilities that X1 or X2 would be selected by a random approval ballot drawing (restricted to our pair of candidates), and let p0 be the probability that a random ballot would approve neither X1 nor X2. Elect the pair with the greatest value of min(p1, p2) - p0.<br><br></div><div class="gmail_quote">This actually gives two methods, since there are two natural ways of selecting a candidate by random approval ballots. <br><br>The first way is to select ballots at random until the approval set for one of them has non-empty intersection with the set from which we are to select a winner. The names of the candidates are drawn randomly from a hat. The first name drawn of a candidate in the intersection set is the name of the winner.<br><br></div><div class="gmail_quote">The second way starts out as above, but once the first non-empty intersection set is determined, additional ballots are drawn as needed to narrow down the intersection to one candidate, the winner.<br><br></div><div class="gmail_quote">More later ...<br></div><div class="gmail_quote"><br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
From: Forest Simmons <<a href="mailto:fsimmons@pcc.edu" target="_blank">fsimmons@pcc.edu</a>><br>
To: EM <<a href="mailto:election-methods@lists.electorama.com" target="_blank">election-methods@lists.electorama.com</a>>, Andy Jennings<br>
<<a href="mailto:elections@jenningsstory.com" target="_blank">elections@jenningsstory.com</a>><br>
<br>
<br>
How about thie following ideas?<br>
<br>
Elect the pair that covers the most voters (i.e. that leaves the fewest<br>
voters with nobody that they approved elected). In case of ties, among<br>
tied pairs elect the one whose weaker member has the most approval.<br>
<br>
Or this variant: If no pair covers more than 70 percent of the voters,<br>
elect the pair that covers the greatest number of voters. Otherwise<br>
consider all pairs that cover at least 70 percent of the voters to be<br>
tied. Then among tied pairs, elect the one whose weaker member has the<br>
greatest approval.<br>
<br>
<br>
From: Andy Jennings <<a href="mailto:elections@jenningsstory.com" target="_blank">elections@jenningsstory.com</a>><br>
> To: Election Methods <<a href="mailto:election-methods@electorama.com" target="_blank">election-methods@electorama.com</a>><br>
> Subject: [EM] Approval ballots. Two to elect. Best method?<br>
><br>
> SPAV?<br>
> 1. Candidate with most approvals wins.<br>
> 2. That candidate's voters have their voting weight halved (or multiplied<br>
> by 1/3).<br>
> 3. Remaining candidate with most points wins.<br>
><br>
> STV-like?<br>
> 1. Choose quota Q = one-third (or one half) of voters.<br>
> 2. Candidate with most approvals wins. (T = # of approvals)<br>
> 3. That candidate's voters have their voting weight multiplied by<br>
> max(1-(Q/T), 0)<br>
> 4. Remaining candidate with most points wins.<br>
><br>
> Monroe-like?<br>
> 1. For each pair of candidates, find the voter-assignment which maximizes<br>
> the number of voters assigned to a candidate they approved, such that no<br>
> more than half the voters are assigned to one candidate.<br>
> 2. Elect the pair which satisfies the most voters.<br>
><br>
> Others? Toby, what are your favorite PR methods at the moment? Can you<br>
> give a short explanation of how Phragmen/Ebert would work with only two to<br>
> elect?<br>
><br>
><br>
><br>
> Specifically, I'm worried that in practically every approval-ballot PR<br>
> method, if there is a candidate you really like, but are sure that she can<br>
> get elected without your vote, you gain an advantage by not approving<br>
> them. Is there any method that minimizes that incentive?<br>
><br>
> ~ Andy<br>
><br></blockquote></div></div></div></blockquote></div><br></div></div></div></div></div></div></div></div></div></div>
</blockquote></div><br></div>
</div></div></blockquote></div><br></div>
</div></div></blockquote></div><br></div></div></div></div></div></div></div></div></div></div></div></div>