[EM] Fwd: Thoughts on Burial

[Warren Smith]

---------- Forwarded message ----------
From: Warren Smith <warren.wds at gmail.com>
Date: Sun, 18 Jul 2010 11:08:57 -0400
Subject: Re: [EM] Thoughts on Burial
To: Jameson Quinn <jameson.quinn at gmail.com>

On 7/17/10, Jameson Quinn <jameson.quinn at gmail.com> 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