Dilemma about TEA
Say two candidates have enough 3s to meet the quota but not enough 4s to meet the quota. What should act as a tiebreaker? The original version suggests number of 3s but I would argue that having a quota of 3s is already 'enough' and that the candidate with more 4s is closer to having a quota of 4s so that candidate should be preferred.
@matija I'll put a link to the method for reference. I'm not sure either tiebreaker is ideal. Number of 3s doesn't seem to make sense anyway - surely it should be 3s, 4s, and 5s together in that case. I think this seems logical anyway because the approval threshold has moved down to 3, so it's just the most approved candidate at that point. Or maybe the total score from the ballots that count as approving it - so you add up the 3s, 4s and 5s.
I don't know really. It's not a method I've spent too much time thinking about. It's a quota method, and I think they're crude and outmoded.
One option would be 3s/2+4s+5s