Least-bad ranking method

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Steve Cobb

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Feb 15, 2015, 6:36:42 AM2/15/15
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For people who insist on ordinal voting methods, should we not recommend Borda Count? 

It is clearly the best among them, based on this chart comparing common voting methods according to the criteria of simplicity, fairness (voter satisfaction efficiency), and vulnerability to strategic voting:

http://www.electology.org/#!tactical-voting/coiw

Dick Burkhart

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Feb 15, 2015, 5:55:34 PM2/15/15
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I commonly recommend the “top 3”  Borda Count:  Rank your top 3 choices only and compile by the Borda method. This both simplifies voting and minimizes tactical voting. However in some situations you may need a primary vote by proportional representation to eliminate candidates who’ve been added only for tactical purposes.

 

Dick Burkhart, Ph.D. mathematics

4802 S Othello St,  Seattle, WA  98118

206-721-5672 (home)  206-851-0027 (cell)

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Andy Jennings

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Feb 15, 2015, 6:26:54 PM2/15/15
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It's important to ask who controls the candidates.

There is much that I like about Borda, but it is not clone-proof.  (http://en.wikipedia.org/wiki/Independence_of_clones_criterion#Borda_count)  Factions have an incentive to nominate as many candidates as possible that are similar to their preferred alternative.  So I think Borda is my favorite when the set of candidates arises naturally, but am wary of it when the candidates come from some nomination process.

If I recall, Warren analyzed the rank-order methods once and concluded that Schulze beatpath method was his favorite.  It is clone-proof, so it should be fairly resistant to strategic nomination.

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Bruce Gilson

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Feb 15, 2015, 7:29:59 PM2/15/15
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On Sun, Feb 15, 2015 at 6:26 PM, Andy Jennings <abjen...@gmail.com> wrote:
It's important to ask who controls the candidates.

There is much that I like about Borda, but it is not clone-proof.  (http://en.wikipedia.org/wiki/Independence_of_clones_criterion#Borda_count)  Factions have an incentive to nominate as many candidates as possible that are similar to their preferred alternative.  So I think Borda is my favorite when the set of candidates arises naturally, but am wary of it when the candidates come from some nomination process.

If I recall, Warren analyzed the rank-order methods once and concluded that Schulze beatpath method was his favorite.  It is clone-proof, so it should be fairly resistant to strategic nomination.

​The problem I have with Schulze is that it is so complex that I think it is just about impossible to explain to the average voter (it took me a long time to understand, and I'm a lot more knowledgeable than most), so it would be hard to convince tsomeone that his vote is being fairly counted.​
 

Steve Cobb

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Feb 16, 2015, 7:45:05 AM2/16/15
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It was Warren who produced that simulation data showing that Borda is the best of the ranking methods. So we could recommend it with procedural qualifiers, e.g. limiting stacking the ballot?

Jameson Quinn

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Feb 16, 2015, 8:17:10 AM2/16/15
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I strongly disagree with recommending Borda. In my opinion, Borda is the only method which could, under reasonable strategy, give results that are worse than plurality. I think that Borda's famous defensive quote that it's a method "for honest men" about sums it up. That is, with honest voters, it's nearly as good as score; but with strategy, it could be unboundedly bad.

In particular, it's the only method where I think the "dark horse 3" (Warren's DH3) pathology would be a real threat. In my VSE simulations (my redo of Warren's BR), this was evident: several of the "top/bottom" strategic scenarios had negative VSE for Borda (that is, worse than random ballot); and Borda was worse than anything but Plurality using one-sided strategy (where the supporters of the honest winner are honest, but anyone who prefers the honest second-place is strategic). In the latter case, the VSE's in a representative voter utility model are: 

scenarioshonBallot0stratBallot1stratBallot2OssChooser3OssChooser4ProbChooser5ProbChooser6ProbChooser7ProbChooser8ProbChooser9LazyChooser10ProbChooser11DeltaChooser12
Score00.92784962440.82246588750.79718699770.68316841840.83332451780.91600404780.90082074690.87179892040.92371525530.93187296270.81830360430.82888839680.8125142468
Score10.57470019440.78304570920.75680445580.60436172890.66862935170.81486703940.82478673220.81392343010.78295718120.74107346870.77677939870.77868401250.799286268
Approval20.91337099970.7924905590.74688095980.67554513880.83362532420.91944411060.88018240490.84833804530.90864920860.92320294930.8015536560.78385368420.8006959849
Plurality30.70420324730.3819663610.30161079360.38648707790.6326111690.68237541920.48394702640.37127163610.61806185430.7061909030.37887144880.36932261610.3679137522
Irv40.86567687720.67393999150.67659196290.76271785570.81442486040.86928214640.77307448120.71051232240.80468679460.84218582710.6732094070.66785206340.6742136402
Minimax50.87537751130.40385902530.37138529540.58719706020.7892714160.84875450510.69977553280.54734044140.76141018340.83219692560.40316891640.40819386740.4050216189
Borda60.8790139305-0.8500994391-0.66004921040.5210188830.76853870080.84124095590.6326718822-0.15350718880.68602573930.8582263523-0.8529629248-0.8562396573-0.8697301604
MavBy70.87050561850.8819862340.86275427540.86769740960.8705917520.87623868430.89526952760.88128596080.87788348850.88505180090.82008084310.88087489730.8395890512

