Also, most party systems tend to clump together in ideology making the 'paradoxical' choices rare. For instance in Canada we have two leftist main parties and one conservative. Ranking the conservatives in the middle would be virtually unheard of.
Also, most party systems tend to clump together in ideology making the 'paradoxical' choices rare. For instance in Canada we have two leftist main parties and one conservative. Ranking the conservatives in the middle would be virtually unheard of.
I thought the same thing about party rankings in the US. If we had ranked voting, and the Green and Libertarian Parties ran serious campaigns, it seems that the vast majority would rank G/D and L/R together. I suppose if we had many parties, though, we could see a 'paradoxical' choice.
In this case your rankings for the candidates constitute a Markov matrix where you specify that your vote should "flow" from these candidates over to those candidates. For example someone voting for Alice over Bob over Carol might have a Markov matrix which states their vote is:
1.0 1.0 0.5
0.0 0.0 0.5
0.0 0.0 0.0
saying "All of the A votes I make should ideally stay for A, all of my B votes I would like to flow to A, all of my C votes I would like to flow to B or A." Part of the reason that I haven't figured this out yet is that I don't know how one should pick these numbers, exactly, to be robust against Arrow's problems (like adding another candidate not affecting the relative preferences of two existing ones).Anyway, based on the election, we average everyone's Markov matrices together! Then we start from a state where each candidate has an even share of the vote and everyone collectively determines the flow of the votes in successive cycles. This basically selects out the eigenvector of the Markov matrix with the highest eigenvalue, which represents some probability distribution among the candidates. Now to resolve the pesky problem of cyclic preferences, we use a fair random number generator to choose from the resulting probability distribution. Thus everyone has an even say in who they want to win, someone gets chosen at the end, and if someone really is strongly dominated by someone else then they will generally get eliminated by the flow of votes from the other to them.
[1] While I have you excited about randomness I am also a fan of choosing a leader at random out of the entire population, educating them during a transition period, then handing them the reins. If you happen to disbelieve in the Buddhist metaphysics then an example is the tulku system for obtaining leaders, where they follow somewhat-ambiguous instructions from the previous holder of the seat to seek out a child born at a specific place at a specific time who is believed to be the specific reincarnation of the previous holder of the seat. If you don't believe that metaphysics is right then the search essentially chooses a child at random to be the leader. Seems to usually work pretty well for them either way. There's another example from Muslim history where the leader of the dynasty did not go to one's child because one did not have children; instead it was transferred from eunuch slave-soldier to eunuch slave-solder.
You did mention giving the person training. The problem with that is this will result in a fully self-perpetuating system that finds ways to not give power to people who disagree with it.
It still sounds better than the current system, so I'm all for it.
> If you don't believe that metaphysics is right then the search essentially chooses a child at random to be the leader. Seems to usually work pretty well for them either way.
A cynical take on this might be that those educating the next ruler get a chance to shape him/her to their specific desires.
I'd be interesting in how you would explain this system to an average voter. I didn't really follow it myself, but I've at least heard all the words before (except probably tulku).
Also, interesting point about averages and ranked voting. By the values, Strawberry would have won out, however as stated in the video, a majority does prefer chocolate to strawberry—so how would that be dealt with?
I look forward to the discussions around our [Canada's] new electoral/voting system this December.
I am not sure which part of the video you were looking at where Strawberry would have won, but it seemed by my quick averaging, that the 'right' choice always wins under simple averaging (unless there is a tie which is fine in Canada, not so much in the US).
I was referring to where In ranked voting where the paradox was cyclic preferences, Paul mentions that strawberry would win on averages, but yet the majority of voters would have preferred chocolate over strawberry.
Here's the most basic problem: the video that we're looking at made an implicit assumption when it proposed ranked voting: it said that "here are peoples' preferences!" and then it copied those preferences over to the ballots. Given the way that they're doing ranked voting, this equivocation makes sense, of course you're going to rank in your actual preference order. But with the Borda count this assumption is totally wrong and a lot of the confusion from first-past-the-post systems reoccurs: there is no reason to vote your actual preferences. Consider the 2016 US presidential election as done via Borda count, if we assume peoples' present polling is honest. Think about someone who supports Jill Stein; their preferences are typically Jill > Hillary > Gary > Donald. Will Jill lead their ticket? Given the very lackluster showing of Jill, they will probably narrow down the race mentally to "Hillary or Donald" and turn in the ballot "Hillary > Jill > Gary > Donald," to try to avoid a Trump presidency. It's the same spoiler effect as we had before.
