Wikipedia talk:Requests for comment/Arbitration Committee Elections December 2011

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Tactical voting – who cares?
In the RfC there's a good bit of talk about this supposed "tactical voting" and how we need to take that into account. But why is this something that should matter? Non-formally, if I vote "Support" for Candidate 1 and "Neutral" for Candidate 2 what I'm saying is "I think candidate 1 is ok, and I don't care about candidate 2". If I vote "Support" for Candidate 2 and "Oppose" for Candidate 2, even though I don't care about candidate 2 - this evil tactical voting thing - then what I'm saying is "I *really* like candidate 1 and while I don't care about candidate 2 in and of itself, I like candidate 1 so much that I'm willing to throw 2 to the dogs just to get my most preferred candidate elected". What's wrong with having the second preference? At the end of a day any voting system is just a mechanism for aggregation of individual preferences and with this "tactical voting" what I'm doing is expressing not just the *ordering* of my preferences, but also their *intensity*. Since that signals more information (a=2*b, say, rather than just a>b) it would seem like a good thing, not a bad one.

And does it really have an effect? Think of a simplified set up. I take myself to be the pivotal voter. If I don't see myself as the pivotal voter then, voting doesn't solve anything, if voting accomplished anything Jimbo would've made it illegal, what better way than to enslave a Wikipedia editor than to give him a vote in the ArbCom election and call him free (bonus points for identifying that one), yadda, yadda, yadda - there's no point in voting anyway. More precisely there might be a point in voting to express one's preference but since by definition I don't see myself as actually affecting the outcome, I might as well vote my 'true' preferences and not engage in any sneaky "tactical voting".

Ok, so I'm pivotal (DJ, cue the music...) This means that we can think of it as a 2 player game between myself and a second player called "RoW" (Rest of Wikipedia). To keep things simple suppose there's only 3 candidates, 1, 2, and 3 (if you know your voting theory then you know that pretty much all voting systems can be analyzed in terms of just 3 alternatives (2 don't work)). Suppose that I would like to see candidate 1 on the ArbCom, I would like to NOT see candidate 3 on the ArbCom and 2 I just don't care about. Given how Mr.RoW votes, what are my incentives for "tactical voting"?

Well, the only choices I'm going to consider are either (S,N,O), (S,O,O) or (S,S,O) (where (S,N,O) is "Support 1, Neutral on 2, Oppose 3", etc.). Forget the last one. Option (S,N,O) is me "not voting tactically" (since I don't care about Mr.2) while option (S,O,O) is me "voting tactically" (since I'm opposing 2 even though I don't really give a shit if it makes it or not). Suppose further that there's two systems - the "low Support threshold" system (Lo-St) under which a candidate needs at least one Support vote to be elected (so a "Oppose" vote by the other person does not nullify a "Support" vote) and a "high Support threshold" system (Hi-St) where a Support vote is canceled out by a Oppose vote - basically one S and one N, or two Ss gets you elected but one S and one O means you go to the dustbin of history.

Under what plausible choice made by "Rest of Wikipedia" do I have an incentive to choose (S,O,O) over (S,N,O)? Basically none, but for sake of addressing possible objections to this simplified analysis we can think of a situation where it's possible that all 3 candidates can get elected and a system where at most 2 can get elected. If it's possible for all 3 to get elected then no matter how rest of Wikipedia votes, there's no incentive for me to "vote tactically" - choose (S,O,O) over (S,N,O). The worst thing that can happen if I fail to vote tactically is that Mr.2 will get elected and since by definition I don't care about it, it's no skin of my back.

Same thing is actually true under a situation where at most two can get elected. The only wrinkle is that if I also consider voting (S,S,O) and rest of Wikipedia votes (O,O,S) (completely opposite to me) then, I would be better off voting *either* (S,N,O) or (S,O,O) - but I'm still indifferent between the two (I'm assuming that in this case if all three get elected then the two who actually make it on to the ArbCom are chosen at random).

You can play around with this some more and, for example, assign some precise numerical values to possible outcomes. If, however, the value I place on 2 getting elected is 0 (and + for 1, - for 3) then everything I just said is the same. Alternatively you could assign a really high positive value to 1 getting elected but a small, either + (very slightly like'em) or - (very slightly dislike'em) value to 2 and whatever - value you want to 3 and then you get that you should vote (S,O,O) over (S,N,O) only if ... you very slightly dislike Mr.2. Which is what you'd do anyway, tactical voting or not.

Basically "tactical voting" is nothing but an expression of slight opposition to a candidate (the one you might consider voting "Neutral" but end up voting "Oppose"). Now that opposition of course may be less than a heartfelt robust "No! Basta!" but it is still opposition. And so should be respected.

So don't worry about "tactical voting" and just choose a "Support" threshold which makes sense to you for other reasons - in this, I support Sandy Georgia's proposal, and same for the number of members of ArbCom.

(Calculations, game theoretic matrices and proofs of above theorems available upon request (not really)))

 Volunteer Marek  22:22, 27 September 2011 (UTC)
 * But which candidate "sings the right songs"? That's what I want to know.  And which one is "a blue pill for the editors that have lost their interest"?  Neutron (talk) 23:10, 27 September 2011 (UTC)
 * I should probably start reminding people now that the (almost) same joke told twice is not as funny second time around. This year I might just let everyone know what I *really* think!  Volunteer Marek   23:18, 27 September 2011 (UTC)
 * Tactical voting changes everything—in particular, it stops people reliably assuming that all oppose votes are active opposes ("I think you'd be a bad arb"). Some opposes may still be that, but many are tactical opposes ("I wouldn't mind you as an arb, or I just don't know; but I want the candidates I support to have the best chance"). You can't know the proportions of A and B, and nor can you know whether you're a "pivotal" voter, which is healthy. Sorry about the colours, but here are the three pie graphs, published last December. 2008 was the last manual election; 2009 and 2010 no longer required voters to go through an elaborate click-and-scroll process to register an oppose, so it's no surprise that SecurePoll brought with it a huge increase in voting intensity (i.e., a drop in no-shows on candidates' pages in 2008, and in "no-click leave it at default neutral" in 2009 and 2010). And apart from this, the graphs show how arbitrary this fixation on a 50% S/(S+O) threshold is: in terms of support from all voters, only five of the 71 candidates represented in these three graphs ever got the tick from half or more of the voters. That is why the threshold should be binned, now, and never revisited. Tony   (talk)  01:46, 28 September 2011 (UTC)

Useful reference. Count Iblis (talk) 16:15, 28 September 2011 (UTC)

You are looking at the wrong possible moves for the RoW player, which means you're not actually analysing the game that voters are playing! :D These elections are a mixture of a race between candidates and an absolute ranking. You're correct that the absolute threshold element limit your possible moves to SSO, SNO and SOO. The moves the RestOfWikipedia can make determine the order of the candidates for the ranking phase, and it is here that there is a very small advantage in tactical voting. RoW's moves are actually E, X+, X- and U for each candidate, where E is that the candidate is given votes such that they are certainly elected whatever you do (they are high enough in the ranking that you cannot cause them to fall into the N+1th place), and U is equivalently that they are certainly not elected. X is borderline, meaning that the candidate is in the Nth (X+) or N+1th (X-) place, and is less than one Support or Oppose vote from swapping places with the other borderline candidate. That is to say, where there are some combinations of votes that you as pivotal voter can cast which alter the order of the bottom candidates and cause one of them to be elected and not the other; it is actually irrelevant how many candidates there are in total, and how many there are above or below the borderline (and also where the border between Nth and N+1th lies in relation to the absolute threshold, if there is one).

You're also right that the election can be considered in terms of only three candidates; if the voter does not have any candidates about which he feels ambivalent then tactical voting is clearly a non sequitur. Obviously only moves by RoW which involve borderline candidates are of any interest, which means there can be at most one such point of interest in any possible ranking of candidates, since there can be at most one X+/X- pair on the borderline (although the majority of possible rankings do not have any such pair at all). So in terms of who out of your three candidates A/B/C are on each 'side' of the border, we have nine possible options: AA, AB, AC, BA, BB, BC, CA, CB, CC; although none of the options with similar candidates are of interest since we do not care which of the 'identical' candidates wins! Equally we do not care about AC or CA since we have already expressed the maximum possible difference between those candidates by supporting A and opposing C. So the only interesting cases are AB, BA, BC and CB.

Note that voting identically for both candidates will certainly not change their order. In the case AB, neither of our available options (neutral or oppose for the X- 'B' candidate) alter the ranking; the same is true for BC where our possible options are only support or neutral. For BA we have the possibility of changing the order, and also for CB.

Now here is an interesting thought. The CB scenario (a candidate you oppose is currently Nth, and a candidate about which you are ambivalent is N+1th, and you have the ability to alter the order) is precisely the mirror image of the BA case. In the CB scenario, your cause is furthered if you support your ambivalent candidate, as by doing so you prevent a candidate you oppose from being elected. In the BA scenario, your cause is furthered if you oppose them, as you prevent them from taking a place from your preferred candidate. There is certainly no benefit to voting neutral. If you are voting tactically, how you rank your ambivalent candidates should depend just as much on how strongly you want people not to be elected: if you are one of the people who are picking N candidates to support but actually are reasonably positive about the middle ground, but vehemently oppose several candidates, you probably receive a better payoff by supporting your ambivalent candidates. This is an element of the game theory that I think a lot of voters don't really work through.

Although it doesn't affect whether you should adopt a particular strategy, it's worth thinking also about how vanishingly small the effect the tactical voting has. Your tactical vote can make a difference only if the results of the 1000-odd other voters produces a borderline pairing where one vote separates the outcome, and that pairing is on the N/N+1 boundary, and the candidate on the appropriate side of the boundary is one of your ambivalent candidates, and the candidate on the other side of the boundary is not an ambivalent candidate. For people who feel tactical voting is an important procedure it is a valuable insight: the odds that your tactical vote will have an effect are utterly miniscule. In game theory terms, since in all the other scenarios your tactical vote has absolutely no effect the fact that the tactical scenario is such a tiny edgecase is irrelevant.

TLDR: there are scenarios where voting support or oppose rather than neutral for candidates in which you are ambivalent can affect the outcome; indeed there are no scenarios in which it is best to vote neutral for them. However, there are equally valid arguments for supporting ambivalent candidates as there are for opposing them, and which you should do depends on which give you a greater satisfaction: seeing a candidate you support elected instead of a candidate about which you are ambivalent, or seeing a candidate about which you are ambivalent elected instead of a candidate which you definitely oppose. Happy‑melon 23:43, 1 October 2011 (UTC)


 * Damn you! Of course my real tactic the whole time was to convince *everyone else* that tactical voting didn't matter, so that I could get all the full benefits of tactically voting myself!
 * Slightly more seriously, you're right - when I wrote the above I did it just on a basis of a simple example with payoff matrices where I signed +x to having favored candidate (A) being elected, and a -x to a disfavored candidate (C) being elected, so the fact that the case for tactical voting is symmetric (If you are voting tactically, how you rank your ambivalent candidates should depend just as much on how strongly you want people not to be elected matters as much as wanting some people to be elected) washed out (I'm pretty sure this is correct then). Am I right in saying that if you value "my most hated candidate not getting elected" more than "my most favored candidate getting elected" I should "vote tactically" by supporting my "ambivalents", and if I value "my most favored candidate getting elected" more than  "my most hated candidate not getting elected" then I should "vote tactically" by opposing all my "ambivalents", and if I value these two equally, it doesn't matter (so it's not "best" to vote neutral, it just doesn't matter)?
 * Ok now, harder. Suppose there are N-1 other voters. What % of the possible combination of RoW outcomes is made up of cases where tactical voting can make a difference (given asymmetric valuations per above). How small is that % compared to the % of these possibilities in which just voting by itself (voting support for a candidate you support, say) can make a difference? I mean, voting sort of doesn't matter anyway, but tactical voting matters even less. So, how much less?  Volunteer Marek   03:49, 2 October 2011 (UTC)


 * Hehe.
 * In general, yes; although it should also be a weighting over all the candidates, not just your most- and least-favourite. It should also be affected by the probability of each candidate appearing in a borderline situation.  In 2010, for example, Newyorkbrad was a much more certain candidate than Giano, so if you supported Brad but opposed Giano (and were ambivalent about everyone else, to keep things simple), you should have supported your ambivalent candidates as the probability of Giano appearing at the top of a borderline was greater than the probability of Brad appearing at the bottom of one.
 * Calculating that percentage is hard. As I said before, it's made up of four components.  The last two are fairly straightforward: If from a field of M candidates you support A of them, oppose C of them, and are ambivalent about $$B=M-A-C$$ of them, then the last two terms are
 * $$\frac{A}{M}\frac{B}{M-1}+\frac{C}{M}\frac{B}{M-1}=\frac{B(A+C)}{M(M-1)}$$.
 * The second element (probability that, given that a borderline exists, it is on the boundary between elected and unelected) is also fairly straightforward, although a bit obscure. The outcome of the election must be to produce a ranking: of the M candidates, at least one of them has to come first, at least one of them must come after that, etc.  If we take the probability that more than N candidates rank equally to be negligible, then in the scenario where 2 of the M candidates are participating in a borderline, then the M-2 remaining candidates can be allocated into two bins E and U, above and below the border (equivalently certainly-elected or certainly-not-elected); the second criterion is fulfilled if there are N-1 candidates in the upper bin.  This is a simple binomial distribution with probability
 * $$\tbinom {M-2} {N-1} Q^{N-1} (1-Q)^{M-N-1}$$
 * Where we have had to introduce Q as the absolute probability of the candidate being elected (more precisely, the probability of a random other candidate being ranked above the borderline candidate we are considering).
 * The first term (probability that a borderline scenario exists) is seriously hard to analyse. The condition on the number of support and oppose votes for the upper (unprimed) and lower (primed) candidate can be easily expressed as:
 * $$\frac{S}{S+O}>\frac{S'}{S'+O'};\frac{S}{S+O+1}>\frac{S'+1}{S'+O'+1}; S+O < K; S'+O'\frac{S'}{K-N'};\frac{S}{(K+1)-N}>\frac{S'+1}{(K+1)-N'}$$
 * Now here is another interesting result. We have already seen that there is no scenario where voting neutral is the best option.  So if all voters were to vote tactically, no one would ever vote neutral!  But if we set $$N=N'=0$$ in that equation we get (requiring K nonzero):
 * $$\frac{S}{K}>\frac{S'}{K};\frac{S}{K+1}>\frac{S'+1}{K+1}\equiv S'<S<S'+1$$
 * Which has no solutions where S and S' are both integers: if no one votes neutral, there are no possible scenarios in which tactical voting has any impact! If everyone votes tactically, then everyone's tactical votes have no effect.  You being able to vote tactically depends on there being enough 'dumb' voters who do not vote tactically to allow your tactics to have an impact.
 * I could not find an analytical approach to determining the region in which the conditions are satisfied for N not equal to zero. However, since all the values are constrained to be integers, the problem is within reach of an exhaustive search by computer.  Such a program is very simple to write:

