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

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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.