28 Sep 01:28 2014

### Re: General PR question (from Andy Jennings in 2011)

Toby Pereira <tdp201b <at> yahoo.co.uk>

2014-09-27 23:28:37 GMT

2014-09-27 23:28:37 GMT

I was thinking recently again about Andy Jennings's PR question (below) and available here http://lists.electorama.com/pipermail/election-methods-electorama.com/2011-July/093278.html, which is about the trade of between proportionality and having candidates with strong support. Warren Smith (http://lists.electorama.com/pipermail/election-methods-electorama.com/2011-July/126111.html) gave the extreme example of a 500-member parliament where two candidates each get 50% approval, and the others each get 0.2% approval. Perfect proportionality could be achieved by electing 500
candidates with 0.2% approval, but in many ways this would seem a perverse result.

But the more I think about it, the more I think there isn't a non-arbitrary solution to the problem. What's the exchange rate between proportionality and support? There isn't an obvious answer.

I proposed my own proportional approval and score system a few months ago (http://lists.electorama.com/pipermail/election-methods-electorama.com/2014-May/098049.html http://lists.electorama.com/pipermail/election-methods-electorama.com/2014-June/130772.html), and it purely bases result on proportionality, so would elect CDE in Andy's example but would also elect 500 candidates with 0.2% support in Warren's example.
However, this also assumes that every possible winning set of candidates would be looked at and the most proportional one found. In practice, the system might be used sequentially. This would force through the most popular candidate, and then each subsequent candidate would be elected to balance it proportionally. This would elect the two most popular candidates in Warren's example, but would fail to elect CDE in Andy's example. But given that there may be no non-arbitrary solution, electing sequentially may be the simplest and least arbitrary way around the problems we have. It is also a solution that would likely be forced upon us due to limits on computing power when it comes to comparing all possible sets of candidates. Necessity may force the pragmatic solution upon us.

Toby

>Forest and I were discussing PR last week and the following situation
came

>up. Suppose there are five candidates, A, B, C, D, E. A and B evenly

>divide the electorate and, in a completely orthogonal way, C, D, and E

>evenly divide the electorate. That is:

>One-sixth of the electorate approves A and C.

>One-sixth of the electorate approves A and D.

>One-sixth of the electorate approves A and E.

>One-sixth of the electorate approves B and C.

>One-sixth of the electorate approves B and D.

>One-sixth of the electorate approves B and E.

>It is obvious that the best two-winner representative body is A and B. What

>is the best three-winner representative body?

>CDE seems to be the fairest. As Forest said, it is "envy-free".

>Some methods would choose ABC, ABD, or ABE, which seem to give more total

>satisfaction.

>Is one unequivocally better than the other?

>I tend to feel that each representative should represent one-third of the

>voters, so CDE is a much better outcome. Certain methods, like STV, Monroe,

>and AT-TV (I think) can even output a list of which voters are represented

>by each candidate, which I really like.

>I also note that if there was another candidate, F, approved by everybody,

>it is probably true that ABF would be an even better committee than CDE. Is

>there a method that can choose CDE in the first case and ABF in the second

>case?

>Andy

>up. Suppose there are five candidates, A, B, C, D, E. A and B evenly

>divide the electorate and, in a completely orthogonal way, C, D, and E

>evenly divide the electorate. That is:

>One-sixth of the electorate approves A and C.

>One-sixth of the electorate approves A and D.

>One-sixth of the electorate approves A and E.

>One-sixth of the electorate approves B and C.

>One-sixth of the electorate approves B and D.

>One-sixth of the electorate approves B and E.

>It is obvious that the best two-winner representative body is A and B. What

>is the best three-winner representative body?

>CDE seems to be the fairest. As Forest said, it is "envy-free".

>Some methods would choose ABC, ABD, or ABE, which seem to give more total

>satisfaction.

>Is one unequivocally better than the other?

>I tend to feel that each representative should represent one-third of the

>voters, so CDE is a much better outcome. Certain methods, like STV, Monroe,

>and AT-TV (I think) can even output a list of which voters are represented

>by each candidate, which I really like.

>I also note that if there was another candidate, F, approved by everybody,

>it is probably true that ABF would be an even better committee than CDE. Is

>there a method that can choose CDE in the first case and ABF in the second

>case?

>Andy

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