 
 * Testing for K=1000, this program took about 25 minutes to run on my home computer. The conclusion: the fraction of all possible scenarios where a borderline position exists is a shade under 0.2% (493,756,881 out of 250,500,250,000 searches).
 * The grand conclusion, then: $$P=\sum_i P_b \times \tbinom {M-2} {N-1} Q_i^{N-1} (1-Q_i)^{M-N-1} \times \frac{A+C}{M(M-1)}$$, where $$P_b$$ is approximately 0.002, $$Q_i$$ is the probability Q for the ambivalent candidate i, the sum is over all the candidates about which you are ambivalent, and the other symbols have the meanings previously assigned; $$P$$ is the probability of at least one of your tactical votes having an impact, remembering that since there can be no more than one such situation this is also the probability in aggregate.
 * So let's try and put this all in perspective with some reasonable figures. Take all the $$Q_i$$ to be 0.5 (this roughly maximises the binomial part of the probability); take $$N=12$$ and $$M=21$$, as at the ACE2010 election; assume you are adopting the common tactic of choosing as many prefered candidates as seats ($$A=12$$); you fiercely oppose maybe five candidates ($$C=5$$) and are ambivalent about the remaining four ($$B=4$$).  The overall probability then becomes:
 * $$P=\sum_4 0.002 \times \tbinom {19} {11} 0.5^{11} 0.5^{8} \times \frac{17}{21 \times 20} = 0.0047\%$$
 * That is, in a real situation comparable to a typical ACE10 vote, the probability that your tactical vote has the possibility to cause an improvement in the outcome of the vote is less than 1 in 20,000. In this scenario you will want to use your tactical vote to help your prefered candidates rather than hinder your disliked candidates, so you will choose to oppose your ambivalent candidates, meaning you can only make an impact on 12/17 of those scenarios ($$P=0.0033\% \equiv 1/30000$$).  Then you need to weight it by the actual benefit you gain from having your A candidate instead of the ambivalent candidate.  Say your actual percentage preference for your A candidates average 80% and for your ambivalent candidates average 40%; the actual payoff from all these tactics is that difference, normalised by your hapiness of the whole committee; or $$(80-40)/(80\times 12) = 0.042$$.  Throw that into the mix and bake for three hours on a medium heat: in a typical example, tactical voting can be expected to increase your happiness with the final outcome of the elections by $$0.0033\% \times 0.04 = 0.00013\%$$, or one part in 725,000.
 * My conclusion? Tactical voting - who cares?  :D Happy‑melon</b> 14:36, 3 October 2011 (UTC)
 * I just reran the exhaustive-search with $$K=850$$, as in the ACE10 election. Since the search is roughly $$O(K^4)$$ it was a fair bit faster than the $$K=1000$$ case.  Since the distribution of borderline scenarios is concentrated at low values of S, O, etc, I expected the $$P_b$$ probability to be higher, and indeed it was: 0.23% rather than 0.197%.  An 'improvement', but hardly one worth writing home about. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 15:01, 3 October 2011 (UTC)
 * It's gonna take me some time to work through this but looks right. This 1 in 725,000 probability is calculated for .012% of voters voting neutral for the borderline candidates (i.e. "dumb" voters who don't vote tactically), right? How does this change with N?
 * I'm also wondering a bit if this wouldn't be easier, or at least interesting, in some way to analyze by making it continuous, and finding the candidates right at the boundary. So you'd have a measure M of candidates, which you order from least favored to most favored, and your vote is a piecewise function (-1,0,1) on that support. Then also define another function (this one could be continuous) which summarizes your preferences over the candidates (which could be normalized somehow since at the end of the day it's the relative ranking that matters, which here would be just the slope of your preference function/distribution - though some kind of piecewise function with values (a,b,c) might be easier to work with). You then find the two "indifferent" candidates - the guy that you get the same payoff for voting "Oppose" as voting "Neutral" and the guy that you get the same payoff for voting "Neutral" as voting "Support". This would remove the restriction that S and S' have to be integers (since you can vote "Support" for say 3.789 candidates. I'm guessing that would turn that set of inequalities S'<S<S'+1 into an equality S=S' (though looking quickly at it it looks like those inequalities cannot be satisfied strictly for any S and S', not just integers). You'd then have to somehow add up all such preference and voting functions. Sorry, just thinking out loud.
 * Anyway, this may be worth writing up in a more serious manner.  Volunteer Marek   15:55, 3 October 2011 (UTC)
 * Partial answer my own question about change with N: The binomial coefficient part of the probability is maximized at N=M/2, so assume that M is even (so we don't have to worry about the integer restriction). With Q=.5, ln((1-Q)/Q)=0 so maximizing P is same as maximizing the binomial coefficient --> N*=M/2, which is close, but not exactly the same, as the example you picked with N=12, M=21. Changing N to 11 (both 10 and 11 are max) would increase the binomial coefficient by *1.22 or 22%. Raising N to 13 or lowering it to 8 (9 would give same answer as 12) would lower the binomial coefficient by *.66.  Volunteer Marek   16:30, 3 October 2011 (UTC) (Reasoning's slightly wrong since it's log of a sum not sum of logs, but with Q=.5 the "Q part" is independent of N so the solution's still true. With Q ne .5 it's bit of a pain).   Volunteer Marek   16:44, 3 October 2011 (UTC)
 * Q is also a particularly difficult parameter to assign a value to; its value depends on the overall outcome and is probably nonlinear in S and/or O. It's probably most informative simply to try and maximise that part of the expression, as you've done, and thereby get an upper limit on the strength of the tactical vote; which is pretty much what we're interested in anyway. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 16:57, 3 October 2011 (UTC)


 * With Q(i)=Q for all i (same for all) I get that the N which maximizes the P is given by
 * $$N^*=\left[ \frac {M} {2-ln \frac {Q} {1-Q} }\right]$$
 * which is slightly weird because for high enough Q this would become negative. So for it to make sense there has to be some upper bound on Q or something (or again, N*=0). Is Q also a function of N? The more general case is that N* solves
 * $$N^*=M \frac {\sum Q_i^{N^*-1}(1-Q_i)^{M-N^*-1}} {\sum Q_i^{N^*-1}(1-Q_i)^{M-N^*-1} \left[2-ln \frac {Q_i} {1-Q_i} \right]}$$
 * This is the part which makes me want to make this continuous and turn the those sums into integrals.  Volunteer Marek   17:27, 3 October 2011 (UTC)


 * Ok, playing around some more, half calculations and half just simulating the parts I don't feel like working through, for symmetric Q's I get that:
 * $$\lim_{Q \to 1} P=\lim_{Q \to 0} P=P_b \frac {(A+C)(M-A-C)} {M(M-1)}$$
 * or in other words the binomial part just goes to 1. This also maximizes P. So for M=21, A=12, C=5, the upper bound (in fact a supremum, not a maximum) on the probability that at least one of your tactical votes will have an effect is .000324, or 1 in 3088. This is quite a bit better (though honestly still really small) than the 1 in 20000 above. Of course here I'm still not exactly sure on the Pb part so I'm keeping it constant - likewise I'm treating Q as a constant but if it is a function of other stuff then that will imply bounds on Q which will again decrease the probability of tactical vote having an effect.  Volunteer Marek   18:41, 3 October 2011 (UTC)
 * That can't be right, or certainly can't be useful. The limits of Q correspond to the ambivalent candidate certainly ending up either at the bottom or top of the ranking (as all candidates become certain to place on the same side of them).  $$\lim_{Q \to 0}P$$ must go to zero for $$N\neq 0$$ and $$N\neq M$$ for the other limit; both of which are academic. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 20:03, 3 October 2011 (UTC)
 * Hmm, I don't think I made any mistake here though it could be the integer restrictions or something. And it's a double limit (in N and Q) so that gets funky (I'd have to dig out my Rudin to see when you can reverse the order). Finally, like I said, I'm treating N and Q as parameters here (like you did in your simulation) but they could be related to each other. But given all that, the problem is "pick N and Q to max P". Take N=1. Then the binomial coefficient ( (M-2) (N-1) ) is 1. The other part is then (Q^(N-1))*((1-Q)^(M-N-1))=(1-Q)^M. Obviously Q=0 then means that that stuff equals 1. So the whole binomial function part is 1. The binomial function is a probability so it can't be more than one so N=1, Q=0 is a max. Likewise for N=M-1 and Q=1. If you restrict N>1 and N<M-1 then the max in Q occurs in interior and the binomial won't quite end up equaling 1 but since we're looking to make that P as big as possible it doesn't do much harm to ignore that restriction. I'm not sure what the intuition behind this result is.  Volunteer Marek   23:22, 3 October 2011 (UTC) A quick stab at a guess of what's going on - I'm thinking the Pb is also a function of Q. So as Q goes to the boundary one of the four probabilities you mention goes to 1, but another, Pb goes to zero, so P ends up as zero. And then we're both right.   Volunteer Marek   00:36, 4 October 2011 (UTC)


 * The maths is excellent, as far as I can tell; but the psychology is wrong. People don't vote assuming they'll cast the deciding vote—it's a socially constructed motivation. Tony   (talk)  15:10, 3 October 2011 (UTC)
 * Well, if people don't vote assuming they'll cast the deciding vote ... then presumably they're also assuming that their tactical votes aren't going to be deciding either (otherwise you're saying that they're assuming that ONLY their tactical "Opposes" or "Supports" will make a difference, but their non-tactical "Opposes" and "Supports" will not make a difference). So why would anyone tactically vote then?  Volunteer Marek   15:55, 3 October 2011 (UTC)
 * Indeed. "Assuming they'll cast the deciding vote" is merely a mathematical construct: the election result is the aggregate of a thousand votes, the maths applies equally to any one of them.  The deciding vote scenario comes only from assuming that the vote we are analysing is your vote.  There is no reason we couldn't analyse one of the very many possible scenarios where 'your' vote does not alter the outcome; those scenarios are not wrong, they are just uninteresting. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 16:53, 3 October 2011 (UTC)
 * Why bother voting at all, then? Tony   (talk)  04:02, 4 October 2011 (UTC)


 * Well, that's a good questions. Economists, political scientists, sociologists, psychologists, mathematicians, anthropologists, and theologians have tried to answer it and there's a large literature in each of these areas addressing this quandary. I've read all of that, so let me ellulaminate you. Here's the break down:
 * Economists: the chance that a particular vote will decide an outcome of a particular election is minuscule. Math tells us so. Hence, people vote because they're irrational (yet, economists continue to assume that people are rational in all their other models)
 * Political Scientists: even though a single vote has in all probability no effect on the outcome of an election, people vote because they realize that what matters it is the size of the *mandate* a candidate gets, as expressed by the visibility of their "base", and not just whether they win or loose (yet, if a single vote is unlikely to have an effect on an outcome of an election, it will have a likewise insignificant effect on the size of a "mandate", whatever the fuck that is)
 * Sociologists: people vote because societies develop social norms which say that it is your social obligation to vote (which is true, but also people foondangle dangle diggle diggle because societies develop social norms which say that it is your social obligation to foondangle dangle diggle diggle. Also, people don't foodle moodle bricky brack brack because societies develop social norms which say that you shouldn't foodle moodle bricky brack brack. And if you're wondering why some people goop da da woop while others moopy moopy mope mope, it's because societies develop social norms which say that some people will... you get the point)
 * Psychologists: These guys fall into two camps. The first camp says you vote cuz you want to kill your father and have sex with your mother. This explains the viability of the Bachmann candidacy. Also, Coren (though not sure which one of the two that is). The other camp says it's because you get a "psychic thrill" from voting. These guys are like the economists and the sociologists mixed up in a blender.
 * 'Mathematicats: The reason people vote is because for every $$\delta$$ there exists an $$\epsilon$$ such that $$P=\sum_i P_b \times \tbinom {M-2} {N-1} Q_i^{N-1} (1-Q_i)^{M-N-1} \times \frac{A+C}{M(M-1)}-\delta>\epsilon$$. You have obviously not been following the discussion above. Learn to keep up please.
 * Anthropologists - Some societies vote and they do it by simple majority rule. Some societies vote and they do it by ranked voting. Some societies vote but they have liberum veto. Some societies vote but they give Big Chief veto power. Some societies vote with First-past-the-post voting (Such systems are used by democratic countries). Some societies vote with A ranked system (Ranked systems captures preference. Such systems are used by democratic countries. ) Some societies vote with The "oppose voting" system (Unique to Wikipedia. It aims to capture preference although it isn't used by any country and has no references). Some societies vote with Approval voting (see HJ Mitchell's proposal below). Wait... what was the question? (Actually I can sort of respect this kind of answer).
 * Theologians - people vote because... because it's your fucking DUTY to do so. Yet they don't consider that... wait. They're actually right. If you don't vote you just go to hell (or in Kantian terms, you at least deserve to). That's all there's to it. Don't vote = eternal suffering. If you don't vote in the ArbCom election you will at some point (it might be today, it might be tomorrow, you cannot know either the time or the hour ... wait, that's the same thing ... the hour or the, um, umm, particular edit war you're in at that moment - anyway, you just don't know) ... you will at some point be BANNED! And then you will appeal. And your appeal shall be turned down. And then someone shall come along and they shall speaketh: if you agree to vote in the next ArbCom election, oh Israel, we will unban you, and we will unban your children and your children's children and your sockpuppets as well. And you will say "yeah, ok". And then they will say "after further consultation we do not think that is enough at this time, haha, tough luck sucker, for your God is a vengeful God". And then you will start an account at Wikipedia Review and it will be ... no good. Ye shall start posting there about how unfairly you were treated, and the children of Edom will mock thee, and the sheep will poop on your sandals. And then people there will say "we don't care", and thou will thirst of water and be hungry of just being able to make some minor gnomish edits, that's all ye shall ask for. And then Ottava Rima will start an argument with you there, in the land of Mordor, out there in the wilderness of Joseph. And you will be drawn into this argument. And it will be pointless. But thou shall continue in this argument unable to stop, like a wage laborer who grew a vine around his porpoises but was not prudent to prune it before it became putrescent, and covered the pillars of his place. And then they will have the "Dick of the Year Award Poll". And thou shall remember your experience on Wikipedia and promise to thyself that you will vote for (Coren) in the Dick of the Year Poll on Wikipedia Review. Yet thou will have something better to do and constantly put it off. And thus ye will neglect your DUTY to vote one more fucking time, on Wikipedia Review, and ye shall enter the next circle of Dante's Inferno 2.0 (I don't know where it goes from there, never had the guts to go that deep myself, but I'm pretty sure it goes on and on and on... it's like Borges' labyrinth but stupid). Ay, thou shall suffer.
 * Volunteer Marek: Me? I vote for the simple reason that the way that Wikipedia ArbCom voting is structured you always get to see who voted already. So if I vote, I'm pretty sure all the people on Wikipedia who hate me see that I've voted and then they all go scrambling to vote too, just to neutralize my vote. Given the number of my enemies around these parts, I figure that if I waste ... 10 seconds of my time voting (many many more seconds writing pointless stuff *about* voting here - but that's sort of fun) then all the peeps who hate me will waste at least 10 seconds of their time voting too, at least, each. By a modest estimate I think there's probably a few dozen freaks like that. Say 60. So for my ten seconds of wasted time, I get to waste ten hours or so of their time. In a way I see it as my DUTY to vote, though not really in a Kantian sense (in some weird twisted Shopenhaurian sense or something) - by voting I save the encyclopedia about ten hours worth of useless edits these people would have collectively made otherwise. Ya'll are welcome.  Volunteer Marek   05:20, 4 October 2011 (UTC)
 * You know, a one-sentence statement of Eliezer Yudkowski's Timeless Decision Theory (TDT) sounds weirdly like a generalization of the Kantian categorical imperative. (I am perusing the whole thing right now.) You know, playing the Prisoner's Dilemma under the assumption that the other guy will do the same thing as you results in a superior outcome than the Nash equilibrium. I wonder how that generalizes to the case of elections, but I guess the answer won't be “Stay home unless there's a non-negligible chance of your favourite party to win by one vote.” (As for me, the reason I vote – when I do vote – is that I freakin' want to, which I guess means the psychologists of the second camp are kind-of right. Of course, that's not much more than a tautology unless you have a model of why I get a psychic thrill from some things but not from others, but still...) <span style="background:#00ae00;white-space:nowrap;padding:3px;color:black;font:600 1em 'Gentium Book Basic', serif">― A. di M.​  14:22, 23 October 2011 (UTC)