It's not just an isolated problem like that suggests. Suppose an election where we have three candidates, Alice, Bob, and Carol. There are nine voters with the preferences,
4xABC, Alice > Bob > Carol
1xACB, Alice > Carol > Bob
2xBAC, Bob > Alice > Carol
2xCBA, Carol > Bob > Alice.
Alice is the majority's first choice and only 2/9 of the peoples' last choice, so you'd figure she'd be a shoe-in to win the election. Indeed, she'd win a ranked-runoff election and first-past-the-post election, while the pairwise election has the following ballot counts: A beats B 5 - 4
A beats C 7 - 2
B beats C 6 - 3
The pairwise election therefore also decides Alice > Bob > Carol. But while these other systems all have Alice winning for relatively easy-to-understand reasons, if Alice's supporters turn in honest ballots in the Borda election, then under a Borda election, Alice LOSES!This is because the remaining four voters see it as down to an Alice-vs-Bob election and want Bob to win, so they turn in ballots saying BCA. The resulting total of points (3 for first place, 2 for second, 1 for third) is:
(ballots: 4x ABC, 1x ACB, 4x BCA)
Alice: 4 * 3 + 1 * 3 + 4 * 1 = 19 points
Bob: 4 * 2 + 1 * 1 + 4 * 3 = 21 points
Carol: 4 * 1 + 1 * 2 + 4 * 2 = 14 points
Alice however is a savvy politician and convinces her own supporters to rally against Bob to get Carol to look much better than Bob, so that they hand in ACB ballots. This is a measured tactic to turn it into an Alice-vs-Carol race which gets the BAC voters back on her side, since now Bob is no longer a strong contender and they vote Alice. It gets complicated! (ballots: 5x ACB, 2x ABC, 2x CBA)
Alice: 5 * 3 + 2 * 3 + 2 * 1 = 23 points
Bob: 5 * 1 + 2 * 2 + 2 * 2 = 13 points
Carol: 5 * 2 + 2 * 1 + 2 * 3 = 18 points
Except Carol has a trick up her sleeve. She sees that Alice is doing this and so runs with her friends Dwayne and Erica, who agree with her on everything but are slightly less preferred by the voters in general. Since they are so similar our original setup is 4xABCDE, 1xACDEB, 2xBACDE, 2xCDEBA. Now if the ballots are cast with the smear campaign against Bob in full force you have: (ballots: 5x ACDEB, 2x ABCDE, 2x CDEBA)
Alice: 5 * 5 + 2 * 5 + 2 * 1 = 37 points
Bob: 5 * 1 + 2 * 4 + 2 * 2 = 17 points
Carol: 5 * 4 + 2 * 3 + 2 * 5 = 42 points
Dwayne: 5 * 3 + 2 * 2 + 2 * 4 = 27 points
Erica: 5 * 2 + 2 * 1 + 2 * 3 = 18 points
Suddenly Carol wins! Or, if Alice smears Carol almost as much as she's smeared Bob, then Dwayne just takes Carol's place and Dwayne wins; remember, they're largely identical! (You might wonder whether the effect of adding Dwayne and Erica to the voting has meant that Alice doesn't need to smear Bob anymore and whether Carol becomes her enemy #1; the answer is that if we just substitute C -> CDE in the above "Bob wins" tally then Bob gets 34, Carol gets 32, and Alice gets 29, at which point we need to re-evaluate the strategic voting because you KNOW that after that mock-poll comes out, the news will classify Alice as a 3rd-place finisher and her own majority voting block might rebalance around the assumption that it's largely a Bob-vs-Carol race.)With Nanson it seems that the second to last scenarios (Carol wins) is cleansed of the strategic voting leading to the 'correct' Alice win.