(Score1 is actually a version of approval but with a sillier "honest" cutoff.)

Imágenes integradas 2
Here's the chart version. Borda is the tiny yellow bubble under minimax.

I think that this measure is too kind to IRV, because it doesn't account for centralization issues or issues of how slippery the strategy slope is. That latter issue also might edge approval over score. But it's clear that Borda is not in the running.

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Toby Pereira

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Feb 16, 2015, 3:09:06 PM2/16/15
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I've wondered about the Borda results on Warren's Bayesian Regret charts before, given that BR has been used to promote score, but not used to promote Borda over the other ranked methods to anywhere near the same extent.

However, even if we take the graph at face value, it doesn't mean that Borda is better than Condorcet. It means it's slightly better with honest voting, and slightly better with full strategic voting. It doesn't say anything about how much voters would feel encouraged to vote strategically by each system. So it doesn't say where on each bar the most likely BR level is for Borda or Condorcet, and so doesn't give a conclusive answer as to which is better.

Jameson Quinn

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Feb 16, 2015, 3:24:33 PM2/16/15
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Note again that my graph actually shows negative VSE for Borda with full strategic voting. The reason that's different from Warren's BR numbers, I think, is that my strategy model is more sophisticated, using honest outcome as a proxy for "chance of winning" and maximally strategizing against the candidates which have high chance of winning and lower utility than the better of the two frontrunners. Warren's strategy was just, pick two arbitrary candidates, and everybody strategizes for one, against the other, and is honest about the remainder. Which, since it's only strategizing about two candidates, is immune to DH3.

2015-02-16 15:09 GMT-05:00 'Toby Pereira' via The Center for Election Science <electio...@googlegroups.com>:
I've wondered about the Borda results on Warren's Bayesian Regret charts before, given that BR has been used to promote score, but not used to promote Borda over the other ranked methods to anywhere near the same extent.

However, even if we take the graph at face value, it doesn't mean that Borda is better than Condorcet. It means it's slightly better with honest voting, and slightly better with full strategic voting. It doesn't say anything about how much voters would feel encouraged to vote strategically by each system. So it doesn't say where on each bar the most likely BR level is for Borda or Condorcet, and so doesn't give a conclusive answer as to which is better.

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Clay Shentrup

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Feb 20, 2015, 11:15:48 PM2/20/15
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On Monday, February 16, 2015 at 12:09:06 PM UTC-8, Toby Pereira wrote:
It doesn't say anything about how much voters would feel encouraged to vote strategically by each system.

That's mostly irrelevant, since virtually no one understands the mechanics of their ranked voting system. Come to San Francisco and I'll show you.

William Waugh

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Apr 4, 2015, 9:55:26 PM4/4/15
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How do "ordinal" voting methods treat the case where there is more than one candidate not ranked by a given voter?

On Sunday, February 15, 2015 at 6:36:42 AM UTC-5, Steve Cobb wrote https://groups.google.com/d/msg/electionscience/o2wrmqnsu4c/fcwHSEPAzBwJ

Dick Burkhart

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Apr 4, 2015, 11:37:39 PM4/4/15
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This depends on the voting method. For the Borda Count the most common method is the “Modified Borda Count” of Peter Emerson. If ‘n’ candidates are elected, then the top candidate gets n points, the next n-1 points, etc., whereas a full vote for m > n candidates would be (m,m-1,m-2,…,1,0,…). The undervote sums to n(n+1)/2 points whereas the full vote sums to m(m+1)/2 points, so the undervote is penalized by the factor n(n+1)/[m(m+1)]. This is a very heavy penalty if the voter selects only his or her top candidates (n = 1).

