[EM] rainbow lottery

[Simmons, Forest]

Jobst and All,
 
So far I don't have any takers on this topic, but I am going to forge ahead a little, and then do some examples that show how well it works at providing fairness in some sticky cases that have been discussed from time to time on this list.
 
In order to compare with most of the single winner methods commonly discussed on this list we need a version that doesn't rely on approval cutoffs.
 
Here it is:
 
1. Without loss in generality assume that the alternatives are colors.
 
2.  Ballots are ordinal.
 
3.  For each color find the maximum pairwise opposition, i.e. the largest "votes against it" in any of its head-to-head contests with another color.
 
4.  Make a rainbow where the colors are in order of max pairwise opposition, and the width of each band is the number of ballots (or fractions thereof) on which that color was ranked first. [If  k colors are ranked equal first on some ballot, then that ballot contributes only 1/k to the width of each of those colors on the rainbow.]
 
5.  Slice the rainbow in half  (parallel to the edges of the outside bands), and throw away the bad half of the rainbow, i.e. the half with the colors that have larger max pairwise opposition.  Typically there will be one color that is partially excised and partly saved.  This color is considered "remaining" along with the other remaining colors.
 
6.  Use random ballot to pick a winner from among the remaining colors.
 
Here are some instructive RGB (red, green, blue) examples:
 
9 R>B>G
6 G>B>R
5 B>R>G
 
The respective max PO's are 9,  11, and 14  for B, R, and G.
 
The respective faction sizes are  5, 9, and 6, so the rainbow is
 
BBBBBRRRRR|RRRRGGGGGG,
 
where the divider  is the half way mark for the cut.
 
The good side of this rainbow is the left side,
 
BBBBBRRRRR,
 
which has only R and B represented, so the winner is chosen by random ballot between R and B.
 
This yields odds of  11 to 9, or in other words, 55 to 45,  in favor of  B over R.
 
Condorcet would have given 100 percent of the probability to B.
 
But the best Condorcet methods would give 100 percent of the probability to R in the following modification obtained from the first example by burial of B on all of the R faction ballots, which therefore would give that faction an incentive to carry out that insincere order reversal:
 
9 R>G>B
6 G>B>R
5 B>R>G
 
But let's see how the rainbow method compares:
 
The respective max PO's for  R, G, and B are 11, 14, and 15.
 
The respective faction sizes are 9, 6, and 5, so the rainbow is
 
RRRRRRRRRG|GGGGGBBBBB,
 
where the divider is the halfway mark.
 
Since the good half has both R and G represented, the winner is decided by random ballot between R and G.
 
This yields odds of  14 to 6, i.e.  70 to 30,  in favor of  R over G.
 
How likely is it that the 9 R faction would prefer this rainbow result (based on insincere ballots) to the rainbow result based on sincere ballots?
 
Remember the sincere order is R>B>G
 
Sincere case lottery probabilities are 45% for R and 55% for B.
 
Insincere case lottery probabilities are  70% for R and 30% for G.
 
To see which is better let's assign hypothetical utilities for the R faction voters:  100% for R,  x for B, and 0 for G.
 
The expectation based on sincere ballots is better when
 
45 +55x  > 70, that is when  x > 5/11 .   
 
In particular,  when B is halfway between G and R, i.e. when x = 50%, which is greater than 5/11,  then sincere strategy yields greater expectation.
 
[The original rainbow method, which takes into account approval cutoffs, is even better in this regard.]
 
 
Let's finish with another well known thorn in the side of both Condorcet and Approval:
 
49 R
24 G
27 B>G
 
The respective max PO's for R, G, and B are 51, 49, and 49.
 
Since blue and green are tied for first place (i.e. smallest max PO) we blend them together to get  A (for aqua) in the rainbow getting
 
 A...A|AR...R,
 
where the divider separates the two halves of the rainbow.
 
When we throw away the bad half   A*R^49,  we are left with  A^50, which has B and G, but no R in it.
 
So random ballot will pick G or B, with odds in favor of B at 27 to 24, which are the same odds as 9 to 8.
 
Note that if  the 27 B>G faction truncates to 27 B, then R's max PO reduces to 27, which makes the Red end of the rainbow  A*R^49 the good end.  Since all three colors are represented in this half, the random ballot lottery yields the respective probabilities of
 
49%, 24%, 27% for R, G, and B.
 
This is worse from the perspective of the B faction, so it has no incentive to truncate.  Similarly, if the 24 G faction really prefers B to R, it is to their advantage to express that preference, thereby reinforcing the exclusion of R from the lottery without compromising G's odds relative to B.
 
This rainbow method can be thought of as residing halfway between MMPO (MinMaxPairwiseOpposition) and Random Ballot.  [To get the two extremes just slide the slice marker away from the halfway point toward either extreme of the color spectrum.]
 
Since Random Ballot satisfies the strong FBC, and MMPO satisfies ordinary weak FBC,  I wouldn't be surprised if the rainbow version satisfies the FBC, or even the strong FBC.
 
As I mentioned before it does satisfy Clone Independence and Monotonicity.  Also it is clear that this version satisfies Independence from Pareto Dominated Alternatives, since any Pareto Dominated Alternative will have a max PO of 100 percent, which cannot be exceeded, which puts it in last place (or tied for last place) with a band width of zero (because of no top rank on any ballot) on the bad end of the rainbow.
 
I hope you like it  :-)
 
Forest
 
 
 
 
 
 
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