[EM] Monotonic Burial Resistant Method

[Forest Simmons]

On Thu, Jul 6, 2023, 10:46 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> On 7/6/23 19:33, Forest Simmons wrote:
>
> >     This produces the following example:
> >              2: A>B>C
> >              1: B>C>A (honest is B>A>C)
> >              2: C>A>B
> >
> >     It's an A>B>C>A cycle: A has 2 first preferences out of 5, hence >1/3
> >     first preferences. The probabilities of a ballot having a last
> >     preference for
> >              A is 1/5
> >              B is 2/5
> >              C is 3/5
> >
> > Since these events are mutually exclusive, the probabilities should not
> > surpass 100 percent.
>
> You're right. I had originally written an example with more voters and I
> forgot to update the numerators. The right numbers are:
>
>         A: 1/5
>         B: 2/5
>         C: 2/5
>
> which can be seen by counting last preference votes.
>
> So the penalties should be:
>
> A: (C's probability): 2/5
> B: (A's probability): 1/5
> C: (B's probability): 2/5
>
> B>A=C
>
> Does that seem right?
>

Yes.

>
> -km
>

Let's compare with the MMPO analysis:

: 2 A>B
1 B>C
2 C>A

A>B 4 to 1, B>C 3 to 2, C>A 3 to 2

MPO's are 3, 4, and 3, respectively for A, B,and C.

So A and C are tied for MinMPO. The tie is broken in favor of A, because
its second highest MPO is only 1.

B has the fewest losing votes against this MMPO candidate, so the sincere
final is between A and C ... which is won by the sincere CW, namely A.

The new method is based on the idea of finding the candidate X that needs
the least supplematation to form a pair that covers the candidate set.

If X is A, then Y=C is not yet covered because C beats A. The cost of
supplementing with C is C's antifavorite count, which is 2.

If X=B, we need a candidate (like A) that beats B to complete our covering.
The cost of adding A is only 1, its antifavorite count.

This is clearly the least costly ... so X=B is the candidate most cheaply
built up to a covering.  Therefore B (the burier in this example) is the
method winner.

The burier is rewarded (frown!)

Now suppose that sincere had been

2 A>C
1 B>C
2 C>A

Then  the MMPO method would have detected & restored C as the sincere CW in
the A vs C sincere final ... while our new method would fave still elected
B ... punishing the burier A by electing its sincere antifavorite.

The equality in the B>A=C cycle confounds things a little with two equally
weak links B>C and C>A.

How about the monotonicity and Landau Efficiency of the new method?

BTW in the formula for S(X) the sum of over all Y defeating X of RBAFP(Y)
could be replaced by
Max over all Y beating X of  BottomCount(Y).

This makes it easier to explain and compute ... no need of
lottery/probability concepts, etc.

-fws

>
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