User:R.H./Condorcet, IRV, travel through time and dimensions

Condorcet Voting is like interdimensional travel. (like in the TV show "Sliders"). You know that Plurality Voting with only 2 candidates shows clear results that are easy to grasp so you travel to many different dimensions that are identical except that all candidates but 2 (a different pair in each dimension) are somehow missing (nudge, nudge).

Instant Runoff is like time travel (like in the "Back to the Future" movies) where those voters that vote for the least popular candidate according to Plurality are the first to get back in time to the point when voting is. However their favourite disappeared (wink, wink). That happens repeatedly until only 2 candidates are left.

Everybody knows that interdimensional travel is less dangerous than travel backwards in time, especially if you meet yourself there. This is why I prefer some Condorcet methods to Instant Runoff (and this is also why "Sliders" is more boring than "Back to the Future").

New (?) single winner method without proper name yet: Even if you prefer time travel, would it not be more fun to allow those voters that voted for the Plurality winner to be the first time travellers so they can get rid of the candidate they despise most (measured for example by the most last place ranks in the traveller group) and repeat that until one is left?

Take a look at PWVGBITAKTCTDMMBMLPRITTGRV (Plurality winner voters go back in time and kill the candidate they despise most, measured by most last place ranks in the traveller group - repeat voting): 1. situation 39 A>B>C 35 C>B>A 26 B>C>A

The A-Team goes back in time and kills C. B (Condorcet-Winner) wins.

1.b) reversal which shows that Plurality fails Reversal Symmetry. 39 C>B>A 35 A>B>C 26 A>C>B The followers of A go back in time and kill C. A (Condorcet Winner) wins. Hmm, now I ask myself whether the travelling group should only shoot one or if it is also okay to pepper a bit the less-but-still despised in preparation of rounds to come. I will come to a conclusion after I am sure how the version without peppering works.

2. situation: a Condorcet Cycle 39 A>B>C 35 C>A>B 26 B>C>A The A followers go back in time and kill C. A wins.

2. b) reversal which shows Instant Runoff fails Reversal Symmetry. 39 C>B>A 35 B>A>C 26 A>C>B The C supportes go back in time and kill A. C wins.

3. situation: IRV failing Mono-Raise failure example 39 A>B>C 35 B>C>A 26 C>A>B

The A group goes back in time and kills C. A wins (also wins in IRV).

3. b) part 2 of IRV mono-raise failure example 49 A>B>C 25 B>C>A 26 C>A>B

A wins again (but in IRV C wins which shows a Mono-Raise failure).

Looks good so far. Given that this ends with a duel, the Condorcet Loser Criterion is satisfied. The First Place Majority Criterion is also satisfied. What else?

Let's try this junk: 10% FarRight>Right>Centrist>Left>FarLeft 10% Right>FarRight>Centrist>Left>FarLeft 15% Right>Centrist>FarRight>Left>FarLeft 16% Centrist>Right>Left>FarRight>FarLeft 15% Centrist>Left>Right>FarLeft>FarRight 13% Left>Centrist>FarLeft>Right>FarRight 11% Left>FarLeft>Centrist>Right>FarRight 10% FarLeft>Left>Centrist>Right>FarRight Round 1: Centrist voters travel back in time and kill FarLeft. Result: 10% FarRight>Right>Centrist>Left 10% Right>FarRight>Centrist>Left 15% Right>Centrist>FarRight>Left 16% Centrist>Right>Left>FarRight 15% Centrist>Left>Right>FarRight 13% Left>Centrist>Right>FarRight 11% Left>Centrist>Right>FarRight 10% Left>Centrist>Right>FarRight Round 2: Left voters travel back in time don't kill the centrist (his crew killed far left after all) but FarRight. 10% Right>Centrist>Left 10% Right>Centrist>Left 15% Right>Centrist>Left 16% Centrist>Right>Left 15% Centrist>Left>Right 13% Left>Centrist>Right 11% Left>Centrist>Right 10% Left>Centrist>Right Round 3: Now voters for Right have the upper hand and kill Left 10% Right>Centrist 10% Right>Centrist 15% Right>Centrist 16% Centrist>Right 15% Centrist>Right 13% Centrist>Right 11% Centrist>Right 10% Centrist>Right Round 4: Centrist wins the duel. 65% Centrist 35% Right

40 B>C>A 25 C>A>B 35 A>B>C B kills A and B wins. Reverse and A kills B and A wins. 40 BUVWCXYZA 25 UVWCABXYZ 35 AUVWBXYZC Round 1: B kills A: 40 BUVWCXYZ 25 UVWCBXYZ 35 UVWBXYZC Round 2: U kills C: 40 BUVWXYZ 25 UVWBXYZ 35 UVWBXYZ Round 3: U kills Z: 40 BUVWXY 25 UVWBXY 35 UVWBXY Round 4: U kills Y: 40 BUVWX 25 UVWBX 35 UVWBX Round 5: U kills X: 40 BUVW 25 UVWB 35 UVWB Round 6: U kills B: 40 UVW 25 UVW 35 UVW Round 7: U kills W: 40 UV 25 UV  35 UV Round 8: U wins 100% FLAWLESS VICTORY. Reverse: 40 AZYXCWVUB 25 ZYXBACWVU 35 CZYXBWVUA Round 1: A kills B: 40 AZYXCWVU 25 ZYXACWVU 35 CZYXWVUA Round 2: A kills U: 40 AZYXCWV 25 ZYXACWV 35 CZYXWVA Round 3: A kills V: 40 AZYXCW 25 ZYXACW 35 CZYXWA Round 4: A kills W: 40 AZYXC 25 ZYXAC 35 CZYXA Round 5: A kills C: 40 AZYX 25 ZYXA 35 ZYXA Round 6: Z kills A: 40 ZYX 25 ZYX 35 ZYX Round 7: Z kills X and then wins against Y. 100%

