Voting for the location of the CIS-2107 tournament.

Yesterday, Dirty and I discussed the voting algorithm. So, I'll share what I came up with. And let's not start another argument – we've already had three in different threads this week, which is more than we had in the past six months on the entire forum. 1. We've slightly modified the Scorpa system. Here's how it works: - Didn't participate in any CIS tournaments – 1 point. - Participated in 1 CIS tournament – 2 points. - Participated in 2 CIS tournaments – 3 points. - Participated in 3 or more CIS tournaments – 4 points. (I would also suggest awarding an extra point to those who attended all 7 CIS tournaments of the New Era – there are only a couple of them: Cobra, Napalm, Preds, and Bayazet?) 2. Dirty insists on the possibility of multi-voting. However, if a person wants to vote for one city, they have, say, 4 votes. But if the same person votes for two cities, they then have 4 + 4 votes, which creates an imbalance. Therefore, I propose that each person has a set number of points. If a person votes for one city, they spend all their points on it. If they vote for two, their points are divided equally between them. 3. The criterion for winning a bid. We have three candidates. And if the votes are distributed roughly equally (34%-33%-33%), then 34% becomes the winning criterion. In my opinion, this isn't ideal because the argument will likely continue after such a result (since 34% is, to put it mildly, a little less than 66%). Therefore, in this case, I don't see anything better than the Presidential election system. The winning criterion should be more than 50% of all votes, either in the first round or the second (where the candidates with the most votes in the first round proceed). The only weak point here is the possibility of a tie. In this case, the votes of the CIS tournament champions from previous years should be decisive. If you like, a quick poll can be conducted among them in controversial situations. There are exactly 7 of them (Stinger has two votes).