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Round 1 vote count

• Bartlet: 1

• Dwayne Elizondo Mountain Dew Herbert Camacho: 2

• Curtis: 0

• McCord: 1

• Kirkman: 1

• Orlean: 0

• Simpson: 3

• Underwood: 0

• Whitmore: 0

• Meyer: 0

• Garrison: 0

Curtis, Orlean, Underwood, Whitmore, Meyer, and Garrison eliminated.

@TheAllMemeingEye

Round 2 vote count

• Bartlet: 1

• Dwayne Elizondo Mountain Dew Herbert Camacho: 2

• McCord: 1

• Kirkman: 1

• Simpson: 3

Bartlet, McCord, and Kirkman eliminated.

@TheAllMemeingEye

Round 3 vote count

• Dwayne Elizondo Mountain Dew Herbert Camacho: 2

• Simpson: 3 + 2 = 5

Dwayne Elizondo Mountain Dew Herbert Camacho eliminated, the winner is Simpson

@TheAllMemeingEye excellent!

• 4 of the 7 voters preferred Bartlet over Simpson and she still won.

• Two voters could have gotten better results by betraying their favorites and voting strategically.

• Different tie breaking rules, one arguably better than the other, produce different winners.

• Removing Camacho from the election altogether shows that either he was a spoiler, or Simpson would have won with 3 out of 8 votes - depending on the tie-breaker.

Hey, at least I can't find violations of monotonicity, participation or consistency (using the current tie-breaker). Great success!

@BrunoParga Thank you for your service lol 🫡

@TheAllMemeingEye I'd love to hear from die-hard supporters of this system why they support it rather than, you know, something that doesn't have these problems, like list-based PR.

No, Arrow's Theorem does not apply to multi-winner elections. Yes, there is the theorem that shows divisor methods may fail quota; that is very hard to happen with Webster's, I don't think it has ever happened in any real-life election because it requires an absolute landslide, and the winner losing a seat in that scenario hardly matters - unlike ranked choice failing to elect the Condorcet winner.

@BrunoParga lmao like I said I supported it purely because it was the only single winner alternative to fptp I had heard of, and at the time it seemed to solve the problems

Needless to say I now get why it doesn't work as planned either

Unfortunately I don't understand much of the terminology in the second paragraph, is Wikipedia the main resource you used to learn or are there other more efficient ways?

@TheAllMemeingEye oh, to be clear I didn't mean to ask you specifically, I understand that you hadn't heard much about the flaws ☺️ my question was more if anyone else who voted wants to explain why it is good and proper that Bartlet lost.

@TheAllMemeingEye and I appreciate your curiosity about other systems! Now that the election is over I kinda want to de-emphasize that in my life a bit, but I expect GPTs to do a pretty good job at deciphering my ranty comment ☺️ I think it'd save you time over Wikipedia

@BrunoParga Thanks :)

Nice, we have favorite betrayal! The users who ranked McCord and Kirkman first in their ballots got a worse result (pending the final hours of voting and tie-breakers) than if they had betrayed their favorite and ranked Bartlet first. This is exactly what happens under FPTP, where you can't vote for your preferred third-party candidate or else you risk getting the major party you hate winning.

I also need to confirm tomorrow, or whenever the hangover from the real election wears off, but I believe in either tie-breaker scenario (Simpson or Bartlet winning) I might be able to show a break of consistency. Think like Maine picking its four EVs: one candidate wins both congressional districts, another wins the state. With the same fucking ballots. Makes no fucking sense.

My vote reinforces Bartlet as the Condorcet winner, while also reducing the chance he actually wins.

First, honorary mentions that are edge cases I wasn't sure I could include in my ballot (they wouldn't necessarily be at the top or at any particular position): Arnold Schwarzenegger in Demolition Man - the movie mentions an amendment was passed to allow him to run. AFAIK he's only mentioned, not shown. Then also Sarah Susan Eckert, the fifth face on Mt. Rushmore in Star Trek V: The Final Frontier.

Without further ado, my vote:

1. Dwayne Elizondo Mountain Dew Herbert Camacho

2. Whitmore

3. McCord

4. Bartlet

5. Garrison

6. Kirkman

7. Orlean

8. Curtis

9. Meyer

10. Underwood

DEMDHC, Whitmore, McCord, Simpson and Kirkman are the only candidates whom 2 of the 7 ballots (this count includes mine) prefer over Bartlet; he's clearly the Condorcet winner, no doubt about that, since all other candidates have at most 1 voter preferring them over old Jed.

However, he only has one first vote; Simpson and Camacho have two each. Kirkman and McCord are also at one. Everyone else has zero first preferences and is eliminated.

In principle, one should only eliminate the candidate with the second-fewest first votes if they couldn't pass the third-fewest even by receiving all of the votes of the fewest. This condition does not hold here, so a principled count should eliminate Bartlet, Kirkman and McCord one at a time.

If Bartlet is eliminated, there is still a tie between Kirkman and McCord; regardless of which is eliminated next, the other one goes out third, and Simpson defeats Camacho 4 votes to 2.

However, if either Kirkman or McCord is eliminated before Bartlet is, the other is as well and their votes go to Jed; the count is Bartlet 3, Simpson 2, Camacho 2. None of the ballots for Simpson or Camacho go to the other; Bartlet defeats Camacho 5:2 and Simpson 4:2.

However, one might choose to eliminate the three candidates who have just one vote all at once, like we did with the 0-vote guys. In that scenario, Simpson beats Camacho 4:2.

In short: this system sucks balls and any resolution is problematic.

@BrunoParga Thanks for the excellent deconstruction of my system haha

@traders Yo within the next 3h someone needs to cast whichever vote is needed to undo Bruno's master plan haha

My vote:

1. Simpson

2. Orlean

3. Curtis

4. Underwood

5. Meyer

6. Whitmore

7. McCord

8. Kirkman

9. Garrison

10. Bartlet

@SentientTree Bartlet is still the condorcet winner, 4:3 against most of them and wider margins against a few.

@SentientTree in the scenario where the three 1-voters get eliminated at the same time, you just made it 5-2 for Simpson against Camacho. Same if Bartlet is eliminated first. The only difference your vote makes is if McCord or Kirkman lose first; then Simpson and Bartlet are tied. One vote – mine – is completely disregarded in its preference for Bartlet over Simpson.

You literally disenfranchised me, bro.

Edit: don't do complicated graph theory problems in your head when you're supposed to have gone to bed forty minutes ago. If McCord or Kirkman is eliminated before Bartlet is, we get to a point where you have S3, B3, DEMDHC2, and my vote breaks the tie for Bartlet 4:3.

1. Dwayne Elizondo Mountain Dew Herbert Camacho

2. Mr Garrison

3. Frank Underwood

@FunkyTikiGod BRUH

@FunkyTikiGod I love how everyone knows you have to refer to Dwayne Elizondo Mountain Dew Herbert Camacho by his extremely long full name.

I'm really surprised that Janie Orlean has been at the bottom of the list the whole time. Even Camacho and Selena Meyer were voted up but nobody wants to vote for the most realistic depiction of a modern politician on the list?

@becauseyoudo lmao this is about what you want not what's likely irl under fptp lol

1: Tom Kirkman
2: Jed Bartlet
3: Lisa Simpson
4: Thomas Whitmore
5: Andre Curtis

@Kraalnaxx you should have betrayed your favorite. The way the election is right now, Simpson wins*, but if you had ranked Bartlet first he'd win. Ranked choice supporters are lying when they tell people that voting according to their honest preferences can never hurt their interests.

*depending on how the market creator decides to break ties