I've just come across this section, a few days belatedly. It is amazing. I move that we cancel the election and appoint the people who contributed to this thread as the new arbitrators. If nothing else, they've shown they have the appropriate interest level and attention to detail. Newyorkbrad (talk) 03:33, 11 October 2011 (UTC)
 * I'd certainly much rather have as arbitrators people who with the analytical skills to go into a matter in detail, rather than those who just dismiss anything a bit hard or complicated with a sarcastic comment or a group insult ("you're all being very silly"). Unfortunately, as in real-life politics, the electorate tends to take a different view.--Kotniski (talk) 06:18, 11 October 2011 (UTC)
 * Well, I stand by my words. It is silly to expect human beings to act within a mathematical formula, although the mathematics above is a fascinating intellectual exercise. I hold that it's probably more valuable to be able to parse huge volumes of written text (the average arbcom case runs about 200 pages if printed out, and that doesn't count following all the links and actually reading the entire discussions they lead to, which can easily increase the reading factor by 5). Mathematical analysis is rarely called upon in determining behavioural patterns.  Risker (talk) 06:29, 11 October 2011 (UTC)

This sections is, indeed, awesome. (And I'm not saying that sarcastically, I really think it's awesome.) I noticed a dearth of discussion on rational ignorance's effect on tactical voting, though. Okay, that last part was a bit facetious. But just a bit. --Philosopher Let us reason together. 06:50, 20 October 2011 (UTC)

Ranked voting
As just added to the RfC, this seems fairly close to the Schulze method proposal from last year, which was decidedly shot down. I would figure that any similar method of "ranked voting" would be the same, though it's possible that other systems could have a smaller "black box" than the Schulze method. –MuZemike 16:06, 30 September 2011 (UTC)


 * A motivation for 'oppose voting' instead of 'first past the post' is that people want to capture preference. Unfortunately, bolting a negative vote option onto a positive vote system (‘first past the post’) destroys it and creates a bizarre hybrid. Furthermore, 'oppose voting' devalues positive votes that aren't supported by negative votes for for non-supported candidates.


 * One of the sticking points for people is that there isn’t currently any clarity on the detail of the ranked voting system. Whatever we do, we need to look seriously at the disadvantages 'oppose voting' compared to the other options. There are better systems that are discussed/used in the outside world and you've just mentioned one. Ironically, the ‘oppose voting' system isn't clear to voters, *and* it keeps changing. I was astonished to find that my good faith positive votes haven't been given the highest value.
 * Lightmouse (talk) 17:33, 30 September 2011 (UTC)
 * Response to MuZemike: "Ranked voting" is a generic term and is used in methods that range from the moderately simple (single transferable vote) to the very complex, totally non-transparent system that you mentioned, which so far I have been treating as the Voting-Method-that-must-not-be named. And these methods do not always accomplish the same thing.  According to the people who supported it last year, the Schulze Method is a "majoritarian method," although quite frankly, after reading a couple of articles about it (both on WP and elsewhere) I could not understand it well enough to know whether that is correct or not.  If I correctly understand what "black box" means in this context, then it is the ultimate black box of voting systems.  Or as I would put it, after ranking the candidate and pushing the "vote" button, the voter really has no idea how his or her vote affected the outcome, if at all.  This is in stark contrast to any of the other methods that have been proposed (including Support/Oppose, which seems as clear to me as first-past-the-post, although Lightmouse disagrees.)  Another ranked system, STV, is not majoritarian but rather is a proportional representation system.  Presumably, therefore, Schulze and STV would have different results, perhaps very different results.  Additionally, STV, being a PR system, would fly in the face of those editors (a very large majority so far) who have supported requiring at least 50 percent support for a candidate to be elected.  STV and other PR systems assume that some candidates will be elected without majority support; in fact, that is their very purpose.  (The "quota" and "vote transfer" aspects of STV do get a little black-box-like, but not nearly as much as the Schulze Method.)  Overall, I don't think anyone has made a good case that any of these methods are better (in a WP context) than what we have already.  Neutron (talk) 19:01, 30 September 2011 (UTC)
 * Let me clarify as far as the context is concerned: many people opposed the Schulze method last year because the complexity of the system creates a sort of "black box" in which people cannot readily follow the results. While it's plausible to implement a "ranked voting" system that is simple to use and follow, I don't think that the Schulze method would be preferred because of what I had explained above. –MuZemike 20:00, 30 September 2011 (UTC)
 * I agree. I just think it also needs to be recognized that some of the "ranked voting" systems that are much simpler and more transparent than the Schulze Method have a different drawback:  They do not require that a candidate have even 50 percent support or the support of any particular number of editors in order to be elected.  I personally would give up the 50 percent requirement in order to give STV (or perhaps cumulative voting, which Volunteer Marek has proposed at the RfC) a try for a year, but I think such a thing is so unlikely to win a consensus that it isn't even worth talking about.  As for the Schulze Method, unlike all the other methods being discussed, all I see are disadvantages, without a single reason why it would be better than the current system.  Neutron (talk) 20:53, 30 September 2011 (UTC)
 * From a more purely mathematical/sociological standpoint, the Schulze method is good – however, the problem is its inherent complexity. –MuZemike 22:09, 30 September 2011 (UTC)
 * One question each to the previous two posters: Neutron, could you spell out what you see as the disadvantages of Schultze? MuZemike, is there not already an automated voting system for Schultze? I seem to remember that it's been used for at least one Foundation election. Tony   (talk)  02:22, 1 October 2011 (UTC)
 * Certainly there is a Schulze tallier built into SecurePoll; indeed that's the system used for all the Foundation Board elections which is what SecurePoll was written for. Neutron and MuZemike are, quite rightly, pointing out that whether the software can be programmed to follow the tally rules, is a quite separate question to whether voters can build an acceptable level of conceptual understanding of those rules in their own minds. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 17:29, 1 October 2011 (UTC)

Here is my opinion, as requested by Tony, on the Schulze Method. First, Happy-melon is correct, my opinion is not based on the feasibility of employing that method this year. For purposes of this RfC I have been assuming that the Wikimedia staff can throw together whatever we ask them to in whatever time we give them, though I obviously realize that this might not actually be the case, and if it's not, we'll deal with it then. Tony asks what I see as the "disadvantages" of this method. I will answer that in a moment, but I would first point out that nobody, either last year or this year, has stated what the advantages are. If we are going to have a serious discussion about it, I would like to see that first. Can someone explain why that system is better than what we have now? Or why it is better than any of the other proposed systems? I have looked at Wikipedia's Schulze Method article to try to find the answer, and I find that it is basically written in gibberish. So all I can see at this point is that we would be substituting one majoritarian method for another -- and the one that we would be replacing has been in use for ArbCom elections for about five years, is consistent (for better or for worse) with the method used to select administrators and other "officials", and seems to have generally retained the support of the community. So what's the point in changing? (That doesn't necessarily mean I think Support/Oppose is necessarily the best system; if I saw any major support by other editors for approval voting, or cumulative voting, or STV, I would probably conclude that it would be worth a try for a year; but I am not seeing it. If that makes you wonder why I proposed "Support/Oppose" (with some options as to details) in the RfC, here's why:  Based on last year's RfC, I thought the choice this year was going to be between Support/Oppose and the Schulze Method, and I wanted to make sure Support/Oppose (basically, the status quo) was one of the "candidates."  (Last year's RfC was a mess in this regard; there was support for Schulze at the beginning, but I don't think the status quo was made an explicit option until close to the end, so I decided to fix that for this year.))  As it turns out, nobody has even "nominated" the Schulze Method or any other specific ranked method. There has been some support for "ranked systems" generally, including by you, Tony, but I don't see this as meaningful because nobody has proposed any of the specific ranked systems, which vary widely.)

So, okay, finally, here is what I see as the primary disadvantage of the Schulze Method: It is completely non-transparent. (I have probably said elsewhere that it is too complex, but that to me is only an element of the real issue, which is that it is non-transparent.) By that I mean that as a voter, I would have no idea how the result was achieved, and I would have no idea how my vote affected the result, if at all. I voted in the last WMF board election, and I looked at the results, which included a table of numbers that had absolutely no meaning to me. Although someone has said in one of these pages that Support/Oppose is complicated, in fact it is the very model of simplicity -- as is first-past-the-post and approval voting (HJ Mitchell's proposal, which after some digging I have found was used in the first two ArbCom elections in 2004, but replaced for the 2006 election by the current system) -- and cumulative voting, which VolunteerMarek has proposed but doesn't seem to be getting much attention. The impact of every single vote in any of these systems is very clear. None of these, by the way, involve rankings; once you start to get into rankings, you start getting some complexity and non-transparency, from the mild (such as STV) to the extreme (Schulze.) But that basically is my opinion of Schulze, based on what I know of it at this point:  Votes go into the box, and a table with a bunch of incomprehensible numbers, along with a result, comes out, with no understandable connection between the two. And its still a majoritarian, non-proportional system, so nothing really has been gained. Neutron (talk) 00:55, 3 October 2011 (UTC)


 * Yes, oppose voting is simple to count. It allows voters to rank candidates on three levels. It's a kangaroo system dreamed up in Wikipedia's barrack room. I can't find documentation of oppose voting outside Wikipedia. Why? Lightmouse (talk) 17:11, 3 October 2011 (UTC)

Consequences of reduction and expansion of ArbCom
One of the consequences of expansion or reduction of ArbCom is that the number of seats up for filling in any one election varies from year to year. I'm a strong supporter of the principle that the number of seats up for election in any one year should remain stable, as it imbalances things to have nine people selected one year and six the following year. You usually end up with a situation where the chances of those on the borderline being elected depends greatly on the year in which they chose to run. I would much prefer the top x to be selected each year, with x staying the same (or roughly the same) each year. Switching and changing the numbers selected each year fails to provide stability and continuity. At the moment, as far as I can tell, reducing to 15 would mean that six seats need to be filled this year, and then nine the next year, and so on in a 6-9-6-9 cycle. This would need to be balanced out somehow. Probably by having all the seats filled this year being 2-year terms and splitting the nine seats elected in 2012 in the ratio of seven 2-year seats to two 1-year seats. It is usually suggested that resignations and retirements help balance things out, but that didn't happen to a great extent this year (which is good) and no system should rely on that. The suggestion that all arbitrators be elected for 2 years would work as well, but only if you make the number of seats an even number and ensure the same number of seats are elected each year, otherwise you get the 9, 6, 9, 6, 9, 6 cycle again as above. However, I don't now support a mix of 1-year and 2-year terms (though I did in the past) and I now agree with the proposal that all terms should be 2 years. My reasoning is that having 1-year terms allows semi-gaming of the system in the following way: (i) Arbitrators are elected to 1-year terms due to receiving lower support than those elected to 2-year terms; (ii) Those arbitrators on 1-year terms take decisions and stances during their first term that aim to build their support base in a run for a subsequent 2-year term; (iii) They are subsequently elected to a 2-year term on the basis of being a reasonably good arbitrator in their first term and getting more support in the election following their first term; (iv) The arbitrators originally elected to 2-year terms take a longer-term view during their time on the committee, also taking some unpopular but needed decisions, but do not stand again at the end of two years (not wanting to face another two years). The end result is that the arbitrators who received less support in the inital elections end up being on the committee for three years, while those who received more support in the initial elections end up being on the committee for two years. This seems perverse to me, and is why I now support 2-year terms for all arbitrators. So my proposal would be for a committee of 16 arbitrators, with 8 seats elected annually, all for 2 years. One way to achieve this would be to: (a) Reduce from 18 to 17 and have only 8 seats up for election this year, and then reduce from 17 to 16 the next year by stating that of the nine seats up for election one is dropped and 8 are filled. All terms to be 2 years. Then in subsequent years stick with 8 seats elected per year (don't try and fill empty seats unnecessarily) and keep it that way for several years and stop fiddling with the system each year. Unfortunately, the way this RfC is set up, there is nowhere really to place this holistic proposal (which tries to consider everything rather than splitting up the components and commenting on them separately). I'll try and lay it out somewhere though, and see what people think. Carcharoth (talk) 11:01, 2 October 2011 (UTC)
 * Part of this was discussed a few sections up, about a week ago. ("Fifteen does not divide by two.") You came up with the same solution for "rebalancing" from 9-6 to 8-7 that I did, i.e. electing two people for one-year terms next year.  I think that's the way to go if the size of the committee is reduced to 15 (which I have opposed on the RfC page, but if it is done, I think it can be done in a reasonable way.)  I do not think there is anything wrong with electing some people for one-year terms to keep the groups even.  Your argument about people elected for one-year terms actually serving for three years is interesting, but I do not think that is a big deal because in order for them to serve the "last" two years, in their second election their support must increase from one of the lowest (which is why they won only a one-year term) to at least the middle of the "winners" group.  I see nothing wrong with such a person serving for three years.  On the other hand, if someone who has served for two years decides that that is enough, that is their decision -- you really can't design a system around how long some hypothetical person might decide is long enough to serve.  By that logic, all terms should be one year, and I don't think that's a good idea.  The bottom line is that if you are going to have staggered terms, it makes sense to keep the groups roughly even with each other, and that is accomplished by electing some people for one year to fill vacancies.  (By the way, I agree with you about the RfC process not being amenable to "holistic" approaches, but I think that is one of the inevitable results of trying to create a new "Constitution" every year, starting two months before elections.  If we want to really get organized about this, we need an official "Board of Elections" -- either an elected body solely for this purpose, or a group appointed by some other elected body.  And since I don't think the ArbCom should be the appointing body, the elected body ("GovCom"?) that should be doing the appointing does not exist. I think that's going to be a continuing issue, but it doesn't help us right now.)  Neutron (talk) 21:57, 2 October 2011 (UTC)
 * Would you mind re-reading what I wrote? You state above that I: "came up with the same solution for "rebalancing" from 9-6 to 8-7 that I did, i.e. electing two people for one-year terms next year." As far as I can tell, my proposal avoids 1-year terms and rebalances from 9-9 to 9-8 and then 8-8 by reducing by one each year over the next two years, and maintains 2-year terms for everyone appointed. This is entirely different from what you proposed. Please also see what I wrote here and here. The number of arbitrators appointed last year was 12. This year it may end up being 6. It could end up being 10 next year if (as is likely) at least one candidate this year gets a degree of support so low that they are only appointed to a 1-year term. The fluctuations are amplified or damped entirely by the number of 1-year terms decided upon, something that is in itself awkward as it is, as far as I can tell, something that Jimbo decides (possibly in consultation with others) and not the community. Some flexibility is needed, but I would suggest the candidates are asked to opine themselves on the merits or otherwise of a 1-year or 2-year appointment. They need to be aware of those possibilities. Having said that, I'm not totally opposed to an odd number of seats, but would suggest that the 15th seat be a genuine full-time co-ordinator or secretary (rather than a sitting arb juggling co-ordinating with voting and case work and sometimes dropping both balls), or a junior arb training for the role, or something even more creative (such as a reserve to be activated if needed at some point in the year). Carcharoth (talk) 23:28, 2 October 2011 (UTC)
 * I understood what you wrote. You mentioned the proposal that I had suggested, and then you rejected it, because you don't like one year terms.  I am just looking at the reality of the situation:  Coren's proposal to retain 1-year terms (based on the order of voting) has 36 supports, and Risker's proposal to eliminate 1-term terms has 8 supports.  So, so far, it seems as if 1-year terms are going to be retained.  If that is the case, and Risker's proposal to reduce the committee size to 15 is accepted (which so far also seems to be the case), achieving the reduction while re-balancing the groups should be easy (though not quick.)  As I said in the previous section, however, by the time the 2013 election comes along, it's likely that the committee size will have changed again.  So I wouldn't make too many specific plans for a committee size of 15, because it will be as transient as the committee size of 18.  As far as your "creative" ideas, my sense of the mood of the "community" right now is that anything "new" is perceived with a great amount of distrust.  For that matter, anything "old" is perceived with a certain amount of distrust as well.  Maybe that's just my perception.  Neutron (talk) 00:09, 3 October 2011 (UTC)
 * Carcharoth, yes, the successive vacancy rates do form an unfortunate wave motion, back and forth, if unchecked (e.g., 6–9–6–9, etc.). That's why Jimmy long ago established the principle of ironing that out with different term-lengths. But note that 2011 has been the only year in living memory in which there have been no early departures of arbs (before their last year of service). Shell left this year, but her seat was always going to be vacant in December, so that doesn't count. Doesit matter where the instability is located, since instability is inevitable, and both ways have disadvantages? Tony   (talk)  15:32, 3 October 2011 (UTC)

