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this post was submitted on 11 Aug 2026
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Mildly Infuriating
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Please forgive any ignorance on my part, but would it be required that we'd know what the 2nd (and 3rd and 4th and so on) preferences of those who voted for E in the first choice?
If so, and if such could be done, what if of those who voted for E on their first choice, in their second preference, half of them (i.e. half of 10% being 5%) voted for C, a third of them voted for B, and a sixth of them voted for A;
meanwhile two-thirds of those who voted in their first preference for A, half of those who voted in their first preference for B, and a third of those who voted in their first preference for C, voted for D in their 2nd preference? —i.e. (⅔ x 26%) + (½ x 25%) + (⅓ x 24%) = 16⅔% + 12½% + 8% = 36⅚%.
Yes. Please look up "How ranked choice voting works" instead of asking randos on social media to explain it to you. The information is readily available.
wp:Ranked voting
(my bold)
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Glad you were able to find the answers on your own for the potential options.
Seeing as Lemmy is not in charge of implementing it I didn't see the point of having Randos argue over which specific version of ranked choice voting to use, as I'm sure we can all agree that any version is better than what is currently being used.
I didn't find "the answers" just some info about it.
I'm not so sure of that. Please be more specific in the type ranked choice voting system you'd favor.
I suppose I rather half a parliament (or similar) be decided by proportional representation, and in the others, perhaps it's a run-off if the winner of a plurality doesn't get a majority.
The one you suggested I suppose is okay too. 😁 🙂
also
Here's a brief explanation of how preferential voting works in Australia: https://www.chickennation.com/voting/
Thank you. 🙂
1st count
Bamboot got 39.
Floundra got 35.
Aggy-Waggy got 20.
Gort got 6.
But why did Bamboot get 4 of Gort's votes while Aggy-Waggy got only 1, even though the latter got over half the number of the former?
Shouldn't it be 3+2+1 instead of 4+1+1?
and why in the 3rd count did Floundra get 16 of Aggy-Waggy votes, while Bamboot got only 5 or less than 1/3rd?
Presumably in a run-off, it would be Bamboot versus Floundra;
while in proportional representation, the Complete Scumbag Party, Discreet Scumbag Party, the [Gort Butterson Party], and the Nice Party, in a parliament of 100 might get 39, 35, 6, and 20 seats respectively, with probably the Nice Party being in a coalition.
Even if it was 1/2 PR and 1/2 FPP, each system electing 100 candidates (hence a parliament of 200), and in the FPP vote, the Complete Scumbag Party won 51 of the FFP seats, the Discreet Scumbad Party won 38 seats (i.e. less than 75% of the former's number of seats, or conversely the former got over less than 35% the latter's number of seats), the Nice Party won only 8 (2/5ths of what they won in the PR count) seats, and Gort won 3 seats it would still be:
We could have 3 possible coalitions:
Because Bamboot isn't getting Gort's votes. Now that Gort is eliminated, all of the voters that voted Gort #1 now have their #2 vote counted as if it's their #1 vote.
This keeps happening down the choices until someone has a majority. People that voted Floundra #1 don't end up having their #2 vote counted because their candidate wasn't eliminated in the end. This is why it's not possible to ~~throw your vote away~~ have your vote taken away - if your favourite candidate is eliminated, your vote goes to your second favourite, then your third favourite, etc until someone has the majority.
Not sure what it looks like in the US, but Australian ballot slips have all the candidates and you number them 1, 2, 3, 4, etc.