From Center for Range Voting puzzle page (http://math.temple.edu/~wds/crv/PuzzlePage.html)

15 people 5 	A>D>F>E>C>B 4 	B>E>F>D>C>A 3 	C>B>E>D>F>A 2 	D>C>F>A>E>B 1 	E>C>F>D>B>A A wins plurality, B wins plurality plus top-2 runoff, C wins sequential runoff (IRV), D wins Borda count, E wins Condorcet, F wins Approval (assuming the top-three candidates are "approved" by each voter).

Testing with my method 5 	A>D>F>E>C>B 4 	B>E>F>D>C>A 3 	C>B>E>D>F>A 2 	D>C>F>A>E>B 1 	E>C>F>D>B>A First round: The A voters go back in time and kill B: 5 	A>D>F>E>C 4 	E>F>D>C>A 3 	C>E>D>F>A 2 	D>C>F>A>E 1 	E>C>F>D>A Second round: Draw between A and E. I hadn't considered this. Possible  solutions: a) The more experienced travellers go back again. If the groups in the draw travelled as often, whoever travelled first. b) The less experiended... c) random d) something else I'll take a) without any reasoning whatsoever! So Third round: The A voters go back in time and kill C: 5 	A>D>F>E 4 	E>F>D>A 3 	E>D>F>A 2 	D>F>A>E 1 	E>F>D>A Fourth round: The E voters who now have the majority of first place  votes go back in time and kill A: 5 	D>F>E 4 	E>F>D 3 	E>D>F 2 	D>F>E 1 	E>F>D Fifth round: The E voters go back in time and kill D: 5 	F>E 4 	E>F 3 	E>F 2 	F>E 1 	E>F sixth round: E wins (Condorcet Winner!)

100 voters, 4 candidates 49 A>B>C>D 26 C>B>D>A 25 D>B>C>A First round: The A voters go back in time and kill D: 49 A>B>C 26 C>B>A 25 B>C>A Second round: The A voters go back in time and kill C: 49 A>B 26 B>A 25 B>A Third round: B wins (Condorcet Winner!)

Voting example from Wikipedia (http://en.wikipedia.org/wiki/Condorcet%27s_method): 100 voters decide between Memphis, Nashville, Knoxville, Chattanooga 42 M>N>C>K 26 N>C>K>M 15 C>K>N>M 17 K>C>N>M Round 1: The M crowd goes back in time and kills K: 42 M>N>C 26 N>C>M 15 C>N>M 17 C>N>M Round 2: M kills C: 42 M>N 26 N>M 15 N>M 17 N>M Round 3: N (Condorcet Winner!) wins.

I hope this method is independent of Tideman Clones and satisfies Mutual Majority. I need to somehow solve ties.

The method does not satisfy the Condorcet Criterion :-/ I think there are no problems with electing the Condorcet Winner if there are strict complete rankings and only 3 options. But if the Condorcet Winner does not have the most first preferences and the voters of the option with most first preferences have the Condorcet Winner in last place the Condorcet Winner is the first to be excluded. Example: a) 100 voters, 7 options 44 A>B>C>D>E>F>G 41 B>A>C>D>E>F>G   5 C>B>A>D>E>F>G  4 D>B>A>C>E>F>G  3 E>B>A>C>D>F>G  2 F>B>A>C>D>E>G  1 G>B>A>C>D>E>F

Exclusion order: G, F, E (after that B has most first prefs), D, C, A. B (Condorcet Winner wins.) Those that put A in first place change their rankings so B is in last place: b) 44 A>C>D>E>F>G>B 41 B>A>C>D>E>F>G  5 C>B>A>D>E>F>G  4 D>B>A>C>E>F>G  3 E>B>A>C>D>F>G  2 F>B>A>C>D>E>G  1 G>B>A>C>D>E>F

B is still the Condorcet Winner but now the exclusion order is B, G, F, E, D, C and A wins. So the method does not satisfy the Condorcet Criterion and has Burying problems.

Condorcet Criterion complying variation: How about PWVGBITACTCTDMMBMLPRITTGTADWIAVVRV (Plurality winner voters go back in time and challenge the candidate they despise most -- measured by most last place ranks in the traveller group -- to a duel which involves all voters' votes - repeat voting)? So this is like the method above, with the additional step that a group can only kill a candidate if that candidate is pairwise beaten by their fav. Hmm, it is a depressing thought to go back in time without killing anybody, so I guess I would give the group a good chance by ignoring the candidates in their ranks that their fav does not pairwise beat. Only in a deadlock when no more candidates can get thrown out the plurality winner of the remaining group (who is in that case also the Condorcet Loser of the remaining group) gets thrown out.

Summary so far:
 * 1) Plurality winner voters vote anti-Plurality (i.e. most last place ranks in that group) for a candidate to challenge.
 * 2) If the challenged candidate loses the duel (=pairwise comparison by all voters) with the Plurality winner throw him out of the game, recalculate rankings and go back to step 1. If he wins take another anti-Plurality vote by the Plurality winner voters which ignores the former challenged candidate(s).
 * 3) If no more candidates can be thrown out, throw out the Plurality winner of the remaining set of candidates unless the remaining set is =1.

I bet this always selects from the Smith Set.