Oppose voting doesn't have the outcome intended
The oppose voting system doesn't have the outcome intended. Some voters, myself included, think/thought it's like 'First past the post' and a support vote is an effective way of getting a candidate elected. It isn't. A support vote has less value if it isn't associated with oppose votes. Any voter that fails to vote oppose is not getting the benefits available to those that do. That seems perverse to me.

It may be one of the reasons why this system has no support in literature about voting systems. It would be interesting to look at previous elections and see how many voters failed to vote oppose. Lightmouse (talk) 11:07, 2 October 2011 (UTC)
 * I do not know whether it is possible to look at the "public" statistics and determine what percentage of voters left at least one button at "neutral." I suspect that this information would be available to the "scrutineers" and the in-house computer people at Wikimedia, but not to us.  Essentially it would require looking at each individual "ballot."  What is publicly available is the information used to make up the pie chart posted above by Tony.  What it shows is that in 2010, of the total possible votes cast by people who voted, 34.7% were supports, 37.5% were neutral, and 27.8% were opposes.  (Interestingly, the number of supports went up significantly from 2009, the number of neutrals went down significantly, and the number of opposes was roughly the same.  2008 doesn't provide a useful comparison because voting was in public then.) This means that of the votes that could have been cast as opposes (in other words, the non-supports), only 42.6 percent were actually cast as opposes, with the majority being neutral.  (In my view, a neutral does not actually represent a "cast" vote, or a vote at all -- it means the person left the "button" for that candidate where it was on the ballot form, which to me means nothing one way or the other.) This suggests that a significant number of people do not vote strategically, in the sense of opposing all the candidates who they don't support. (And of course, many (and probably a majority, maybe even a large majority) of those opposes are sincere opposes, cast by a voter who really objects to that candidate.) Beyond that, we don't really know what these numbers mean with regard to individual voters, because we don't know whether those 37.5% neutrals were spread evenly across all voters or concentrated among just a small portion of "non-strategic" voters.  But I think the numbers tell us that there is a significant amount of "non-strategic" voting going on.  It may even be that a majority of voters only "oppose" those who they actually don't want to see elected.  That, by the way, is how I have been voting -- I have not voted "oppose" on candidates about whom I was truly neutral.  Apparently I am not the only one.  I do not think that is "perverse," as you say -- it is just a way of voting that many (possibly a majority) are comfortable with. Neutron (talk) 21:32, 2 October 2011 (UTC)

It's not sufficient to vote 'support', you must also vote 'oppose' for all the others. Do the statistics show voters fail to vote 'oppose'? Lightmouse (talk) 08:45, 3 October 2011 (UTC)
 * Neutron, the way I voted last was support for 11, by coincidence the number of vacancies; neutral on 2 (which was consciously a partial boost to them in not reducing their % result, and at the same time a slight disadvantage to those I did support, through boosting the competition to the prospects of my supported candidates), and oppose for the rest to reduce the chance they'd vie with my supports. I might just as well have voted 11 support and 9 oppose to give my 11 supports an even better chance. Any voter who ignores the power of opposing to benefit their supports needs to go to a psychiatrist. Tony   (talk)  09:23, 3 October 2011 (UTC)
 * Tony's graphs above show greater % support than % oppose. Does that mean voters are failing to vote oppose? Lightmouse (talk) 10:33, 3 October 2011 (UTC)
 * Maybe voters have subconsciously worked through the 7kB of ugly maths I just posted above and realised how academic the tactical voting 'benefits' are. :D <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 14:40, 3 October 2011 (UTC)
 * The maths aren't ugly, but they hold true for any voting unless the voter already knows their vote will lie on the boundary between the success or failure of one or more candidates. Fine for an independent parliamentarian or congressional member who hold the balance of power, but somehow a lot of us ordinary people are still motivated to put in our twopence worth knowing it almost certainly won't make any different to the outcome. Same for the ArbCom vote. Tactical opposing is one of the few ways open to voters to strengthen their hand in supporting their favoured candidates. Tony   (talk)  14:48, 3 October 2011 (UTC)
 * Tactical opposing is one of the few ways open to voters to strengthen their hand in supporting their favoured candidates - no it's not. That's the whole point of the maths above.  Volunteer Marek   23:15, 5 October 2011 (UTC)
 * 'Oppose voting' isn't mentioned outside Wikipedia. Why? Lightmouse (talk) 16:55, 3 October 2011 (UTC)
 * Given the fact that you have been active on Wikipedia for more than four years, I would have thought that you would have learned by now that most Wikipedians view the fact that something exists only on Wikipedia as being a good thing. Not me, but many people.  Neutron (talk) 21:56, 3 October 2011 (UTC)
 * Paradox of voting. Tony   (talk)  10:11, 5 October 2011 (UTC)
 * Tony, earlier you said "a lot of us ordinary people are still motivated to put in our twopence worth knowing it almost certainly won't make any different to the outcome." I don't know that at all.  I vote (generally, not just here) in the knowledge that there is a possibility that my vote may very well decide the outcome.  I can recall at least three local elections in my area that have been decided by one vote, and at least one tie.  The possibility exists in any election.  I also have seen a number of elections decided by less than 100 votes, in electorates ranging from 5,000 to 25,000 people or more.  These elections could very easily have swung the other way.  I also think that if I looked back at the records of past ArbCom elections I would find at least one election (and maybe more) in which the gap between the lowest-ranking winner and the highest-ranking loser was less than 10 votes.  I think many people vote at least in part because the possibility exists that their vote will decide the election.  Neutron (talk) 21:45, 5 October 2011 (UTC)

Rough timetable for ACE2011
Working out the planning for the election, I came up with the following rough timetable for the stages of the election:


 * Nomination period: Sunday 00:01, 13 November – Tuesday 23:59, 22 November
 * Inbetween period: Wednesday 00:01, 23 November – Sunday 23:59, 27 November
 * Voting period: Monday 00:01, 28 November – 23:59, Sunday 11 December
 * Scrutineering period: Monday 00:01, 12 December – ???

This includes 14 days for voting, 5 days inbetween, and 10 days for nominations. Note that the scrutineering period ends when the Stewards assigned have endorsed and verified the results, which is normally around 2 days. –MuZemike 02:26, 3 October 2011 (UTC)
 * This looks much better than some of the bloated periods suggested elsewhere: already it's more than a month. Isn't 12 December just a little late to start the stewards on the scrutineering? (BTW, the stewards have to be asked very nicely to do this—it's a tiresome job, and some believe it's outside their stated policy.) Tony   (talk)  02:37, 3 October 2011 (UTC)


 * Thanks for that, MuZemike. Is it possible to add days of the week so people know where weekends fall in that? It is probably also best to use 00:01 or 23:59 instead of 00:00. When 00:00 is used together with 23:59 it is usually clear what it means, but if people see 00:00 on its own, they may be unsure whether that refers to the start or end of that 24-hour period. There is also a list around somewhere (probably made last year) of what actually needs doing, including finding a developer to switch on SecurePoll and stuff like that. Is Happy-melon able to do that or not? Some of that and this might be best discussed at WT:ACE2011 (which I just created), rather than here. One final comment is that it may be worth starting the publicising of the election a little bit before 13 November, as that will give those who might not have realised they need to start thinking about nominations so early, a chance to think about it. Some people, if they only become aware of this a bit into the nomination period may only have five days or so to think about it and prepare a nomination statement. It would be good if people potentially committing to a 2 year role would think about it for at least a week, as we often get candidates withdrawing when they realise what they might have let themselves in for... Carcharoth (talk) 05:12, 3 October 2011 (UTC)
 * Very good ideas. As for publicising, I suppose I can whip up a buzz article in The Signpost a few weeks before the opening of nominations. I'm not sure I'm going to be a coordinator this year. Tony   (talk)  05:31, 3 October 2011 (UTC)

Why wait until Nov 13 to start the nomination period? The sooner nominations start, the longer people will have to ask questions and generate informed opinions. In the discussion about the length of the RFC, Risker suggested nominations start in October, I would suggest that nominations start Nov 1 or as soon as the RFC is closed and summarized, which ever comes later. Monty <sub style="color:#A3BFBF;">845  23:37, 3 October 2011 (UTC)
 * I'll leave it up for the others to decide if they disagree on the timeline, but I believed Risker said that this RfC should close in October and not have nominations start. Moreover, do we want the ArbCom elections to last for over 2 months (nearly taking up the last third of the year)? –MuZemike 02:05, 4 October 2011 (UTC)
 * The statement was at, I missed that it was conditioned on the voting being more then 14 days. Its pretty close between 14 and 21 at the moment. Still, going from 5 and a half weeks to 2 months and then calling it almost the last third of year seems like a bit of a stretch. Monty  <sub style="color:#A3BFBF;">845  02:34, 4 October 2011 (UTC)
 * I put down 14 because, to say the least, we have a consensus for 14 days minimum – more than the previous years voting window of 10. Currently, I don't see how we can get a consensus out of 21 when it's literally split up the middle like that. That was why I went with the "safe number" of 14. –MuZemike 02:39, 4 October 2011 (UTC)

I've changed the start times to 00:01 to avoid any possible ambiguity and added in days of the week, as I'm sure some will be interested in that. Thoughts? –MuZemike 00:32, 5 October 2011 (UTC)
 * Thanks, MZ, one minute past is what we agreed on after the near-catastrophe caused by a misdunderstanding based on 0000 in ACE2009. Any period longer than 14 days I will oppose strongly. Tony   (talk)  10:10, 5 October 2011 (UTC)

I think the nominating period should begin sooner. Among other things, it would give those candidates who choose to self-nominate early more time to answer the large amount of questions that will be asked. It also seems a little odd to have an election that is longer than the nominating period. Is there any reason why it cannot begin three or four days after the RfC is closed? Neutron (talk) 16:47, 5 October 2011 (UTC)
 * While I don't have an inherent objection to that notion, the one thing that a long nomination period leads to is more and more questions being asked, to the point that potential candidates may decide not to run based on the fact that early candidates already have (for example) 50 questions to answer. There is a 5-day period between close of nominations and start of voting which helps in that regard. Risker (talk) 17:08, 5 October 2011 (UTC)
 * I think the number of questions asked is a different issue, and one that needs to be tackled directly, not by limiting the nominating period. Personally I thought the number of questions asked last year was ridiculous.  However, I just read the part of the RfC regarding questions to candidates (I had been staying away from that part of the page) and it does not appear that a consensus exists for any kind of reasonable limitation.  But I don't think the solution is to keep the nominating period so short.  Neutron (talk) 21:30, 5 October 2011 (UTC)
 * What are we looking at so far as far as questions are concerned? I know Tryptofish recommended we make Requests for comment/Arbitration Committee Elections December 2011/Candidate questions so we can iron that out; perhaps one should so that we can take a closer look, as that section of the RfC is rather disorganized and lacks any sense of direction. –MuZemike 02:26, 7 October 2011 (UTC)
 * I went ahead and created the page I mentioned above and preloaded with User:Sven Manguard's revised set of questions from this RfC; I'm assuming, though, that these questions are an improvement over the general set of questions asked last year. People are free to edit it, of course, making changes/additions/deletions when needed (as with any other wiki page). –MuZemike 02:31, 7 October 2011 (UTC)
 * Urgent that the dates be firmed up, and the tech approached with the likely times their assistance will be required. Tony   (talk)  04:05, 7 October 2011 (UTC)

Non-math summary on tactical voting
I'm going to basically summarize in words some of the above discussion where some of us got too math happy and then per my previous comment above throw some ad-hoc psychology into it to reach a conclusion which actually, if you've been following along, reverses my previous stand on this whole issue of "Support threhsholds".

Basically:
 * The chance that your own vote will have a meaningful effect on the outcome of the election is very very small. Of course very very small things add up to big things so, sure, go ahead and vote (or you'll burn in hell).
 * BUT, the chance that your "tactical vote" (where you vote "Oppose" on a candidate that you're really "Neutral" about, in order to help the candidates that you actually "Support") is ... much much much much much much much much much much (much^8) smaller than just the chance that your vote will matter. For all practical purposes there's no point in "voting tactically".
 * Even if you were to "vote tactically" the rational way to do it would be actually to in some cases vote "Oppose" on your "Neutrals" but in other cases to vote "Support" on your "Neutrals" in order to prevent people you don't like from getting elected. Given that other people don't have the same preferences as you, a lot of these "tactical votes" are very likely to cancel each other out. This adds a couple more "much much much much smallers" to what's above already.
 * So basically, if everyone voting took some time to think about "tactical voting" they'd all realize it's pointless and nobody would do it.
 * But for whatever reason, it seems that a lot of folks (including some arbitrators, like Sir Fozzie) DO THINK that a "tactical vote" will help them get the outcome they most preferred. This is really not the case but people think that.
 * Considering the above, if at least the people who think that "tactical votes matter" would realize the symmetry in the impact of tactical votes in aggregate (that it makes sense to vote "Support" for your Neutrals in some cases, as much as it makes sense to "Oppose" them in others) then this really really wouldn't matter. It'd be just a bunch of noise in the result, which would average out and have very little impact on the outcome of the election.
 * BUT BUT - it seems people are only thinking "If I Oppose my Neutrals that will really help my Supports" but they are not thinking "If I Support my Neutrals that will really screw my Opposes". To me that's weird and it says something about the mentality of your average ... person, or Wikipedia editor, but it is what it is. See Behavioral economics. I suspect that people just enjoy "Opposing" people and that's why they do it. This suspicion of mine says something about MY own mentality.
 * Given that people, wrongly, believe that "Opposing" their Neutrals helps their Supports, yeah, you're gonna get lower %s in terms of the S/(O+S) ranking.
 * At that point, I dunno what you wanna do with that. You could say "all those tactically voting people are stupid and are just messing up things, so let's have a low threshold, with an understanding that some of the people on the arbcom will really be "Neutrals"". On the other hand it could very well be that a lot of people voting "Oppose" really DO "Oppose" a particular candidate. And that shouldn't be discounted either. So at that point the whole thing becomes a judgment call ... which reflects YOUR mentality and beliefs about what the average voting Wikipedian is like.
 * So. Before I said that I supported SandyGeorgia's high threshold for the S/(S+O) percentage (in fact I demanded 100%, as a joke) but reconsidering things ... a lower threshold might make more sense. Not because "tactical voting matters", but because a lot of people think that "tactical voting matters" AND (let me stress that AND) they don't consider that to the extent it does matter, they should be Supporting their Neutrals in order to screw their Opposes.

