[EM] Exploring Friendly Voting

[Forest Simmons]

I have used the fractional Top and Bottom counts as a basis for clone
independent Friendly voting variants.

A much better solution (imho) is to replace these fractional Top a d Bottom
counts with my adaptation of the Martin Harper lotteries to the Implicit
Approval and Bottom Disapproval counts:

Let's start with the Bottom Disapproval counts. In our example ...
48 C
28 A>B
24 B

The respective bottom disapprovals for A, B, and C are
48+24=72, 48, and 52.

Since A's disapproval is greatest, according to Martin Harper's rule all of
the ballots that disapprove of A contribute to A's  bottom disapproval
lottery probability of 72 percent. The other 28 percent of the probability
goes to C.

My version of the Implicit Approval Lottery probabilities are a little
trickier because in my version only Top or Equal Top get full approval ...
while rankings between Too and Bottom count for half approval.

On this basis the respective approval counts are 28, 24+(1/2)28=38, and 48.

C gets 48 percent of the Simmons/Harper lottery probability. The next
highest approval value is 38 for B. So acording to the Harper rule, all of
the B ballots contribute to B's probability.

But what about the A ballots ... they supported both the 38 percent
approval of B and the 28 percent approval of A.

Well their approval of B was only fractional, so we compare (1/2)38=19 with
full 28. Since 28>19, all of the A ballots go to A's lottery probability of
28 percent.

Harper's original purpose was to appease the IRV vote transfer mindset by
specifying which approval ballots transferred to which candidates in an
approval election. I extended that to the general Range context, and in
particular the implicit approval context where equal top gets 100 percent
approval, ranked between bottom a d too 50 percent approval, and bottom
gets zero.

Any way ... the idea works great in this context ... each ballot ends up
supporting exactly one candidate in each of the two lotteries ... the
implicit approval lottery and the disapproval lottery.

Back to our example:

The approval lottery probabilities are
28, 24, and 48, while disapproval probabilities are 72, 0, and 28.

The method one scores for a candidate are the disapproval probabilities
minus approval probabilities of the corresponding hostile candidates:
S(A)=d(C)-a(C)=28-48=-20
S(B)=dA-aA=72-28=44
S(C)=dB-aB=0-24=-24

The method 2 scores S' are the friendly approval probabilities minus the
friendly disapproval probabilities:
S'(A)=aA+aB-dA-dB=28+24-72-0=-20
S'(B)=aB+aC-dB-dC=24+48-0-28=44
S'(C)=aC+aA-dC-dA=48+28-28-72=-24

So B wins again.

The method 5 scores (to be minimized) are the friendly disapproval
probabilities plus the hostile disapproval probabilities...
A-->dA+dB+aC=72+0+24=96
B-->dB+dC+aA=0+28+28=56
C-->dC+dA+aB=28+72+24=124

B minimizes the score.


