User:BetterVotingAdvocacy/Big page of ideas

Some ideas:

Here are some how-to guides on using different voting methods for your own elections.

A slightly modified version of Schulze:

In general, any Condorcet method can be done with either rated or ranked ballots. That means you give every voter the ability to number all of the candidates either on a scale or from 1st to last.

When trying to find the result, start by, for each ballot, adding one vote in support of whichever candidate the voter preferred in every possible pairwise matchup. So, for example, a voter who voted A>B>C is treated as giving one vote to A>B, one to B>C, and one to A>C. This means that you'll have, for every pair of candidates, two values to store: the number of voters who prefer the first over the second, and vice versa.

Once you've done this, you'll want to find the margin of victory for every matchup. This tells you which candidate got more votes than the other in that matchup. Once you've found the margin, look for a single candidate who beats all others. If there is none, look for two candidates who beat all others (except possibly each other). Keep looking at more and more candidates until you find a beats-all group. This is the Smith set.

Now, eliminate everyone not in the Smith set. If there was only one candidate in the Smith set, they win, and if there are multiple candidates who tied (i.e. they don't beat each other), then they're tied. Otherwise, decide on your preferred measure of defeat strength (winning votes will be used for this example). Find the candidate who got the fewest votes in their favor in one of their victories against another candidate. Now, ignore this victory, treating it instead as a victory for both candidates. Now, repeat the process of finding the Smith set; you may be lucky and find that the group can be shrunken because some candidate who earlier lost to another now "wins" after ignoring their loss. Repeat this process until the winner is found.

Smith//Score:

Find the Smith set. Whoever has the most points in it wins.

Smith//Approval:

Find the Smith set. Whoever has the most approvals in it wins.

Some examples of finding the Condorcet winner faster:

Take this example: http://web.math.princeton.edu/math_alive/6/Lab1/Condorcet.html

First, it can be observed that there are 55 voters, with all but 18 of them ranking Molson 5th (last), so ISDA allows us to say that Molson can be eliminated because a (mutual) majority prefer anyone but him. So the simplified preferences are:

18 S>MB>G>K

12: K>MB>S>G

10: G>K>MB>S

9: S>G>MB>K

4: MB>K>S>G

2: MB>G>S>K

It should be noted that Samuel Adams has 27 out of 55 1st choice votes, almost a majority, so it'd be prudent to check his matchups first.

S vs. MB: 27 vs. 28, a loss for Samuel Adams. So now it might be best to check MB's matchups.

MB vs. G: 30 in the top 2 lines alone, a majority, so MB is known to win just with that information alone.

MB vs. K: 18 + 9 + 4 = 31, a majority, so MB beats K. So Meister Braum is the Condorcet winner in this example.

A quick note: It can be seen that the top two lines rank only S or K above MB, and the top two lines are a majority. So one way to quickly figure out who won would've been to compare S and K to MB, and if MB pairwise beats them, then it is guaranteed that MB won, because when ignoring S and K, a majority prefer MB over all others. This is a demonstration of how Condorcet methods' attempts to make majority rule maximally comply with IIA helps in analyzing election scenarios.

Miscellaneous

One criterion that might be good for PR methods is the "Duplicated Quotas" criterion: if a PR method elects some candidate in the single-winner case, and the ballots are "duplicated" N times, then if N+1 seats are to be filled, the duplicated winner should win. Example for Condorcet PR:

2 A>B

1 B

2 C>B

2 D>E

1 E

2 F>E

B and E are the duplicated winners, since they're Condorcet winners when ignoring all of the voters who ranked the other.

Here is an example illustrating the difficulty of creating a Condorcet multiwinner method along the lines of RRV:

34 A

35 B>C

31 C

B is the CW, so they'd win the first seat. If their supporters' ballots are reweighted by half, then C pairwise beats A 48.5 to 34 and wins, despite A being bullet voted by a Droop quota. One complicated way of possibly fixing this is to, after electing B, say that if B hadn't been in the election, C would have been the winner, and therefore both B and C voters' ballots should be reweighted by half since they both rank C above all candidates other than B (A), thus allowing A to beat C 34 to 33.

It is likely possible that the tied at the top rule can be made to work with something like Smith//Approval. Example:

4 A>B| >C

3 C| >A>B

4 C| >B>A

There is an A>B>C>A cycle, with C winning with 7 approvals. If 2 of the A-top voters swap A and B, then B becomes the CW. To avoid needing this, if these voters instead vote A=B, then they can create a pairwise tie between A and B according to the tied at the top rule. However, this still means C beats A and A ties B, so all 3 candidates are in the Smith set. To circumvent this, perhaps it's possible to ignore C's defeat against A because A is in a "special tie" with B, who beats C, which then makes A and B the only candidates in the Smith set. An alternative is to, instead of converting the matchup between A and B into a tie, completely erase or drop it. This leaves B as the Condorcet winner with a pairwise victory against C.