For a more coherent explanation than just having someone tell you to look up the answer on Wikipedia (which is what I did), visit Mynnna's blog where he does a pretty good job of explaining things.
Just for the record, this will be my last post on the CSM 8 election until the results are announced at Fanfest. Also for the record, for the first round of the election I endorsed Mike Azariah with all three accounts. I plan on voting for Mike, and only for Mike, in the second round, leaving the other 13 slots blank. If Mike does not qualify for the next round, I won't be voting.
What follows, since I don't believe in sending my mistakes down the memory hole, was my first attempt to understand and explain the new voting system. Caution: after reading Mynnna's explanation, I now know how really bad it is.
I have to admit something. I considered boycotting the CSM8 election. Not just not voting, but not writing anything on the blog about it either. Why? The new election system. Because in many ways this election is now a lottery instead of an election.
So what changed from last year to this? For the election next month CCP will utilize the Wright single vote transfer system. Let's just say I'm not thrilled about its use. Click through the link and the Wikipedia has an example of someone who would win in a "first past the post" election losing to someone who was no one's first choice. But let me give an Eve example that could possibly occur if Goons slip up. Yes, I know the Goons have a master plan so this won't happen, but play along.
Goonswarm has three candidates running this year: mynna, Kaleb Rysode, and Unforgiven Storm. Let's say that turnout this year is a little better than last year. To make the math easy, we'll use 60,000 votes cast. Also assume that Goons and company can gather 10,000 votes. Now assume Goons make a strategic mistake. In their voter guide, everyone is instructed to vote mynnna first. That makes sense since I think the goal is to get mynnna one of the two guaranteed slots to Iceland for the summits. Good so far. But then the choice for second and third position are left to the average Goon's choice, resulting in 5,000 second choice votes for Kaleb Rysode and 5,000 second choice votes for Unforgiven Storm. Then the voters switch their preferences and make either Kaleb or Unforgiven their third choice.
Votes after Round 1 of selection:
mynnna - 10,000
Kaleb Rysode - 0
Unforgiven Storm - 0
Here is where the election gets tricky. Now figure out the quota to see how many votes mynna needs to be elected. The formula is:
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| The Droop Quota formula |
Remember, I am not complaining about who is going to win or who the system favors. I'm just referring to the lottery type nature of who winds up on CSM 8. Of course, in this scenario one of the Goonswarm candidates will wind up with only 1,998 votes. If the Goons want three seats then they may team up with a couple of other blocs to gain the additional votes, but that is another post.
To tell the truth, if I haven't supported Mike Azariah the past couple of years I probably would chuck the whole CSM election and just wait until after Fanfest to start paying attention again. But since I'm voting for Mike I'll pay attention enough to see if he can sneak through the null sec blocs and get a seat.
I suggest you reread the wikipedia article on Wright-STV. There is no lottery involved whatsoever.
ReplyDeleteIf Mynna gets 10,000 votes and only needs 6,000 to be elected, then each of the ballots gets split into two parts, a 60% part use to elect him, and a 40% part used to elect whoever is the second choice on that ballot.
Ballots can be subdivided multiple times until there is no longer any surviving candidates on a ballot. The net result is that for any voter, the maximum amount of his voting power is used to support the candidates he prefers, in order of preference, and with maximum possible efficiency.
I have written a Wright-STV election script to cross-check CCP's software. You can see the output of a sample run at the link below, using data from last year's election with the ballots expanded in plausible ways. It should make the process more clear.
https://www.dropbox.com/s/zhvksyrkf6716gr/bigballot.txt
In Wright, the election is actually reset and run several times, eliminating the weakest candidate each time, until enough candidates pass the threshold.
The tl/dr: you maximize your voting power by listing every candidate you think is reasonable, in your honest order of preference.
One minor correction -- there is one instance where chance comes into play. If at the end of a round, the two least preferred candidates have exactly the same number of votes, then the choice of who gets excluded is random.
DeleteHowever, given that vote totals are usually fractional because of transfers, this is most likely going to happen in the early rounds, and thus will almost never affect the final results.
I don't think that is how it works.
ReplyDeleteAs far as I know, excess votes are split according to the proportion of the next vote down. So, if 6000 votes are released, and the other two split 50/50, they will get 3000 of those transfers each.
The votes just spill over when the candidates reach the vote count to be accepted (200) and no more votes for that candidate are counted, even if they where number 1. The first 14 that reach 200 votes are the CSM council.
ReplyDeleteThe rest of the positions are voted for internally by those who won seats.
This is not correct. The pre-election phase (which I think is largely a waste of time) requires each candidate to collect at least 200 endorsement votes (one endorsement per account) in order to qualify for the actual election, which is run using Wright-STV and ballots where voters can specify up to 14 candidates.
DeleteOk, I got a bit ahead of myself. I stand corrected
DeletePoint is that a STV is not really a lottery vote.
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DeleteSTV >> first past the post.
ReplyDeleteYou might want to read up on the mechanics, no way this Goonswarm scenario would play out like you fear.
Even if the extra ballots *were* released randomly, with a large number of ballots and a halfway decent random number generator the chances of having a different result are vanishingly small.
ReplyDeleteIn your example, second preferences were split 50:50, so the chances of getting each candidate is 50% each time you pick a ballot. So it's like tossing a coin - and if I tossed a coin 6000 times, I'd expect it to be heads about 3000 times. Even though *technically* they could all come up tails, it's mind-bogglingly unlikely.
In your example, the chances of being more than even 100 votes off a 50:50 split is less than 1%, and it gets more unlikely the further you go from the average.