[EM] Exploring Friendly Voting

[Forest Simmons]

Now let's consider another variant that we skipped right over in methods 1
through 5 ... maximizing Friendly Tops minus Hostile Tops.

A consequence of our careful pains to ensure clone independence, we can now
take advantage of the fact that the hostile Top fractional counts or
lottery probabilities must be equal to 100 percent of the ballot totals
minus the friendly totals. Therefore maximizing the difference of the
friendly Tops and hostile Topsbis the same as simply maximizing the
friendly tops... at last we're getting to a version of Friendly Voting
fully worthy of the name.

Now let's combine this idea with the Simmons/Harper interpretation of the
Implicit Approval Friendly Lottery:

In this version we count all implicit approval as full approval in the
approval count, saving the distinction between equal first and intermediate
categories for the conversion to the clone free lottery probabilities. This
policy allows the A and B factions (in our running example) to get full
majority approval when neither defects:

48 C
28 A>B
24 B>A

Approvals are 52 each for A and B with 48 for C.

To allocate lottery support we consider in turn each ballot faction:

The A>B faction approved both A and B ... but to which of these will its
final "vote" go to? To find out we compare the 28 first place votes of
"going it alone" with the (1/2)52 votes of sharing with B. S

Since 28>26 all 28 of the A>B faction votes go to A.

How about the 24 votes of the B>A faction? That faction approved both B and
A, but now that the wave function has collapsed (so to speak) who gets
those 24 votes?

To find out we compare the 24 first place votes of "going it alone" with
the 52/2 votes of sharing with A:

Since 52/2 > 24, all 24 of the B>A faction go to A, according to the
Simmons/Harper score based lottery rule.

So the respective Implicit Approval lottery values are 52, 0, and 48.

A's Friendly Voting score is ProbA+ProbB
= 52+0
B's Friendly Voting score is ProbB only since B itself is B's only friend.
C's Friendly Voting Score is ProbC only since C itself is C's only froend.

So A wins the election.

Now suppose B had defected. Then B would have gotten an approval of 52 a d
A only 28.

However in figuring out who gets A's votes (i.e. lottery probability) the
same comparison is made 52/2<28, so A still keeps all 28 of her votes.

But this time since B approved only B, candidate B gets all 24 of B's
votes in the lottery count.

That's OK because B is still friendly to A, so A gets to count B's
probability in the Friendly Voting totals.
So A gets 52.
B gets 24.
C gets 48 plus 28, since A is a friend of C ... thanks to B's defection.

C's total Friendly Vote of 76 wins the election for C.

B's defection backfired!

So this is the version of Friendly Voting that seems best to me,so far.

Note that by using the Simmons/Harper Implicit Approval lottery instead of
the ballot favorite lottery it avoids the problem of ignoring entiry
candidates with no first place representation ... but by distinguishing
first place from other approval in the approval-to-lottery conversion, it
still filters out sufficiently weak candidates ... that's the beauty of the
Simmons/Harper lottery conversion!

Now I'm blushing to have my name attached to it. It would have been better
for Harper to get credit for the whole thing as a generalization of his
idea. Perhaps we should just call it the "generalized Martin Harper
approval to lottery conversion" or something like that.

On Tue, Feb 14, 2023, 11:53 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> 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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