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So I have some voting data. There were 25 possible things to vote for, and each voter had to fill out a 10 position ballot, ie each user picked 10 from the 25 possible things for each ballot. Additionally, each spot on the ballot had a rank, 1-10. The goal was for the user to pick their top 10 from the 25 things and rank them.

The goal now is determine the 10 winners from the 25 things, and their rank in the winning 10. Some of the 25 things have the most votes for multiple ranks, ie object A has the most votes for rank 1, and the most votes for rank 2.

To determine the winning 10 and their rank, I used a weighted sum for each of the 25 things. Every vote was worth (10/rank), with a vote being a rank 1-10, with each user having 10 votes. (10 votes per ballot)

So my question is, would it be better to use a 10/rank weight? (ie, 10, 10/2, 10/3...ect) weighted system or use a (10,9,8,6....) weighted system? (linear vs log weights)? I used both methods to compute weighted sums for each of the 25 things and then I took the top 10. In both cases, the ones with the top5 weights weighted sums are the same, but the bottom 5 are different. I am not sure which method of weighting is better for this situation.

The votes are pretty evenly distributed, so my logic was to have rank 1 have the most weight relative to everything else. This is so that if a object of the 25 has a lot of votes for 1 rank, but less for 2-10, it would still rank higher.

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2 Answers

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Different voting systems often produce different results.

By only having the top 10 ranked, you are trying to find the most popular and you seem to have done that. Your first method successfully rewards being in first place disproportionately.

The real difference between your two systems is whether you think distinctions at the top are either more important than those slightly further down or as important. The other curiosity is that being a 9th choice rather than a 10th choice is much less significant a distinction on one of your systems than being an 11th choice (i.e. not voted for) rather than a 10th choice, but not on the other. There are ways round this.

The ones near the bottom of your tables are clearly not the most popular, but may also not be the most hated: something that is so bland that it is in the middle (13th) on everyone's list would get no votes.

There is no right answer, but if this matters, then it is generally good practice to decide the counting rules before people rank their choices.

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Optimal control systems engineering says "without a measure of goodness there is no such thing as best" and "if there is such a thing as best it is because there is a measure of goodness".

You ask what is better of the two options for ranking. I prefer logarithmic scale, where the first is weighted by 2^(10-1), the second by 2^(10-2), etcetera. So is the value of two "1" weight votes equal to the value of a "2" weighted vote. What is the distance between nth place and (n-1)th place? If two people gave a single option a vote with value "1" does it have the same rank as if one person gave a second option a vot with a value of "2"?

The measure of goodness is the answer to the question of "how many of those makes one of these". In the logarithmic scale though two of the (n-1)ths make up an nth, it takes 2 of the nths weight to make an (n+1)th.

In your linear scale, 10 of the 10th place values has the same weight as a single 1st place value.

Good luck.

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