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. ------------------------------------------- Nothing is foolproof given a talented fool.