---------- Forwarded message ---------- From: Warren Smith <[email protected]> Date: Sun, 18 Jul 2010 11:08:57 -0400 Subject: Re: [EM] Thoughts on Burial To: Jameson Quinn <[email protected]>
On 7/17/10, Jameson Quinn <[email protected]> wrote: > To clarify my position: > > I think that, because of social dynamics which push voter groups towards > symmetry (ie, B voters like A as much/little as A voters like B), honest > condorcet cycles will be a fraction of what they would be in "impartial > culture"-type models. Since such models usually give somewhere around 10% > cycles, or a little more, I think honest cycles will be somewhere in the low > single digits - 1%-4%. For this, I have little evidence, although it should > be noted that Romania is not at all counter-evidence; one documented > possibility in a large number of modern, polled elections is about what my > proportion would have predicted. It is certainly not evidence against "most" > cycles in a Condorcet system being due to truncation, as we have essentially > 0 data on condorcet systems in public elections. > > I think that the necessary conditions for truncation/burial to be a rational > strategy will be much more common. It depends a lot on the average number of > "serious" candidates per election, but assuming that with a Condorcet method > that number would be somewhere between 2.5 and 5, with a minimum of 2... > well, I don't want to pretend I've done the calculations, but my guess is > that that would lead to somewhere between 20% to 60% of elections having a > rational truncation which would affect the result. I'd imagine that a > possible truncation would actually happen somewhere from 25% to 75% of the > time. So honest cycles should be roughly 1%-4%, and truncated ones roughly > 5%-45%. If these broad ranges are right, then truncated cycles will be > 55%-98% of all cycles - probably 66%-90% - ie, "most". > > This is why I think that system performance relating to truncation strategy > is at least as important as honest performance, at least for decent systems > where the differences between honest performance are not too large. > > JQ --suppose we employ the random elections (aka impartial culture) model of 3-candidate elections where all 6 kinds of vote A>B>C, C>A>B,... are equally likely, and large #voters V-->infinity. Then probability(cycle)=8.7739828045910905%, derived in the CRV puzzles http://rangevoting.org/PuzzlePage.html . Now, suppose each voter-type has access to real-time feedback about the current election results (i.e. the pairwise matrix), and if their favorite candidate is not a condorcet winner but some other (call him X) is, then that voter strategically alters their vote to "bury" X. [These alterations happen in random voter-order. Voters do not act exactly simultaneously.] What happens then? There are three logically-allowed possibilities: (i) the modifications stop because we reach a situation where X is the condorcet winner even though every X-not-favorite voter has buried X. I.e. X has a majority win under plain plurality voting. (ii) the voters continue to modify their votes forever because no matter what they do, a condorcet winner keeps on existing -- but it is not always the same. (iii) the modifications stop because we reach a cycle. Prob(i)=0 in the limit V-->infinity. ii is logically possible because of elections like #voters their vote 3 A>B>C 9 B>A>C 7 C>A>B here A is condorcet winner, but upon burial 3 A>B>C 9 B>C>A 7 C>B>A B becomes the Condorcet winner, but upon burial 3 A>C>B 9 B>A>C 7 C>A>B it becomes A again. It seems to me that under our probability model, GIVEN that A is initially a condorcet winner, the limit probability is 100% that A will no longer be but somebody else (call him B) will be after A-burial. But after B-burial, A will for sure still be the condorcet winner since A had a majority versus all, and still will if we bury B. So, I claim limit probability(ii) = probability(honest CW exists) = 91.226%. Finally, prob(iii) = prob(honest cycle exists) = 8.774%. So CONCLUSION: in this simplified model, "strategic burying" does NOT alter the probability of getting a cycle, it still is 8.774% in the V->infinity limit with 3 candidates. However, with 4 candidates, in (ii) after A-burial there now is a nonzero probability that the result is a top-cycle hence with no condorcet winner. Hence with 4 candidates strategic burying DOES increase the probability of getting a cycle. But not hugely: I compute for 4 candidates in random election model: prob(honest top cycle) = 17.5479656% prob(burial yields a top cycle, given that a condorcet winner exists) = 8.77398280459109% prob(burial yields a top cycle) = 17.5479656% + 82.4520344% * 8.77398280459109% = 24.7822929% -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step) and math.temple.edu/~wds/homepage/works.html ---- Election-Methods mailing list - see http://electorama.com/em for list info