 Volunteer Marek  23:41, 5 October 2011 (UTC)

Here's a summary ... an algorithm on how to vote tactically - given the assumption that it actually matters (it doesn't):


 * Divide all the available candidates into those you really "Support", those you are "Neutral" on and those whom you really "Oppose".
 * 1) If the number of Really Support+Really Oppose = Neutrals or something close, give up.
 * 2) If the number of Really Support is much greater than the number Really Oppose AND there aren't that many neutrals, vote "Oppose" on your Neutrals.
 * 3) If the number of Really Support is much less than the number of Really Oppose AND there aren't that many neutrals, vote "Support" on your Neutrals.

And this is actually a pretty intuitive result.

If it's #1 it doesn't matter.

If it's #2 then it's better to have the the people you Really Support on the AC rather than a bunch of Neutrals. Your vote will have a 1/200000 chance of making that happen.

If it's #3 then it's better to have the Neutrals on the AC rather than a bunch of people you Really Oppose. Your vote will have a 1/20000 chance of making that happen.

QEMFD.  Volunteer Marek  23:54, 5 October 2011 (UTC)
 * Doesn't this all rather depend on what you think the respective candidates' chances of getting elected are? If there are several weirdo candidates (candidates with ideas, say) who you are confident aren't going to get elected whatever happens, you can forget about them (even if you Really Oppose them) and just concentrate on getting your Really Supports elected. Also I'm not convinced by this "tactical voting has a much^8 smaller chance of making a difference than just voting" - surely if I vote to "oppose" the nth best placed candidate, I'm at least as likely to make a difference as if I vote to "support" the (n+1)st (where n is the number that get elected)?--Kotniski (talk) 06:56, 6 October 2011 (UTC)
 * Yes, but you must also apply the same logic to the several rock-solid candidates that probably exist whom you are confident will get elected whatever happens, and you can forget about them in the same fashion. The key principle is that the situation is reversible, and any logic you apply to filter your preferences you must apply to both ends of the spectrum.  For the chance-of-making-a-difference, your common sense (and most people's common sense, for that matter) must give way before the ugly block of maths a few sections above: it is counterintuitive, but it is also true. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 10:51, 6 October 2011 (UTC)
 * On this occasion, I'm going to prefer to believe that there's a mistake in the maths (or more likely, in the assumptions on which the maths is based) than in my intuition ;) Kotniski (talk) 12:09, 6 October 2011 (UTC)

I'm concerned about the following scenario which isn't covered above: For example, Bob the voter likes two candidates and doesn't dislike any. So he votes 'support' for the ones he likes and doesn't vote for anyone else. I've been saying that Bob should have voted 'support' for the ones he likes (as he did) and 'oppose' on the rest (as he didn't do). Is that true? Lightmouse (talk) 09:26, 6 October 2011 (UTC)
 * If the number of Really Support is greater than the number of Really Oppose AND there are many neutrals.
 * It doesn't actually really matter how many neutrals you have. In the scenario you describe, Bob accepts that the election winners are going to include a lot of candidates he does not actively support, simply because he doesn't support enough candidates for it to be otherwise.  So Bob's payoff is maximised by the winners including his two prefered candidates, and then being filled up with neutrals.  By voting Oppose on all the candidates about which he is neutral, he increases the chance of one of his prefered candidates pipping a neutral to the final place (by 1 in 20,000), so his tactical vote should be to Oppose all the neutral candidates.  <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 10:58, 6 October 2011 (UTC)
 * "AND there are many neutrals." - that's just a way of saying here that M is large. I think... I'm starting to get lost in this myself...  Volunteer Marek   18:03, 6 October 2011 (UTC)

Thanks. That's the first time anyone has acknowledged what I've been asserting. Lightmouse (talk) 11:30, 6 October 2011 (UTC)
 * You made a constraint above:
 * ''"AND there aren't that many neutrals".
 * Then you said:
 * ''"It doesn't actually really matter how many neutrals"
 * Is it true that for all optimum scenarios support+oppose=number_of_candidates? Lightmouse (talk) 11:39, 6 October 2011 (UTC)
 * Are you asking whether it's always the right strategy to avoid voting neutral for anyone? Intuitively I would say it is; for any candidate X, you have to judge whether you think it's more likely that they'll be competing for nth place with a candidate you like less than X, or with a candidate you like more than X (where strictly you should weight the likelihoods by multiplying the individual probabilities by the amounts by which you like the respective candidates less or more) . In the first case you should vote support; in the second you should vote oppose. I can't think of a situation where it pays off to vote neutral.--Kotniski (talk) 12:00, 6 October 2011 (UTC)

Yes. I was asking whether it's always the right strategy to avoid voting neutral for anyone. Thanks for confirming that to be the case. I assume *failing to vote* is as bad as voting neutral. Lightmouse (talk) 13:14, 6 October 2011 (UTC)
 * HappyMelon appears to have confirmed in plain language everything I think is bad about S/(S+O). Tony   (talk)  13:45, 6 October 2011 (UTC)
 * But your non-Neutral votes for your Neutral candidates aren't going to matter except at a .0000000000004 level. Failing to vote is different.  Volunteer Marek   18:03, 6 October 2011 (UTC)
 * I'm still far from convinced that this kind of claim is true. According to my relatively simple maths (see what effect you have on S/(S+O) by adding one to numerator and denominator, or to denominator only) your oppose votes for neutral candidates are just as likely to matter (slightly more likely in fact, assuming the border percentage is around 55%) as your support votes for support candidates. So someone who votes tactically is getting more than twice the voting power of someone who doesn't.--Kotniski (talk) 18:27, 6 October 2011 (UTC)
 * The simplicity in your maths is not mistaking anything in that analysis, it is just that what you are claiming as "voting power" is not a quantity which has any useful meaning. Being able to raise the support percentage for a candidate is of no use whatsoever in isolation; you improve the outcome of the election only when your vote causes your candidate to win.  To judge the "power" of a vote you ask the question, "in what fraction of possible cases does my vote cause my prefered candidate to win?", accepting the fact that that number is not very informative in isolation, but only when compared to other figures.  To judge the "increase in strength provided by tactical voting" you would ask "in how many more cases can I cause my candidate to win compared to not voting tactically?"; to which the answer is as described above: an utterly pathetically small fraction of the time.  You are analysing the increase in strength in the voting system where all candidates who score over a certain threshold are elected, however many there are.  The real system is significantly more complicated than that, hence the ugly maths, and the more complicated result. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 23:10, 6 October 2011 (UTC)
 * I'm still not persuaded - if we assume the top n candidates are going to be elected, my vote can make a difference if the difference (d, say) between the current nth candidate's score (=S/(S+O)) and the current n+1st candidate's score is smaller than the amount by which my vote can affect those scores. The amount by which a support vote will increase a score comes out to be (1-s)/v (where s is the current score and v the number of votes including mine); the amount by which an oppose vote will decrease a score comes out to be s/v. So if I'm voting "tactically" I can make a difference if d is less than 1/v (the sum of the last two expressions), whereas if I don't use my oppose vote I need d to be less than (1-s)/v, which (if s is around 55%) is slightly less than half of 1/v. So assuming relatively uniform probability distributions for the S/(S+O) variables over the narrow intervals involved, my vote makes a difference twice as often if I vote tactically. Or, it takes only one tactical voter to neutralize two non-tactical voters.--Kotniski (talk) 07:03, 7 October 2011 (UTC)
 * It's hard to respond to this without it descending into another round of maths, which I think this section is supposed to avoid :D Basically your analysis is as correct as your previous comment, but only in a very tiny fraction of possible scenarios.  Because the election is a race, not an absolute ranking, your tactical vote can only have an impact in the specific case of a borderline candidacy which is also on the N/N+1 boundary, and involves the right pair of candidates.  For instance, your normal vote has an impact when the Nth and N+1th candidates are both ones you support, but a tactical vote does not; ditto for the borderline candidates both being ones you oppose, or any other permutation that is not as I described it above.  Your tactical vote can only have an impact in a very small subset of the scenarios where your ordinary vote is already having an impact; and it is a mistake not to account for that when trying to calculate the "power" of the tactical vote. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 13:50, 7 October 2011 (UTC)
 * I think we know that the chance of my vote's possibly affecting the result is pretty small to start with. But assuming the chance of its doing so is significantly large that I can be bothered opening the form, then that chance will be only half as great if I decline to vote tactically. I don't understand this "very small subset" thing at all - my normal vote may have an impact; my tactical vote may have an impact; put both together and they're twice as likely to have an impact. Still not very likely; but if I were allowed to put two papers into the ballot box at election time instead of one, I'd want to know about it.--Kotniski (talk) 14:37, 7 October 2011 (UTC)
 * We're getting technical again, but never mind. Taking the assumption that we will only consider scenarios where your main vote has an effect is reasonable (indeed that's what I've said all along, you need to be comparing the "power" of the tactical vote to that of your main vote); the mathematical interpretation of that in the formula a few sections above is to throw out the first two terms: you assume that a borderline scenario (where your actions can change the order of the candidates) exists, and that it is on the boundary between a candidate getting elected and not getting elected.  So you are assuming that there are two candidates, for whom you can cast votes, one of whom will be elected and the other of whom will not: the "power" of your vote is your ability to cause people you support to be elected.  In every scenario, your 'regular' vote will have an effect on this outcome: your regular vote is "powerful".  But the vast majority of scenarios do not allow your tactical vote to have an effect, because for that to happen one of the candidates must be one of your 'neutral' (B-type) candidates, and the other candidate must be the right type of 'other' candidate (either positive (A) or negative (C)).  In a scenario where both the candidates are the same (AA, BB, CC) or where there isn't a neutral candidate (AC, CA) or where the candidates are already in the order you would want them (AB, BC) your tactical vote on the B candidates does not improve your happiness with the election outcome: your tactical vote carries no "power".  And of the two remaining scenarios (BA, CB) you can only choose to 'play' your tactical vote card on one of them.  In the example I gave above, this element of the probability equation comes to 4/105, or 3.8%: out of all the scenarios where your 'regular' vote can have an effect, in only 4% of them can your tactical vote also have an effect.  The exact figure varies depending on how many of the candidates fall into each category; the figure actually decreases as the number of neutral candidates rises.  In those instances, yes, it has very roughly twice the "impact" of a regular vote on its own.  But far from you being "allowed to put two papers into the ballot box", you're being offered the chance to put something like an extra 2.5% of a vote in. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 15:17, 7 October 2011 (UTC)
 * But you're deliberately restricting your analysis to cases where my regular vote can have an effect. Of course in those scenarios it's pretty unlikely that my tactical vote will have an effect as well (since any given vote is pretty unlikely to have any effect anyway). However there are also cases where my regular vote by itself will not have an effect, but where in combination with a tactical vote it will produce an effect. I reckon (from my simplified maths above) there will be just as many cases like that as there will be cases where my regular vote will have an effect by itself. (Of course the candidates have to be of the right sort for me to be satisfied with the effect, but that's true of my regular vote as well.)--Kotniski (talk) 11:54, 8 October 2011 (UTC)
 * No, I was just sloppy with my language, which is a cardinal sin in mathematics. The limits imposed in the first two terms of the equation in the previous section, and what I was refering to above, serve to pick out the scenarios where you have the ability to affect the outcome of the election by voting; that is, all possible scenarios where the two borderline candidates are close enough together that some vote combination from you can alter their order and change who wins.  That's what I meant by "your main vote has an effect", I should have said "your vote has the potential to have an effect".  Within that, there are scenarios where you cannot make an effect in a way that pleases you (your regular vote can make an impact, but it would worsen your situation not improve it), scenarios where your 'regular' vote is sufficient to cause a desirable outcome, and scenarios where your 'regular' vote is not sufficient but your tactical vote is.  That last group, as I explained, comprise about 2.5% of all those scenarios. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 16:57, 8 October 2011 (UTC)


 * And at a more macro level, in an extreme case, one tactical voter can outvote a million non-tactical voters. (If the million all vote for A and remain neutral on B, and the tactician votes for B and against A, then B scores 100% while A scores slightly less.)--Kotniski (talk) 07:14, 7 October 2011 (UTC)
 * But if you believe tactical voting is powerful, you are forced to assume that everyone (or at least a majority of people) will do it, in which case its power is greatly reduced; one of my first mathematical discoveries was that there are no possible scenarios in which a tactical vote can have an impact unless there are people who are not voting tactically. Yes, in the scenario you have described your tactical vote has a strong impact, but it is exactly as "powerful" as the scenario where there is only one other voter who votes the same way as the million in your example.  In both cases the tactical vote has exactly the same impact on your happiness with the election result, and that is the only thing that you can assign a value to.  Or if you are of the mindset that 'beating' a million other voters is its own reward, you can do that... but remember to weight that measure of "power" by the probability of that scenario actually occuring: assuming the uniform probability distribution you took above, that's 1.8&times;10-477,121, times whatever satisfaction level you want to assign to it.  Extreme examples are fun and sit easily in the mind, but they do not represent a realistic analysis of the problem. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 13:50, 7 October 2011 (UTC)
 * The practical point is that results are going to be skewed if tactical voters are not evenly distributed among the candidates. So if among candidate X's supporters there are 40% tactical voters, say, and among Y's supporters there are 60% tactical voters, it means X may lose even though he has 20% more supporters than Y. And this can realistically come about either because X attracts a "nicer" class of voter than Y does, and/or because X's supporters are less clued up about how the voting system works than Y's are. (Of course it may also be for the "right" reason - that X's supporters are genuinely happier about Y than Y's are about X.) --Kotniski (talk) 14:37, 7 October 2011 (UTC)
 * But if you're now descending into comparisons with multiple tactical voters, all those tiny percentages are going to be compounded even further. Remember also that the greater the proportion of tactical voters, the less powerful each tactical vote becomes (that was a result from the previous section).  I don't have exact figures for how the "power" varies with tactical-voter-density, and it's a horribly nonlinear analysis, but we can consider two edgecase scenarios.  If the "power" were in fact constant, then assuming that the situation is reversible (which I think it is) the aggregate effect of the two opposing 'teams' of voters would cancel out as far as possible, and what you are actually looking at is 20% of Team Y casting 2.5% of an extra vote, for an overall difference of 0.05%.  That's the best case in terms of impact on that scenario.  In the worst case, if "power" is inversely proportional to the proportion of tactical voters, multiplying the two together produces a constant which will exactly cancel, leaving no net impact.  If the power were actually proportional to a higher negative exponent of the voter density (which I doubt), it would actually produce the counterintuitive result that Team X would win because their tactical votes are so much more powerful than Team Y's tactical votes.  Most likely the relationship is somewhere between the first two scenarios, but assume the best case of 0.05%.  That means that by voting tactically, Team Y can make Y win not only in the 1-in-20,000 scenarios where they could make him win anyway, but also in the scenarios where he is up to 0.05% further behind X.  How many of those do you think there are?  I ran the exhaustive search again to look: for 850 voters, there are 2,421,323 of them... out of 13,080,805,625 total combinations; or 0.0185% of them.  So in the scenario you describe, that balance of tactical voters will cause Y to win 0.0185% more often.  Nothing in your analysis is wrong, per se, you are just considering scenarios which seem 'typical' but actually represent only a tiny corner of the parameter space.  The only 'mistake' is to not interpret the "power" of a tactical vote as being as averaged over the whole space, which it must be since you have no control over where in the parameter space the outcome of the election falls.  <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 16:57, 8 October 2011 (UTC)


