[EM] Using weights to compensate multiple votes
Philippe Errembault
phil.errembault at skynet.be
Fri Aug 20 15:23:45 PDT 2004
Hi all,
Sorry about not reading ALL your mails. It could happen that the question I'm going to ask has already be processed...
... But since english is not my mother tongue and since you write much ;-) and since there is only 24h in a day and since blah blah blah ;-) ...
So...
I'd like to have your opinion about an idea concerning the impact of voting for multiple people :
In any approval-like system, where you can give your voice to more than one candidate, you have a problem since giving voices to multiple candidate raises the weight of vote. Especially if a group of voters give their voice to the same specific group of candidates, the will drasticly raise the chances of all their vote to match with other voters who didn't arrange with them.
So, why not evaluate the amount of information (let's call it [I]) contained in a vote, and weight this vote by 1/[I] !?
first idea : let's give the value 1 to each approval to a candidate and 0 for others. the vote will then be a set of 1 and zero, for wich we can compute the average, variance and standart deviation. Let's use standard deviation for [I] and each vote by 1/[I].
this reduces impact of explicit arrangement, but not for implicit arrangements : let's call implicit arrangement, the fact that a well known has more chances than others to accumulate random votes, so people voting for them will have more chances to match other voters.
second idea : instead of using each approval and disapproval, let's count them as many times as the count of votes received by the candidate to compute the standard deviation.
I have attached a small excel simulation but I'm not sure it will be accepted on the mailing list.
procedure :
for each voter [i], for each candidate [j], the voter chooses [Vij] to approve ([Vij] = 1) or not ([Vij] = 0) the candidate.
then, for one candidate, let's call
[Sj], the sum of [Vij] for all voters [i].
and [S] is the sum of [Sj] for all candidates [j].
then, for one voter [i], let's call
[AVGi] the average of [Vij], each [Vij] counting [Sj] times. ( [AVGi] = SUM( [Sj]*[Vij] ) / [S] )
[VARi] the variance of [Vij], each [Vij] counting [Sj] times. ( [VARi] = SUM( [Sj] * ([Vij]-[AVGi])^2 ) / [S] )
( or [VARi] = SUM( [Sj]*[Vij]^2 ) / [S] - [AVGi]^2 )
(where ^2 means squared)
[SDVi] the standard deviation of [Vij]. ([SDVi] = Square Root([VARi]);
so each votes are weighted with 1/[SDVi] and each candidate receives SUM([Vij]/[SDVi])
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