 

For a less heavy penalty, I specify a minimum and a maximum number of candidates to be ranked for a full vote. For example, if 3 is the minimum and 6 is the maximum, out of 9 candidates, then 6 rated candidates would be (6,5,4,3,2,1,0,0,0) summing to 21 points, 3 rated candidates would be (6,5,4,1,1,1,1,1,1) also summing to 21 points, 2 rated candidates would be (6,5,10/7,10/7,10/7,10/7,10/7,10/7,10/7) scaled by 2/3 so that it sums to 14 points instead of 21, and 1 rated candidate would be (6,15/8,15/8,15/8,15/8,15/8,15/8,15/8,15/8) scaled by 1/3 so that it sums to 7 points instead of 21.

 

Dick Burkhart

4802 S Othello St,  Seattle, WA  98118

206-721-5672 (home)  206-851-0027 (cell)

dick...@gmail.com

 

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William Waugh

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Apr 9, 2015, 6:30:40 PM4/9/15
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Dick Burkhart, do you echo Steve Cobb's question?  Are you interested in other responses to it besides your own?

I am trying to narrow down what constraints a proposed answer would have to meet to satisfy the questioners.  Up to now, the only questioner I know of is Steve Cobb.  I would be particularly interested to know whether Steve Cobb would consider a given system to be "ordinal" if it allows to equal-rank the unnamed candidates on a particular ballot.  And if so, how about equal-ranking some of the named candidates as well?

On Saturday, April 4, 2015 at 11:37:39 PM UTC-4, Dick Burkhart wrote:

This depends on the voting method. ... 

Dick Burkhart

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Apr 10, 2015, 1:22:41 AM4/10/15
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I permit equal rankings in my Borda type voting algorithms. I assign points (possibly fractional) equal to the average of a strong ranking of the equally ranked candidates.  Thus if a ranking of (1,2,2,2,3,4) would convert to points of (5,3,3,3,1,0) instead of (5,4,3,2,1,0), and (1,1,2,3) would become (2.5,2.5,1,0) instead of (3,2,1,0). Thus the point sum remains the same.

 

Dick Burkhart

4802 S Othello St,  Seattle, WA  98118

206-721-5672 (home)  206-851-0027 (cell)

dick...@gmail.com

 

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William Waugh

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Apr 10, 2015, 6:06:05 PM4/10/15
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If the questioners accept equal-ranking ballots as "ordinal", then here's my proposal for the "least-bad ordinal method":

Before I lay out the proposal, I'll give the principle I'm basing it on.  I suspect that if a system doesn't meet the balance constraint mentioned by Frohnmayer, it is vulnerable to vote splitting.  I don't have a proof of that, but suspect it is so.  I would be interested in any line of reasoning to make that connection firmer, if you have such to offer.

So here's my proposed system:

The voter may list any finite count of "grades" [note 1] in an order, from most preferred to least preferred, and may list within each grade any set of candidates, even an empty set.  Where the voter could list a candidate, the voter may also state "all others".  If a voter lists a candidate in more than one grade, which should be discouraged by the instructions, the tally treats the ballot as though the candidate were listed in the highest of the grades in which the voter listed that candidate.  The same resolution applies if the voter lists "all others" in more than one grade.  If the voter does not list "all others" in any grade, the tally creates for that ballot an additional grade below the least preferred grade, and considers it to contain "all others".

Then the tally converts the grades to scores.  The least preferred grade receives the lowest score, and the most preferred grade receives the highest score.  The intermediate grades are assigned scores according to a curve that bunches the grades near the top and bottom, spreads those in the middle, and is symmetric about the middle (the degree of bunching near the bottom matches that near the top).

The candidate with the highest total score wins.

[note 1] I could have said "ranks" instead of "grades".  I think "ranks" might connote strict ranking.  So I offer the term "grades", thinking it does or should connote nonstrict ranking.

Steve Cobb

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Apr 10, 2015, 6:08:10 PM4/10/15
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>I would be particularly interested to know whether Steve Cobb would consider a given 

>system to be "ordinal" if it allows to equal-rank the unnamed candidates on a particular ballot.


I’m not one of the theorists, but it seems to me that an ordinal system could allow equal-ranking. I can imagine that the question is the source of some debate. However, such a system certainly wouldn’t be cardinal.


The motivation for the original question arose from the chart summarizing Warren’s simulation, which we often refer to. The original post included a link to it. According to that chart, Borda is clearly superior to Condorcet and IRV in both fairness and simplicity, and it bears more resemblance to cardinal systems than Condorcet and IRV, so we should recommend it as the best ordinal option. Jameson questioned the simulation that produced the results.


Personally, my favorite ranking system is randomly choosing one among several ranking systems after all the ballots are cast.

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