[EM] The family of "regular" probabilistic (stochastic) electoral systems

[David Catchpole]

Can I call 'em the Metameucil systems, can I? Can I?

Okay- "Regularity" is the name used earlier by Albert Langer (Craig might
recognise the name ;) ) to describe the probabilistic analogue of IIA. It
goes like this-

The addition (removal) of a candidate does not, for any other candidate, 
increase (decrease) the probability of that other candidate winning.

Now, say you know how the system works for a two-candidate contest- for
simplicity, assuming no dependence on the number of voters- that is, you
know the function relating the ratio of probabilities of those two
candidates winning to the ratio of voters ranking one over the other to...
oh, blah, you get the idea.

Such a function g(x)=1/2+f(x-1/2) where f is an odd function f(x)=-f(-x) .
f is also a monotonic increasing function. Obviously this has to be the
case- would you vote if it just meant decreasing your favourite's chances?

Now, let's make it a three candidate election. It's easy enough to see
that in order for regularity to be satisfied, the three-candidate
probabilities of winning for each candidate must be less than or equal to
all the two-candidate probabilities of winning for each candidate.

For instance-

A vs. B vs. C- p3(A) p3(B) p3(C)

p3(A)+p3(B)+p3(C)=1

A vs. B- pAB(A)=g([A>B]) pAB(B)=g([B>A])

A vs. C- pAC(A)=g([A>C]) pAC(C)=g([C>A])

B vs. C- pBC(B)=g([B>C]) pBC(C)=g([C>B])

pAB(A)+pAB(B)=1 etc. etc.

Now, for regularity,	p3(A)<=pAB(A), <=pAC(A)
			p3(B)<=pAB(B), <=pBC(B)
			p3(C)<=pAC(C), <=pBC(C)

All this boils down to saying that in order for regularity to be satisfied
for the two-three candidate transitions,

1<=pAB(A)+pBC(B)+pAC(C)<=2.

Now, remember the Saari cube? It's a graphical representation of the
possible vote configurations in a three-candidate contest. Along the axes
are the ratios of votes. So, we can talk about g being a transformation
from the Saari cube to... Well, let's see what

1<=x+y+z<=2,
0<=x,y,z<=1

looks like- It's the Saari cube! Geez, that makes it easier. So the
question is- what restraints are there on this transformation? Well, this
neoSaari cube has "neighbourhoods," regions from which the transformation
cannot stray. For instance, where x>1/2, y>1/2, z>1/2 and x>y, y>z,
g(x)>1/2, g(y)>1/2, g(z)>1/2 and g(x)>g(y), g(y)>g(z). The problem is to
split the Saari cube into these neighbourhoods and consider the extremes
it has for each dimension. These extremes look like this-

1/2<->1
1/2<->3/4
1/2<->2/3
0<->1/2
1/4<->1/2
1/3<->1/2

Now, because you can't stray from the neighbourhoods, the implications
are-

g(1/2)=1/2
1/2<=g(3/4)<=3/4
1/2<=g(2/3)<=2/3 and the rest just follows...

So, the restraints I've found already for Metameucil systems are above.
In a two candidate contest, 3/4 means at most 3/4 and 2/3 means at most
2/3.

Now, the optimal system it seems is one where, for a two-candidate
contest,

g(x)=	{0 if x<1/4
	{1/4 if 1/4<=x<1/3
	{1/3 if 1/3<=x<1/2
	{1/2 if x=1/2
	{2/3 if 1/2<x<=2/3
	{3/4 if 2/3<x<=3/4
	{1 if x>1/4

Now, this is where I get stuck. Can anyone prove / disprove that a system
exists which has the properties just above and which is regular for _all_
number-candidate contests?

If anyone needs more info, I'll be happy to help, particuarly with
pictures of the Saari cube and th "neighbourhoods," etc.



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Nothing is foolproof given a talented fool.