 * Keep in mind, though, a few people have now spent several weeks, thousands of words and several ugly-looking equations analyzing the flaws of Support/Oppose voting. There has been very little analysis (well, there has been some, by me) of the other voting systems.  In the RfC so far, there does not seem interest in any particular alternative system.  The "simplest" and most common system in the "real world," plurality voting (aka first past the post) has three people in support.  In fact, although there has been talk of a "ranked system," nobody has even "officially" proposed any particular ranked system.  The only discussion of particular ranked systems has been on this talk page, not in the RfC itself, and most of the discussion about ranked systems here has been negative.  So however many flaws you may see in Support/Oppose voting, if it were abandoned it would have to be replaced by something, and there does not appear to be a consensus, or even an "emerging" consensus, that it should be replaced by anything, and even less of a prospect of consensus as to what that something should be.  Neutron (talk) 17:47, 6 October 2011 (UTC)


 * I like my idea of everyone getting 100 points to allocate, and I hate everyone else's ideas. Actually I'm Neutral on yours but I'm being tactical here.  Volunteer Marek   18:03, 6 October 2011 (UTC)
 * A tactically-minded voter would invest all 100 points (or at least 90 of them) into a single candidate. --Michaeldsuarez (talk) 19:07, 6 October 2011 (UTC)
 * Not necessarily. Ever heard the phrase "don't put all your eggs in one basket" or "portfolio diversification"? Roughly speaking you would want to allocate your points to where they make the most difference. So you wouldn't put them into any candidate that is sure to win nor any candidate that is sure to loose. The second one is ok, since that's a self-confirming expectation, but the first one would imply, that if your beliefs are shared by others, the sure-to-win candidate is going to loose (actually there's probably multiple expectational equilibria and the like, but I'm just guessing here).  Volunteer Marek   20:10, 6 October 2011 (UTC)
 * I don't think there's necessarily anything wrong with the current voting system (at least, I believe it's been mathemtically proven that there must necessarily be something wrong with any voting system), as long as we explain it properly (make it clear in the instructions that you can double the power of your vote, should you wish to, by making use of the "oppose" column), and as long as we don't conclude from the fact that someone got less than x% for S/(S+O) that they only have the trust of x% of the community, as we might at RfA.--Kotniski (talk) 19:13, 6 October 2011 (UTC)
 * Arrow's impossibility theorem and Gibbard–Satterthwaite theorem though it all goes back to the Condorcet paradox.  Volunteer Marek   20:10, 6 October 2011 (UTC)

This is all very nice and academic and philosophical and everything, but I feel the need to inject a dose of reality into the discussion: After an initial flurry of interest in this RfC, the "electorate" seems to have pretty much lost interest. The RfC page itself has not been edited at all in three days. The participants in this talk page (at least as far as voting methods are concerned) have dwindled down to about five or six people. As a result, unless something drastic occurs that changes peoples' minds and/or attracts an influx of new participants to the RfC, the result is going to be based pretty much on what appears on the page today, and that means a choice between two things: A Support/Oppose system with a minimum vote of 50 percent, and a Support/Oppose system with a minimum vote of 60 percent. If anybody can interpret the current "voting" to allow for any other possible result, I'd be interested to see your argument, but it seems pretty obvious to me. Now, I have "voted" for 50 percent and I believe that much of the above discussion of "tactical" or "strategic" voting provides justification for 50 percent. I know some of you think it justifies a threshhold of zero, or an entirely different system, but if the RfC closed right now, those options would not be open to the closer. It would be S/O with 50, or S/O with 60. At the moment, a closer could theoretically go either way, although I think that with the current tallies, there are better reasons for keeping it at 50 percent. (Basically because I think the "votes" for zero percent should "default" to 50 as an implied second choice, and the votes for 66 or 70 or whatever (but not 100 or 101) should default to 60, and after trying to untangle the first choices and second choices and not count people twice, I come up with 29 for 50 percent and 25 or 26 for 60 percent. (I'd have to recheck those numbers, and I am not sure how to count CT Cooper anyway.) Another way of looking at it is that there should be a clear consensus to make a change, and it has been 50 percent for at least five years, and even if only express "second choices" were counted, that would make it 21 or 22 for 60 and 19 for 50 (again, subject to rechecking), and that's not a clear consensus, so it stays at 50 percent. But of course, the person who closes the RfC may look at it differently.) So you can talk about Condorcet and strategic voting and equations and things all you want, but the real question is this:  50 or 60? Neutron (talk) 21:00, 6 October 2011 (UTC)
 * Fitty.  Volunteer Marek   21:31, 6 October 2011 (UTC)
 * Perhaps the real question should be: is a competitive election really the best way to select volunteers who are to be allowed to do the work of high-level dispute resolution on a project like ours? Is there any reason we should exclude someone from doing this work just because there are n people who we think will do it slightly better than them? (Or allow someone to do it, even though we think they're not entirely suitable, just because there are only n-1 willing candidates who rank above them?) --Kotniski (talk) 07:25, 7 October 2011 (UTC)
 * The most important thing is that it's plain unfair not to inform voters that they can strengthen their support votes by tactically opposing all candidates they don't support, rather than voting neutral. Irrespective of the proportion of voters who are in the know and do this, it will disadvantage those who don't tactically oppose. Tony   (talk)  14:51, 7 October 2011 (UTC)
 * Again, the amount by which tactical voters can strengthen their impact on the outcome is essentially zero. Sure, in actuality it is very very very slightly positive but essentially zero. And as has already been pointed out before, if anything we should be informing voters that they strengthen (however miniscule-y) their oppose votes by tactically supporting all candidates they don't oppose, rather than voting neutral, since that seems to be the part that people (even here) are not aware of.  Volunteer Marek   16:54, 7 October 2011 (UTC)
 * See some of my scenarios above - it is demonstrably not true (by a very long way) that tactical voters cannot strengthen their impact on the outcome. It's only "essentially zero" in the sense that the likelihood of any individual voter's influencing the result of an election is "essentially zero". But it's true that tactical voting might involve tactical supports as well as tactical opposes, and that information should also be conveyed somehow.--Kotniski (talk) 11:54, 8 October 2011 (UTC)
 * I don't know if that should be part of the "official" voting information. I don't think it is the kind of information that the "entity" running an election (usually the government) typically provides to voters.  I do think this information (it's really more in the nature of "advice") can be shared with the voters by those editors who wish to share it.  You could write a "voter guide" containing this advice, even if it does not endorse or oppose any candidates, and it would (or at least, it should) be linked from the election template.  And/or you could write an essay containing this advice, and I would support linking it from the template, but of course that would apply to other essays about the election as well.  Neutron (talk) 15:04, 7 October 2011 (UTC)
 * All this seems to be part of an ongoing effort to leave casual voters in the dark, and thus strengthen the voting power of those who are intimately involved in the election and are going to read all the bumf. Maybe this is justified, as those who are closely following the election and take an active interest in Arbcom-related matters can be expected to have better judgement as to the best candidates, so maybe it's right that their votes should count higher. But given that this is an area in which Wikipedia attempts to dabble in democracy, it seems we ought to at least try to make sure that all voters are equally informed as to what their votes "mean".--Kotniski (talk) 12:02, 8 October 2011 (UTC)
 * Oppose voting is clearly a ranked preference system: positive, neutral, negative. It's just one that we can't test against other ranked preference systems because there's never been an academic study of it. Since this system is regarded as beneficial and intended to last for many years, why don't we ask a university or other organisation to compare it with other systems? Lightmouse (talk) 17:13, 8 October 2011 (UTC)


 * You're all being silly. Every oppose vote has an equal value whether or not it is "tactical". If people are opposing candidates not because they oppose them but because they would rather someone else get in, that oppose vote is equally counted as if it was an oppose vote because the voter didn't want the candidate to succeed. There is no difference in the value of the vote, and all the theoretical mathematics in the world is not going to change that. We know that some people oppose candidates in the belief that it will increase the likelihood that their favoured candidates will succeed. The opportunity to do this means that the support percentage will be lower, although we cannot tell if it will be lower for all candidates or for some candidates, or to what extent it will be lower.  We cannot control for this, unless we remove the option to oppose entirely, and doing so will require an extensive discussion with the community that cannot be undertaken at this time and still proceed with the election.  So, if someone wants to change the voting system, I recommend they make some proposals in May of next year, so that this can all be thrashed out well in advance, and the sysadmins will have the needed time to make the programming changes. Incidentally, if we want to use SecurePoll, final decisions on dates and format need to be made NOW and provided to the sysadmins, because this will directly impact their other work and must be scheduled. Risker (talk) 17:22, 8 October 2011 (UTC)
 * Yes, totally agree with Risker's comment. Vote in whichever way you want in a manner the you feel your votes will have the greatest affect in support of your favored outcome. Lets get on with it. Off2riorob (talk) 17:28, 8 October 2011 (UTC)
 * Risker is correct that the reason one opposes a candidate is meaningless when figuring out which candidates are elected; but they are very meaningful when attempting to discuss some arbitrary percentage of necessary support to be elected at all as magically "representative of the community's support". That we cannot control for this with the current voting system is indeed inevitable, but this means that any attempt at using the absolute values of the S/O ratio (as opposed to relative ranking) is intrinsically meaningless and destructive.  &mdash; Coren (talk) 21:16, 12 October 2011 (UTC)

Wording of the voting instructions

 * I would certainly be in favour or having only two voting options rather than three, since we've seen that it's irrational to use the middle option, and we've also seen that people only realize this when they start to think about it (and often not even then). So putting the middle option in is like a trap for the unwary - in this way we can tempt people to weaken the value of their vote relative to ours, those of us who are "in the know". Of course I don't expect this to change, but one thing we can do is to word the voting instructions in such a way as not to mislead people - to point out in a sentence or two that in order to give a vote its full potential value, you should mark all the candidates as either "support" or "oppose".--Kotniski (talk) 17:33, 8 October 2011 (UTC)
 * Please let's not tell people how to vote; it is inappropriate and treats our voters as imbeciles. Most people can figure out by themselves how they want to vote, and for whom. Your position that there should only be two options is in the minority, and would require a major demonstrated shift in consensus to be actuated; it also disregards the fact that, if there are only two options, then SecurePoll is going to have to default to one of them, and people will have to consciously change all of their opposite votes. It creates a built-in bias. People are smart enough to figure out the Schulze method when voting for WMF Board members, and it's a lot more complex. Risker (talk) 18:02, 8 October 2011 (UTC)
 * I agree with Risker -- well, except for the part about the Schulze Method, I don't think most people understand how it works, but since there hasn't been a serious proposal to adopt it for the ArbCom election this year, I don't think we need to get diverted by it. This business of two vs. three options is really all semantics.  There are really only two meaningful options, Support and Oppose.  Given the way the voting software works, there has to be a "third" option which the voter sees as being "checked off" before he/she chooses one of the other two options -- or decides to take no action regarding that candidate.  I think that will be even clearer this year because there appears to be consensus (so far) for my proposal to rename "Neutral" to "No Vote."  I think people will understand that failing to choose "Support" or "Oppose" for any given candidate, and leaving the check-mark on "No Vote," means they are giving up the right to affect whether that candidate is elected or not, which includes the right to affect how that candidate fares against another candidate who the voter may support -- or oppose.  As I said above, if people want to write up a voter guide, essay or some other personal-opinion piece that expresses their views on "strategic voting" (either for or against), that is fine.  I also think that if it is in the form of a voter guide or even an essay, it would be fine to link to that from the election template, as long as it is made clear that all of the guides, essays or whatevers are solely the opinions of their authors.  But the "official" election information does not need to include that.  Neutron (talk) 22:55, 8 October 2011 (UTC)
 * But it isn't true that by choosing a particular option you are "giving up the right to affect whether a candidate is elected or not". The only way you can do that is not to vote at all (or to vote "No vote" for all candidates). As soon as you vote for or against one candidate, you are already affecting the chances of all the other candidates. If we don't want to appear to be "telling people how to vote", and we think they're mathematically sophisticated enough to work it for themselves in an instant what it's taken us reams of discussion to partially understand, then we should just give them the formula (a candidate's score will be the number of support votes divided by the total number of support and oppose votes). We shouldn't feed them the meaningless and misleading statement we had last year, that "voting neutral is equivalent to not voting for a candidate" or something like that.--Kotniski (talk) 09:29, 9 October 2011 (UTC)

