[EM] FBC survey & Simmons latest lottery method
Simmons, Forest
simmonfo at up.edu
Mon Nov 21 15:46:17 PST 2005
Warren,
I should have been more clear:
I did not require complete rankings. Indeed, some of my examples incorporated collapse of ranks.
However, I did assume a fair coin, since otherwise the method would not even be monotonic.
With these two points clarified, let's prove that the method does satisfy the FBC:
Think of alternatives F and C as Favorite and Compromise, respectively. [Later we will see that this identification is not essential. However, it is motiviational for the following proof sketch.]
Here is the question: Can collapsing preferences of the type C>F or C>>F to C=F (without decreasing approval for C or F) decrease the probability of F winning or decrease the probability that the winner comes from the set {F,C}?
[We will see that the answer is in the negative on both counts.]
Note that by monotonicity such a change could not decrease F's probability. Furthermore, such a change cannot introduce a majority defeat of F over C, i.e. if there is no majority defeat of F over C before the change, there will not be one afterwards.
We only need to show that if C loses any probability, then F gains all of the probability that C loses.
Case I. Before the change C is the approval winner. In this case after the change no alternative (except possibly C or F) has increased approval, so either [case Ia] C is still the approval winner or [case Ib] F becomes the approval winner, in which case C is the highest approval alternative not majority defeated by F, so that C and F share the probability 50/50, [which takes care of case Ib.]
[Now we finish case Ia] If C retains approval winnership, then the change exerts no effect on the probabilities at all, or else F gains half of it.
Case II. C is the highest approval candidate that is not pairwise defeated by the approval winner A. If A=F, then the change doesn't affect the probabilities, so assume that F is not in the winner's circle before the change. The only way that F can knock C out of the winner's circle is by surpassing C in probability. There are two subcases:
Case IIa. A does not majority defeat F. In this subcase, when F surpasses C in approval, F merely replaces C in the winner's circle, unless F also surpasses A in approval, in which case F will still have at least 50% of the probability, the amount lost by C.
Case IIb. A does majority defeat F. In this case there is no change in probability, unless F or C becomes the approval winner, in which case the highest approval winner not majority defeated by the approval winner will come from the set {A,F,C}, so (after the change) the winning circle is either {C,A}, {F,A}, or {C,F}.
**We have shown that when preferences of the type C>F or C>>F are replaced with C=F (without lowering approval of C), the probability of the winner coming from the set {C,F} is not decreased.
This result coupled with monotonicity is the version of the FBC that this method satisfies.
It does seem possible (in case IIa) that the winners' circle might change from {A,C} to a disjoint set {F, D} where D is so much worse than A that one would rather not have the change; i.e. although the probability of the winner coming from {F,C} does not decrease, the expected utility might decrease. Note that this could not happen with only three candidates. Also note that in this case, A might be a better choice of Compromise than C.
Furthermore, an examination of our proof sketch shows that condition (**) applies to any pair of alternatives {C,F}, not just "favorite" and "compromise," so it is a more general disincentive for insincere ranking than the usual version of the FBC.
Which makes the method extremely difficult to manipulate, even in the presence of perfect information about the preferences of the other voters.
For those who don't like approval cutoffs, one alternative is implicit approval through ranked versus truncated alternatives. Another possibility is to use the Equal Rank Bucklin Whole counts as approval counts, or to use the Ratings Bucklin counts for that purpose.
Forest
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