On Mon, Feb 13, 2023, 4:23 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> Let's add the Method 1&2 calculations for the same example that we did the
> Method 3,4&5 tallies in our previous message (copied below).
>
> Here's a summary of the data from that example:
>
> 48 C
> 28 A>B
> 24 B
>
> For the different sums of differences S and S' we need the following Too
> and Bottom (fractional) counts:
> Top(A)=28, Top(B)=24, Top(C)=48
> Bot(A)=(48+24)/2=36,
> Bot(B)=48/2=24,
> Bot(C)=(24/2)+28=40
>
> Method 2 elects the candidate Y that maximizes S'(Y), the Sum (over the
> candidates X friendly to Y) of Top(X)-Bot(X):
> S'(A)=Too(A)-Bot(A)+ Top(B)-Bot(B)
> = 28-36+24-24=-8
> S'(B)=Top(B)-Bot(B)+Top(C)-Bot(C)
> =24-24+48-40=8
> S'(C)=TopC-BotC+TooA-BotA
> =48-40+28-36=0
> The winner is argmax S'(Y) = B
>
> Method 1 elects the candidate Y that maximizes S(Y), the Sum over the
> hostile candidates X of Bot(X)-Top(X):
> S(A)=Bot(C)-Too(C)=40-48=-8
> S(B)=Bot(A)-Top(A)=36-28=8
> S(C)=Bot(B)-Top(B)=24-24=0
> The winner is argmax S(Y) = B
>
> So B is the winner of all five methods.
>
> Why is this so?
>
> Well, the fractional counts are the same counts we would get by symmetric
> completion of the ballots ... and we know that all five methods are
> equivalent given complete ballots.
>
> -Forest
>
>
> On Mon, Feb 13, 2023, 7:58 AM Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
>
>> Continuing in the same context for a little while we introduce Methods 4
>> & 5, respectively as maximizing and minimizing the mined and subtrahend,
>> respectively, of the difference arising in method 3 ...
>>
>> Method 4: Maximize the minuend, which is the sum of the Bottom totals of
>> the hostile candidates and the Top totals of the friendly candidates.
>>
>> Method 5: Minimize the subtrahend, which is the sum of the Top totals of
>> the hostile (vexing) candidates and the Bottom totals of the Friendly
>> candidates.
>>
>> Note the numerical and psychological advantages of Methods 4&5 due to all
>> terms being positive: ... numerical stability and linguistic congruence in
>> the sense tally=sum, as opposed to difference, in the public mind.
>>
>> That said, let's specialize to our Example 1 context again for Method 4:
>> Here there is only one candidate (C) hostile to candidate A, to whom the
>> other two candidates A and B are friendly ... so the tally for candidate A
>> is ...
>> Top(A)+Top(B)+Bottom (C), which is just fpA+fpB+fpA in terms of faction
>> sizes.
>>
>> For comparison purposes only, let's subtract zero in the form of n-n from
>> this tally, where n is the total number of ballots fpA+fpB+fox. After the
>> dust settles we see that A's tally is numerically equal to ...
>> fpA-fpC+n, the same score for A  as in methods 1 and 2, up to a constant
>> n.
>>
>> Similarly, Method 5 yields (for A) a tally of fpC-fpA+n  to be minimized.
>>
>> So in the special context of Example 1, all five methods agree on the
>> winner.
>>
>> But once we go beyond Example 1, it seems to me that Method 5 is better
>> strategically than Method 4 ... because Method 4, with two terms of foA in
>> its tally, gives too much control to the A faction ... the likely culprit
>> of the burial that created the ABCA beat cycle.
>>
>> So I  propose tentatively Method 5:
>>
>> Elect the candidate that minimizes the total tally TT given by: friendly
>> bottom totals plus hostile top totals.
>>
>> Just one item of business to finish the five method specifications: when
>> rankings are not complete Top and Bottom counts must be done fractionally
>> to preserve clone independence.
>>
>> That's it ... so now let's do an example of Method 5 with incomplete
>> rankings:
>>
>> 48 C
>> 28 A>B
>> 24 B
>>
>> For the different TT tallies we need the following Too and Bottom
>> (fractional) counts:
>> Top(A)=28, Top(B)=24, Top(C)=48
>> Bot(A)=(48+24)/2=36,
>> Bot(B)=48/2=24,
>> Bot(C)=(24/2)+28=40
>>
>> TT(A)=Bot(A)+Bot(B)+Top(C)=36+24+48
>> =108
>> TT(B)= Bot(B)+Bot(C)+Top(A)=24+40+28
>> =92
>> TT(C)=Bot(C)+Bot(A)+Top(B)=40+36+24=100
>>
>> It appears that argmin TT(Y) is B.
>>
>> Let's compare this with the total tallies TT'(Y) for method 4:
>> TT'(A)=Top(A)+Top(B)+Bot(C)
>> =28+24+40=92
>> TT'(B)=Top(B)+Top(C)+Bot(A)
>> =24+48+36=108
>> TT'(C)=Top(C)+Too(A)+Bot(B)
>> =48+28+24=100
>> Argmax TT'(Y)=B, the same as the method 5 winner.
>>
>> The method 3 totals are the respective differences TT'-TT of the methods
>> 4 and 5 ... 92-108=-16, 108-92=16, and 100-100=0. So B wins by method 3.
>>
>> Somebody should check my work on these methods and also calculate the
>> Method 1 and 2 winners.
>>
>> Methods 3, 4, and 5 do not appear to be Chicken Strategy resistant.
>>
>> I hope that these sample calculations provide enough clarity for those
>> who want to explore further.
>>
>> -Forest
>>
>> On Sun, Feb 12, 2023, 9:42 PM Forest Simmons <forest.simmons21 at gmail.com>
>> wrote:
>>
>>> For simplicity we'll stick with complete rankings for now ... with three
>>> methods that are equivalent in that context, though not in general ... like
>>> MinMaxPaiwiseOpposition and MaxMinPairwiseSupport are equivalent in this
>>> context.
>>>
>>> We say candidate X is Friendly to Y if X doesn't defeat Y pairwise ...
>>> otherwise X Vexes  Y.
>>>
>>> Also X has Top or Bottom Status on ballot B according to whether no
>>> candidate outranks it or it outranks no candidate on ballot  B.
>>>
>>> Method 1:
>>>
>>> Elect the candidate argmax S(Y) defined as Sum{Bot(X)-Top(X)| X vexes Y}
>>> where Bot(X) and Top(X), respectively, are the number of ballots on which X
>>> has Top or Bottom status.
>>>
>>> Example 1.
>>>
>>> fpA: A>B>C
>>> foB: B>C>A
>>> fpC: C>A>B
>>>
>>> C is the only candidate that vexes A, so S(A) is merely Bot(C)-Too(C),
>>> which equals fpA-fpC.
>>>
>>> Method 2:
>>>
>>> Elect argmax S'(Y) defined as
>>> Sum{Top(X)-Bot(X)| X is friendly to Y }.
>>>
>>> Method 2 applied to the same ballot profile from example 1 goes as
>>> follows:
>>>
>>> S'(A)=(Top(A)-Bot(A))+(Too(B)-Bot(B)), which equals fpA-fpB+fpB-fpC,
>>> which simplifies to fpA-fpC, the same exact expression given by method 1.
>>>
>>> Method 3. Elect argmax (S(X)+S'(X)).
>>>
>>> Note that S(A)+S'(A) can (by rearrangement of terms) be written as
>>> Bot(C)+Top(A)+Top(B) minus
>>> [Top(C) +Bot(A)+Bot(B)].
>>>
>>> The three positive terms cinsist in the Top counts of friendly
>>> candidates and the Botoom count of the unfriendly candidate.
>>>
>>> The subtrahend consists of the bracketed terms which are the Bottom
>>> counts of the friendly candidates and the Top count of the unfriendly
>>> candidate, as it should be: we want our friends to have high top counts and
>>> low bottom counts ... and vice-versa for those who vex us.
>>>
>>> Does this heuristic make sense as a foundation for a Friendly Voting
>>> Method?
>>>
>>> -Forest
>>>
>>
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