Possibility of having only two options
(Oh, and incidentally, this argument that there "has to be a third option" is utter nonsense - the form can be designed so that no option is marked by default, and that it can only be successfully submitted when an option has been marked for every candidate. Although the ballot papers people have been putting into boxes for centuries have a default option of "no X in the box", and that doesn't seem to have been seen as causing any problems.)--Kotniski (talk) 09:40, 9 October 2011 (UTC)
 * While the form can be designed so that no option is marked by default, most browsers do not expose a way to change the selection back to 'no option' once an option has been selected. As such, we cannot support not having an explicit "third option" unless we require a support/oppose for every candidate. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 10:04, 9 October 2011 (UTC)
 * Well yes, I wasn't suggesting that - I was suggesting that there be only two options, and no third option at all, whether explicit or implicit. (I mean, yes, I do propose requiring a support/oppose for every candidate, though the two options wouldn't necessarily be called support and oppose.)--Kotniski (talk) 10:08, 9 October 2011 (UTC)
 * What would you call them? Alpha and Omega?  <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 10:16, 9 October 2011 (UTC)
 * I would strongly oppose any proposal to require people to either support or oppose every candidate, with no option for "abstain", "neutral", "no vote" or some other "non-counted" option. It's just not reasonable for a number of reasons, but I don't think I need to write a whole essay about it, because I see no chance that you would get anywhere near a majority, much less a consensus, for such a proposal.  Neutron (talk) 16:05, 9 October 2011 (UTC)
 * It would just be like any other election I've ever taken part in - you either vote for a given candidate, or you don't. You wouldn't even need to call the options anything - it would be enough just to have a single checkbox next to each candidate, just like on a normal ballot paper. I don't see anything remotely problematic about it. OR (better solution) don't have a competitive election at all, but just appoint arbs individually, much as we appoint admins individually (though arbs could be appointed for a fixed amount of time) - then the "support" and "oppose" terminology would make perfect sense (if you don't care about a particular candidate, you just don't vote in their matter).--Kotniski (talk) 18:42, 9 October 2011 (UTC)
 * I'm not sure, really, that it *is* a competitive election, at least not in the customary way. It is an interesting point, and I'm not sure that I've really considered it that way; it's kind of hard to when there are usually so many seats open. Given how badly RFA has gone in the past year or two, I'm not sure it's really the way to go, although in many ways these elections are just sort of a "super-RFA" for the candidates (with about 10 times more questions and a much longer period of scrutiny). Perhaps that also explains the limited number of candidates in the last couple of years. Risker (talk) 06:42, 11 October 2011 (UTC)
 * I mean it's "competitive" in the sense that there is a set number of seats to fill, so by voting for any candidate you're necessarily reducing the chances of every other candidate. The conclusion from all the maths that you find so "silly" (at least, my conclusion - not sure if everyone's convinced) is that this kind of competitive election doesn't sit well with a voting system where you get to score each candidate freely on a three-point scale (like we have at the moment), or on any other scale with more than two possibilities. (Because it creates an illusion that you're getting to choose among the candidates with a greater degree of fineness, whereas in fact you're just being tempted to weaken the strength of your vote.)--Kotniski (talk) 07:47, 11 October 2011 (UTC)
 * The tactical voting scenario presupposes the existence of candidates about which you are genuinely ambivalent, which do not naturally fall into any of the other voting 'bins'; the tactical vote premise is that instead of putting such candidates in the neutral bin where they 'belong' you instead put them somewhere else as a tactical measure. As the number of bins increases (ie the scale becomes more fine-grained) it becomes less likely that you have such ambivalent candidates to play with.  On a ++/+/0/-/-- scale (with candidate types A/a/B/c/C to extend our usual nomenclature), for instance, you would naturally give your A candidates and C candidates their usual place, but it does not make tactical sense to place all your ambivalent candidates also in -- because you thereby sacrifice the ability to make them more likely to win than a candidate you genuinely oppose.  You would still not rank any candidate neutral, but you would be best served by sorting your ambivalent candidates into the +/0/- bins according to what you actually think about them: they all still exercise the same limited 'tactical' effect compared to the A and C candidates, but there are also new comparisons within the middle ground which improve your happiness with the result further.  You would then tactically place the candidates that would still have ended up in the neutral bin into + or - according to your tactical voting strategy.  If you extend the scenario to a large enough number of bins, there ceases to be a 'neutral' bin at all, and you find that what you thought was "tactical voting" is in fact just ranking your candidates according to your preferences.  The "tactical voting" scenario is just an artefact of the very limited choice available on the ballot paper. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 12:27, 11 October 2011 (UTC)
 * Hmm, I don't agree. Regardless of how many bins you have, you still do well to vote "tactically" by placing each candidate in one or other of the extreme bins, if that's allowed. (It's a different story, of course, if you're told to rank candidates 1, 2, 3, etc., with only one in each bin, as is done in many proper voting systems - that would be far more satisfactory to me, though for some reason we don't do it here.) To be honest there would still be tactical voting even in my two-bin system (you might not support a candidate whom you actually rate quite highly, because you expect him to be competing for nth place with a candidate you like even more), but at least all voters could be readily expected to understand the system and use it with equal "skill" - saying "the n candidates with the most votes get elected" is hopefully intuitive enough. --Kotniski (talk) 13:12, 11 October 2011 (UTC)
 * No because now you're reducing it to the scenario of "I have some candidates who I absolutely love, and whom I like exactly equally and I don't care which of them get elected as long as as many as possible do", and the rest of the candidates who you care absolutely nothing for; that's not a realistic scenario. Yes, you very slightly increase the chances of that outcome by voting as you describe (it's probably 2.5% again, haven't done the calculations though), but you're getting tunnel vision and neglecting the fact that in an election with more bins you can exercise other voting strategies which increase your happiness in different ways.  For instance, in the SNO scenario we have seen that you can use your tactical vote in one of two ways (either to increase the likelihood of your prefered candidates winning, or to decrease the chances of your disliked candidates winning), but not both at the same time.  If you place your ambivalent candidates in + and - in a ++/+/0/-/-- election, you can exercise both.  Since in reality you also have a continuum of supports when you are not forced to map them onto a binary system, it is also legitimate to say "I support both of these candidates, but if I had to choose between them I'd pick this one"; thus you can make your vote have a positive impact in the case where two A candidates are in the borderline contest; ditto for two C candidates (and indeed two B candidates).  In short, with a great enough number of bins you can make your regular vote have a beneficial impact to you in every scenario where there is a borderline runoff, and in that case you maximise the "power" of your vote by voting exactly as you would be expected to.  As you introduce more granularity you start creating edgecases where there is a legitimate call for tactical voting.  It's not so much that there stops being tactical voting in a smooth system, rather that the strategy for "tactical voting" (as in, voting so as to maximise your happiness with the election) becomes equivalent to the strategy for 'naive voting'. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 19:50, 11 October 2011 (UTC)
 * I don't suppose we'll convince anyone of anything even if we do do the maths, but I still see nothing wrong with my algorithm for voting which I stated somewhere above for the 3-bin case (it generalizes to any case, including 2 bins and continuum bins). For each candidate X, I guesstimate the probabilities P(Y) that X will end up competing for nth place with Y (for each other candidate Y), and determine the amounts W(Y) by which I like X more than each other candidate Y (negative values if I like X less than Y). Then I sum all the weighted probabilities W(Y).P(Y) - if the total is negative I put X in the last bin, and if positive I put X in the first bin. The chance of my vote making any difference at all is already too small for me to want to split it further by putting candidates in intermediate bins.--Kotniski (talk) 18:23, 12 October 2011 (UTC)

Oh, can someone explain or remind me why we don't do ranked voting (i.e. rank candidates in order of preference, down to however many we want to express a preference for)?--Kotniski (talk) 18:27, 12 October 2011 (UTC)
 * And then do what with the rankings? There are moderately simple methods from that point, such as STV, there are very complicated (or at least, non-transparent) methods, such as the Schulze method, and there are other methods and variations as well.  Some have very different results than others, and some would have results that many editors have expressed direct opposition to, such as electing some candidates who have been selected by less than 50 percent of the voters, which is the intended result of the STV method.  To respond to (though not to answer) your question, maybe a better first question is, why should we?  And to answer that question, I think you first have to pick a specific method, and then you have to justify it.  This same discussion has taken place on the RfC page itself, and the entire discussion seems to be going around in circles.  What has not changed is that, at this point, the consensus (as that term is used on Wikipedia) is for a Support/Oppose/No Vote system in this election.  That doesn't necessarily mean I think it is the best system (and therefore I don't have to justify it, for that you would have to find the person who proposed it five years ago and ask him to justify it; I think his name might be Jimbo but I am not positive about that.)  As I have said elsewhere, I would not mind giving either cumulative voting or STV a try for a year, but I don't see any prospect of a consensus for either of them, because they both allow a "minority" to elect a corresponding (very roughly corresponding) proportion of the seats to be elected.  You want to try to get a consensus for that, go ahead.  Neutron (talk) 19:50, 12 October 2011 (UTC)
 * It's perfectly possible to run an STV ballot with a percentage cutoff, you merely run the risk of not filling all the seats in exactly the same way we currently do. The current minimum threshold could be bolted on to an STV ballot in exactly the same way: when the number of votes transfered onto the winning candidate falls below 50%, you stop electing.  Of course, that's not a very good way of running an STV election, and I think (although I haven't verified) that you're more likely to not fill seats on that basis with STV, but it's an easy way to sidestep even the question of 'minority rule'.  Although that's more a question of philosophy than of actual voting mechanics: with an STV election electing twelve candidates, your "minority" has to be roughly a twelfth of the electorate to have anything more than a semantic effect on the vote, it's debatable whether a group that size having a significant impact on the election represents anything other than democracy.  <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 23:01, 12 October 2011 (UTC)
 * I see no point at all in adopting a system which has the purpose of electing the choices of "minorities" and then requiring majority support. It sort of reminds me of Henry Ford's statement that you could have his car in any color you want, as long as it's black.  If we want to require that a winning candidate have the support of some fixed percentage of voters who choose to vote on that candidate (whether it be 50 percent, 60 percent or whatever), it seems to me that we have the best system we can have right now.  I am not necessarily convinced that we need to have that requirement -- as I said, I would support a "pure" STV or cumulative vote election as an experiment, and I don't think anybody too crazy would get elected even if they do not have majority support -- but a large majority seems to think that we should keep that requirement.  Neutron (talk) 23:53, 12 October 2011 (UTC)
 * You misunderstand "minorities". Over the past three elections, only five arbs have ever received support from more than 50% of the voters. Tony   (talk)  03:29, 13 October 2011 (UTC)
 * Actually, all of the elected arbitrators have had more than 50 percent support. Neutral "votes" don't count.  Neutron (talk) 03:40, 13 October 2011 (UTC)
 * Only because you've arbitrarily decided to call the middle option the one that "doesn't count". Tony is correct (or at least, perfectly entitled) to talk about support from more than 50% of the voters. But all in all it makes little sense to consider a "support vs. don't support" dichotomy in a competitive election like this. In this context, support will always be a relative thing. If we want to require the "support of n % of the electorate" in any meaningful way, we need to introduce a voting setup where one candidate's results don't affect those of any other - i.e. no limits on the number of Arbs who get elected.--Kotniski (talk) 07:39, 13 October 2011 (UTC)
 * Excuse me, Neutron. For each election we know the number of voters and we know the number of votes for each candidate. Of the 71 candidates in the past three elections, only five have enjoyed the support of more than half the voters: four of these were in ACE2010. Do the math. Tony   (talk)  09:44, 13 October 2011 (UTC)
 * This is assuming, of course, that to "support" a candidate means to mark the Support button for that candidate. (In fact it makes little sense to ask whether a voter "supports" a candidate in this kind of election. You can support one candidate more or less than another one, that's about all.)--Kotniski (talk) 10:52, 13 October 2011 (UTC)


 * Kotniski, I don't quite understand your previous comment. Neutron, for each election we know the number of voters and we know the number of votes for each candidate. Of the 71 candidates in the past three elections, only five have enjoyed the support of more than half the voters: four of these were in ACE2010. After Kotniski's comment, which which I agree, I say this only to dispel the apparently widespread misunderstanding of 50% benchmark and the moralising that has been going on about it. There's no way of putting a benchmark under what numerical artefact from the voting stats confirms a candidate as a good or a bad arb, or as having sufficient community support to perform the role. There are too many variables and unknowns. As Kotniski says, more or less, you decide how many vacancies there are and you rank the candidates in order of their support to fill all of the seats. That is typical in real-life elections. Tony   (talk)  09:44, 13 October 2011 (UTC)
 * Tony, I am just going by the official calculations as contained in the results tables on elections pages. At WP:ACE2010 I see that in 2010, 13 candidates received at least 50 percent support, of whom 12 were elected, and at WP:ACE2009 I see that in 2009, 14 candidates received at least 50 percent support, of whom nine were elected.  Footnote 3 in each of those tables makes clear that official percentage is calculated by S/(S+O).  I understand that you are calculating it by S/(S+N+O), and of course you can do whatever you want, but I am going with the official calculation.  As for your comments about the 50 percent threshold, the practical fact is that you have a lot of convincing to do, and it really doesn't look like your argument is succeeding in this RfC.  As I said in one of the previous sections on this page (it may have been archived), the choice is coming down to 50 percent vs. 60 percent.  As for "moralising" (using your spelling), I myself wouldn't have a problem with a zero percent threshold, but it looks like I am in a very small minority, so I have chosen to cast my vote for what seems to me to be the more reasonable of the two "viable" alternatives, which is 50 percent.  The amusing thing about this discussion is that the difference between zero and 50 has never actually mattered in an election, since the number of candidates receiving more than 50 percent has always at least equaled (and in fact, has always exceeded), the number of seats to be filled.  In other words, nobody has ever been prevented from being elected by the 50 percent threshold, although last year it came pretty close.  (A 60 percent threshhold, on the other hand, would have changed one candidate from a winner to a loser in 2009 (by one vote), and three in 2010.) But regardless of what you or I think, that is the choice for this year:  50 or 60. Neutron (talk) 15:20, 13 October 2011 (UTC)

Call for statistics
Can anyone produce statistics on previous votes using this system? For example, how many voters gave support or oppose votes but failed to vote for all candidates? Lightmouse (talk) 23:10, 6 October 2011 (UTC)
 * For 2009 and 2010 the data is quite deliberately concealed within the SecurePoll framework, and it while it would be possible to get it out again, it would need a strong consensus for the sysadmins to do so (since you could argue that it's overturning the secret ballot principle). <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 09:55, 7 October 2011 (UTC)
 * Isn't this easily extractable from the tables provided by the stewards, published on the results pages of each election? That's how I made the graphs in The Signpost write-up. Tony   (talk)  10:55, 7 October 2011 (UTC)
 * The information in that table is about votes, not voters. It tells you what percentage of votes (both by candidate and in total) were support, oppose and neutral.  You used the aggregate of those figures to make your graphs, which I discussed in the section above, "Oppose voting doesn't have the outcome intended."  But the table does not tell you what percentage of voters left at least one candidate at "neutral."  Happy-melon says that such information can be unearthed, but that essentially means that the powers-that-be would be digging out the individual votes from the system and essentially providing some information about how each person voted, although presumably not including the names.  I would prefer that that information remain hidden.  In fact, either last year or the year before, I advocated that the individual voting data be completely deleted from the system after some period of time after the election.  That never really caught on.  But I agree with Happy-melon that an argument could be made that publishing that data is overturning the secret ballot principle; in fact I would argue that, at the very least, it would be turning that principle on its side, if not all the way over.  And we really don't need that information.  We know that a significant number of people must leave at least one candidate at neutral, otherwise neutral would not be the single biggest "vote getter."  Whether the 30-something percent of votes that are neutral come from  20 percent of the voters, or 50 percent, or 80 percent, what's the difference?  Neutron (talk) 13:54, 7 October 2011 (UTC)

Thanks. Was 'Oppose voting' ever used without it being a secret vote? Lightmouse (talk) 16:56, 10 October 2011 (UTC)
 * Yes, for all Arbcom elections except the December 2004 one. Risker (talk) 17:20, 10 October 2011 (UTC)

I don't understand. The comments above said it was secret in some elections. Lightmouse (talk) 17:26, 10 October 2011 (UTC)
 * Special:Boardvote (secret voting) was used for the December 2004 election, and was approval voting only. The community found this unsatisfactory, and reverted to on-wiki voting for the 2006, 2007 and 2008 votes, with both support and oppose options. In 2009 and 2010, support/oppose options were used via Special:SecurePoll, combining secret balloting with the support/oppose option that the community desired. Risker (talk) 17:33, 10 October 2011 (UTC)

Thanks. So I presume we can get the statistics we want from 2006, 2007 and 2008 votes. Lightmouse (talk) 17:36, 10 October 2011 (UTC)
 * Category:Wikipedia_Arbitration_Committee_elections. User:ST47/ACE_2008.  Risker (talk) 18:16, 10 October 2011 (UTC)

Please no last-minute rush: timetable?
Could people decide whether MuZemike's draft timetable is it? It's several threads above. Several operatives have to be organised well in advance, for technical support and high-level scrutineering. It's embarrassing to go to these people at the last minute asking for their involvement.


