[EM] The "Fresh Egg" winner - beyond Condorcet's pairs

[Tom Ruen]

Dear EMers,

I've been on this list in the past, and just rejoined in hopes of some feed
back on an idea I've developed lately.

Has anyone else considered looking at subset elections like Condoret, but
looking beyond pairwise preferences?

I've looked at this before, but now I have some new insights.

My primary motivation has been the offensive case of a candidate winning
Condorcet pairs while being in DEEP last place in the plurality count (Like
the horrible case: AC=49%, BC=48%, CB=3%).

I can accept such a winner on grounds of a "fair" compromise, but it doesn't
follow the election tradition that expects a candidate to have a dedicated
core support to win. My idea is to look at competition among all subsets of
candidates, including the full set to pick a winner.

To define my method, I look at the Condorcet winner.

Here's a possible definition of Condorcet's Winner:
Condorcet's Winner is the a candidate that can avoid last place in all
pairwise subset elections. (There are (N^2-N)/2 = N*(N-1)/2 such subsets
among N candidates.)

Now here's my extended definition which I call a "Fresh Egg" winner:
A "Fresh Egg" winner is a candidate that can avoid last-place in all subset
elections of two or more candidates, including the full set of candidates.

I call this the "Fresh Egg" winner by the childhood running game "Last one
home is a rotten egg." By that declaration a candidate that is NEVER last is
never a rotten egg, and therefore a fresh one. :)

Note: There are 2^N-N-1 such subset elections among N candidates. 2^N
subsets allows each candidate to be included or not - a binary switch (2^N
subsets). N subsets only have one candidate, and one subset has zero
candidates, so these noncompetitive cases are removed.

I should point out that just like pairwise Condorcet winner, if a Fresh Egg
(FE) winner exists, there can be only one. (Proof by contradiction - if two
such candidates avoid last place in all subsets, they must both avoid it in
the pairwise subset including both of them. This is impossible. At best they
could tie and therefore both be in last place.)

I would also like to point out that whenever a FE winner exists, the IRV
winner will agree since such a FE winner can never be in last place in any
subset in the IRV elimination process. Therefore when a Fresh Egg Winner
exists, IRV and Condorcet will agree on a winner. I think this is pretty
cool. The FE criterion is a parent or ancestor method to both methods!

This also makes me see the idea of a progression of criteria of what a
majority could mean among more than two candidates:
1. A Top Majority (TM) - If a candidate exists that is never below first
place among any subset of competitors.
2. A Fresh Egg Majority (FEM) - If a candidate exists that is always above
last place along all subset elections with competitors.
3. A Condorcet Majority (CM) - If a candidate exists that is always above
last place among all pairwise subset elections with competitors.

These criteria are progressively less strict downward. A TM winner will
always be a FEM winner and a CM winner. A FEM winner will always be a CM
winner (And a IRV winner).

My goal has mainly been to differentiate between elections where there is a
strong Condorcet winner, versus a weak Condorcet winner. A candidate that
does well in the full-set election AND pair-sets is clearly stronger than
one that only thrives in pair competitions. I dislike that on the surface
Condorcet calls both such winners as equals, even if I were willing to agree
on the declared winners.

Lastly, I found I can offer one more simple step for a more complete
election method. Again, I start from Condorcet.

One way of counting a winner in Condorcet might be:
1. For every pair election, give a point to the candidate in last place. (If
there are more than one candidate in last place, divide that point equally
among them all.)
2. Give victory to the candidate with the least points. (If one candidate
has zero points, there is a unique winner. Otherwise there may be a tie and
a winner remains undefined among the tie.)

This process applies identically to a full-set count in FEM:
1. For every subset election of two or more candidates, give a point to the
candidate in last place. (If there are more than one candidate in last
place, divide that point equally among them all.)
2. Give victory to the candidate with the least points. (If one candidate
has zero points, there is a unique winner. Otherwise there may be a tie and
a winner remains undefined among the tie.)

By these rules, what everyone was calling Condorcet process was really just
a small part of the larger Fresh Egg process! Condorcet unnecessarily
neglected opinions among 3 or more choices.

Example election:
I have data from a fun election I made two years ago at a summer camp -
asking people their favorite seasons with rank preference ballots. There
were 76 voters. Here are the results on the all-sets count:

************************************************
There are 11 subset elections (above) among 4 candidates. 6 are Condorcet
pairs. 4 are triples. 1 is the full set.

Exhausted ballots are counted for completeness as NOTA (None-of-the-above)

Pair subset elections: 6 cases
a1) Spring=39.0 Summer=36.0 [NOTA=1.0]
a2) Spring=41.0 Fall=33.0 [NOTA=2.0]
a3) Summer=43.0 Fall=32.0 [NOTA=1.0]
a4) Spring=57.0 Winter=18.0 [NOTA=1.0]
a5) Summer=59.0 Winter=16.0 [NOTA=1.0]
a6) Fall=56.0 Winter=19.0 [NOTA=1.0]

*** This can be counted as a place count: ***
Name\Place  :     1     2
-------------------------
Spring      :   3.0   0.0
Summer      :   2.0   1.0
Fall        :   1.0   2.0
Winter      :   0.0   3.0

*******************
Triple subset elections: 4 cases
b1) Summer=29.0 Fall=23.0 Spring=23.0 [NOTA=1.0]
b2) Spring=34.0 Summer=32.0 Winter=10.0
b3) Spring=37.0 Fall=25.0 Winter=13.0 [NOTA=1.0]
b4) Summer=41.0 Fall=29.0 Winter=6.0

*** This can be counted as a place count ***
Name\Place  :     1     2
-------------------------
Spring      :   3.0   0.0
Summer      :   2.0   1.0
Fall        :   1.0   2.0
Winter      :   0.0   3.0

*******************
Quadruple (full) election: 1 case
c1) Summer=27.0 Spring=23.0 Fall=20.0 Winter=6.0

*** This can be counted as a place count ***
Name\Place  :     1     2     3     4
-------------------------------------
Summer      :   1.0   0.0   0.0   0.0
Spring      :   0.0   1.0   0.0   0.0
Fall        :   0.0   0.0   1.0   0.0
Winter      :   0.0   0.0   0.0   1.0
*******************

Now looking at the above tables, you can look at the combined last-place
counts:
Spring: 0.5
Summer: 1.0
Fall  : 2.5
Winter: 7.0
-----------
Total  11.0