 * Nominations : Sunday 00:01, 13 November – Tuesday 23:59, 22 November [10 days]
 * Fallow period : Wednesday 00:01, 23 November – Sunday 23:59, 27 November [5 days]
 * Voting : Monday 00:01, 28 November – 23:59, Sunday 11 December [14 days]
 * Scrutineering : begins Monday 00:01,12 December [possibly 5 days, on previous experience]

If anyone objects, please say so now. Tony  (talk)  11:00, 7 October 2011 (UTC)
 * As I said the last time this question was asked (in the section with the draft timetable) I think the nominating period should begin sooner, say on November 7 or perhaps earlier, as soon as the RfC is closed and the "nomination page(s)" can be prepared. The rest of the timetable looks okay.  Neutron (talk) 13:42, 7 October 2011 (UTC)

So, could we make a decision on whether the nom period should be 10 days, like last year, or 14 days? We should be keen to get the timetable sorted soon. Opinions please? Tony  (talk)  14:05, 7 October 2011 (UTC)
 * 10 days. The fallow period after the noms for further questioning and discussion has been expanded from 2 to 5 days. The election circus should not drag on for more than a month. Already, with a 10-day nom period, it's 29 days before we even get to the scrutineering period. Tony   (talk)  14:05, 7 October 2011 (UTC)
 * Longer I don't see a good reason to limit the nominating period. Nominations should start immediately after this RFC is closed and summarized, though not before Nov 1. The rest of the schedule looks fine to me. Monty  <sub style="color:#A3BFBF;">845  15:22, 7 October 2011 (UTC)
 * At least 15 days (No later than Nov. 7, to Nov. 22.) I do not understand the great amount of concern about the total amount of time taken up by the "election process."  Maybe it's because I live in a country where the "nominating process" for the highest office goes on "officially" for about seven months followed by a general election campaign of more than three months, with campaigning that starts about a year before the first primary -- if it ever really stops at all -- so maybe I am just accustomed to people being in "election mode" about half of the time.  I understand that campaigns in other countries are more compressed than that.  And I realize that ArbCom is not the presidency or the prime minister-ship, but it's still an election.  Neutron (talk) 15:37, 7 October 2011 (UTC)
 * 10 days. - seems plenty of time to me. I don't see any good reason for extended naval gazing. Off2riorob (talk) 12:16, 8 October 2011 (UTC)
 * This should be on the project page, not the talk page - You're asking a question of the community, it should be with all the other questions asked during this RfC, on the Project page. If no one objects, I'll cut and paste it over in, say, three or four hours.  S ven M anguard   Wha?  04:02, 11 October 2011 (UTC)
 * Agreed. This is more appropriate for the main RfC page. I'll throw some numbers out there, and people can decide. –MuZemike 00:40, 12 October 2011 (UTC)


 * 14 days for noms, gives those wanting to run enough time to get through that initial batch of questions, and gives those considering a run enough time to mull it over. Wizardman  Operation Big Bear 21:03, 17 October 2011 (UTC)

re, if nominations were to start immediately after the close of this RfC
It depends on how long it would take to close the RfC, but supporting to open nominations immediately after this RfC's close would in effect amount to endorsing a 21-day window for nominations, assuming the consensus for the fallow and voting periods stay the same (at 5 and 14, respectively). –MuZemike 22:23, 17 October 2011 (UTC)
 * Isn't there some work that has to be done by someone (whether "election administrators", "election coordinators" or some other official that I don't even know about) after all the issues are decided in the RfC, but before the nominations begin? For example, I assume a page or pages need to be prepared telling candidates what the eligibility requirements are to run, how many seats are being filled, what kind of balloting it is going to be, and other fun stuff, all of which is being determined in this RfC.  Plus arranging for announcements on the watchlists and wherever else. That's going to take at least a day or so, right?  And whoever is doing that (you?) is a volunteer, so you may not be able to do it right away.  That is why I suggested the nominations begin six days after the RfC closes.  I may also have said somewhere that they should begin "as soon as practicable" after the RfC closes, or words to that effect, but to me that's about the same thing -- "as soon as practicable" can be "capped" at about six days.  The other part of this is that the RfC really should close ON November 1, rather than sometime after that.  I thought there was enough support to close it early, if it had been announced a couple weeks in advance -- in fact there was only one person opposed to that -- but it didn't happen.  On a side note, I can't believe we need an RfC every year on these time periods; they are really kind of routine.  Maybe we should elect, in addition to the ArbCom, an independent "elections commission" (say, seven people) to take care of details like this so we don't have to have an annual vote on all the minor details of the election.  That doesn't help for this year, though.  Neutron (talk) 22:38, 17 October 2011 (UTC)
 * You've got some good points here, Neutron. I think one of the reasons that we keep having these discussions is that in the aftermath of each year's election, there are concerns expressed that then remain unresponded to until it comes time to consider the next election. I would very strongly encourage the community to figure out a lot of these "standard" issues months and months in advance (for example, April or May) so that there is the opportunity to have a proper discussion about alternative methods of voting, standardization of process, and so on for subsequent elections. Some of the participants here have raised some very interesting ideas that can't really be well-explored because of the need to move pretty expeditiously. Perhaps you might be willing to take this on?  Risker (talk) 23:44, 17 October 2011 (UTC)
 * Risker, for now I will regard the operative words in your last sentence as "perhaps" and "might." :)  We'll see after the election.  However, I really don't like this whole annual RfC process for the ArbCom elections.  Every year, we write what is in essence, both a "constitution" (part of one, anyway) and a detailed election law, and we do it by a somewhat chaotic process, and there really are no fixed rules for how we do it.  There are traditions and conventions for how we do it, but as you can see from both the RfC page and this talk page, there is not complete compliance with those traditions and the whole thing becomes kind of a mess.  I am not sure that doing this in April rather than October is really going to solve that problem.  What it might do is, once the RfC is closed in the spring, to allow the "close" to be compiled into a single document such as the recently adopted (I believe) arbitration policy that could then be submitted for community consensus and not go through the annual process again.  Of course, if someone wanted to have part of the "rules" reconsidered in any given year, they could start another RfC.  Whether this would work or not, I don't know.  The "50% vs. 60%" issue seems to arise every year, and the current RfC shows that there is no strong consensus either way, although I am going to have some comments for the closer when November 1 is a few days away.  The question of "Support/Oppose voting" or any of the other possible methods is also a perennial issue, though there seem to be only a few people opposed to the current method, but it seems like more because they oppose it very intensely.  Most of the rest of the issues, such as how long nominations should be and what the "question policy" should be, and even what the eligibility for voting and running should be, really should not require this perpetual discussion.  As I said above, maybe an elected body (because I know what would happen if I suggested an appointed body) should be dealing with those issues, and actually overseeing the election, arranging for the scrutineers, etc.  I would envision that many of the same people who have administered the elections in the past might be good candidates for such a "commission," along with others.  Maybe I will actually propose this after the election.  Neutron (talk) 20:59, 18 October 2011 (UTC)

We have Stewards and a Sysadmin
I have been able to secure three Stewards to help in the scrutineering process: User:Bencmq (originally from zh.wiki), User:Trijnstel (originally from nl.wiki), and User:Vituzzu (originally from it.wiki).

I have been in contact with Developer User:Tim Starling, who said he would be able to help with setting up and troubleshooting the SecurePoll interface for the election. –MuZemike 21:05, 22 October 2011 (UTC)


 * You want Tim in his capacity as a sysadmin, not a developer. Great that these balls are now rolling, though.  <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 10:19, 23 October 2011 (UTC)


 * That's what MuZemike meant. Sysadmins used to be named "developers", as I'm sure you remember, but that term was deprecated and then re-assigned to committers. Regards, AGK  [</nowikI>&bull; ] 09:38, 25 October 2011 (UTC)


 * I know, and the distinction is subtle anyway, although relevant for cases like this. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 14:23, 25 October 2011 (UTC)


 * If (as seems likely) everything is going to be very similar technically to last year, there shouldn't be too troublesome for Tim. He'll need objective criteria to plug into the algorithm that draws up the list of elegible voters, and the number of candidates. And a list of users to make election administrators, Mr.Z-man, MBisanz, Tznkai and myself all have experience in that role; holders have to be identified to the Foundation as it exposes private data. Election Admins are responsible for customising the placeholder options to refer to individual candidates. You can see the sysadmins' notes on the process to see what they need to do, and hence what they need from us. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 15:06, 25 October 2011 (UTC)
 * Whose job is it to ask the four of you whether you would be so kind as to serve in that role again this year? Or can just any random passerby editor (such as myself) do the asking?  Neutron (talk) 16:52, 26 October 2011 (UTC)
 * For fairness, I would personally prefer different people, just as I have asked for different Stewards to participate in the scrutineering process. To me, having the same people gives the impression that we run a "closed shop", which is what we don't want. There are also benefits of having different functionaries participating in the process such as increased diversity and sharing of the experience. –MuZemike 20:15, 26 October 2011 (UTC)
 * I think a bit of both would be preferable. I'd be very happy to do it again (it's not a particularly arduous task), but I'd be equally happy to see someone new in the role.  It would most likely be a Good Thing to have someone with prior experience.  Four people is also overkill; certainly one is insufficient, but there is not enough work to warrant so many people having access to the interface.  Three was a useful number of people to have to agree on things like the discrepancy announcement in 2009, but wasn't necessary in 2010.  I'd say either two or three people. <b style="color:forestgreen">Happy</b>‑<b style="color:darkorange">melon</b> 21:30, 26 October 2011 (UTC)

RFC to wrap up within 24 hours
Well, the RFC is going to wrap up shortly; I'm going to make a call to WP:AN for uninvolved admins to determine consensus for the discussions at hand and determine what we are going to have for WP:ACE2011. –MuZemike 03:26, 31 October 2011 (UTC)
 * It may take a fair amount of work to close, so you may want to warn potential closers of the need for timeliness if they take it on. Monty  <sub style="color:#A3BFBF;">845  03:44, 31 October 2011 (UTC)
 * Fair warning, I'll be summarizing the RfC in detail in the Signpost for the November 7th issue. I'll probably come to the closing admin for a quote or two.  S ven M anguard   Wha?  04:12, 31 October 2011 (UTC)
 * Note that I am not hinging all of this on one single uninvolved admin; multiple admins can close sections of the RfC in order to "divide the workload" and make it easier to manage. –MuZemike 02:05, 1 November 2011 (UTC)

My thanks in advance to whichever administrators take this on. Having been through the election process myself, I agree it is a very good thing to have the election rules and timetables spelled out clearly, in advance of when the process begins. Newyorkbrad (talk) 02:14, 1 November 2011 (UTC)

If this is still unclosed tomorrow I'll take a run at it. Skomorokh  16:04, 1 November 2011 (UTC)
 * I closed some of them, but I left others open for other editors to close. <b style="color:navy;">NW</b> ( Talk ) 03:08, 4 November 2011 (UTC)
 * I've closed a couple. I think I've participated in the remaining ones, so I'll be leaving those for others. Heimstern Läufer (talk) 04:30, 4 November 2011 (UTC)


 * Question - Does the voting period have to be either 14 or 21 days or can it be inbetween? As it stands now the numbers are;
 * 10 days = 8/87 = 9.2%
 * 14 days = 39/87 = 44.8%
 * 21 days = 31/87 = 35.6%
 * 28 days = 9/87 = 10.3%
 * So there doesn't seem to be a consensus for any one option. But it's clear that 14 or 21 days are the most popular. Could the number be something in the middle like 16 days perhaps. I was one of the unfortunate souls that missed last years elections that Sven mentioned. I'm just curious, so I though I'd ask. Thanks. - Hydroxonium (T•C• [//en.wikipedia.org/w/index.php?title=Special:ListUsers&limit=1&amp;username=Hydroxonium V] ) 07:49, 4 November 2011 (UTC)
 * If you're closing it, and you want to make it 17 days, at this point I don't think most people will complain. 17+10+5=32, or just over a month, which won't get too much complaining from anyone, I don't think.  S ven M anguard   Wha?  07:54, 4 November 2011 (UTC)
 * Thanks for the input, Sven. I participated in that discussion, so I don't feel comfortable closing it. Best regards. - Hydroxonium (T•C• [//en.wikipedia.org/w/index.php?title=Special:ListUsers&limit=1&amp;username=Hydroxonium V] ) 08:21, 4 November 2011 (UTC)
 * A majority wants 14 days or less, which is the status quo. A consensus to change is not apparent. Tony   (talk)  13:50, 4 November 2011 (UTC)
 * My understanding, based on years of reading AfDs, is that if there is no clear consensus to change the status quo, the status quo (in this case 14 days) remains in place. Is there a reason why this would be different? Risker (talk) 14:43, 4 November 2011 (UTC)
 * Yeah, on second thought, you're probably right Risker.  S ven M anguard   Wha?  16:11, 4 November 2011 (UTC)
 * Just to mention another angle: 90.8% have supported a length that is a multiple of whole weeks. I'm not sure how much should be read into that regarding the possibility of a middle ground that is not a multiple of 7, but at least some people may prefer a whole week based period so that contributors who edit on specific days of the week are treated equally. Monty  <sub style="color:#A3BFBF;">845  16:30, 4 November 2011 (UTC)

Safe to say, we have a consensus for at least 14 days, which is more than the traditional 10-day window we had in the previous elections. –MuZemike 16:21, 4 November 2011 (UTC)

Closing with numbers like 11 and 13, spread over different opinions?
The "Fundamental mechanics" section has been closed with consensus on the basis of astonishingly low numbers and diffuse opinion. I don't get it. — Preceding unsigned comment added by Tony1 (talk • contribs)
 * It seems pretty clear to me that there is no support for filling the extra seats. On the general question of whether seats should be filled, the split was 11:4 against, with a lot of good arguments on the 11 side. In the following section, all the proposed mechanics to fill seats were opposed by large margins, and the mechanic that would not fill all seats was supported by a wider margin 13:4 then the general question. Personally, I think there is enough there to establish a consensus, as this was a highly publicized RFC. Furthermore, even if I'm wrong, the most you could hope for is a no-consensus close, as there is no way to read that as consensus to fill them, which would default to the status quo of not filling the seats. Monty  <sub style="color:#A3BFBF;">845  15:01, 4 November 2011 (UTC)

Nominations now open for the 2011 Arbitration Committee Elections
Nominations are now open for candidates to run in the 2011 Arbitration Committee Elections.

The role of Arbitrator is important and demanding, there is a perennial need for new volunteers to step forward. This year, 7 arbitrators are expected to be chosen. Nominations are open to any editor in good standing over the age of 18, who is of legal age in their place of residence, and who has made at least 150 mainspace edits before November 1, 2011; candidates are not required to be administrators or to have any other special permissions, but will be required to make certain commitments and disclosures as detailed in the nomination instructions. Experienced and committed editors are urged to Consider standing.

Nominations will be accepted from today, November 12 through Monday, November 21 at 23:59 (UTC), with voting scheduled to begin on Sunday, November 27. To submit your candidacy, proceed to the candidates page and follow the instructions given. Good luck to all the candidates who decide to stand for election, Monty  <sub style="color:#A3BFBF;">845  00:01, 12 November 2011 (UTC)