[EM] Fwd: XA

Forest Simmons fsimmons at pcc.edu
Mon Oct 31 16:40:04 PDT 2016


One nice advantage of XA over Bucklin is that the XA score of a candidate X
varies continuously with the ballot ratings, while the median score does
not:

Profile at time t for 0<t<1:

49 A(100), B(0)
49 B(100), A(0)
2 A(100 - 100t), B(100t)

The respective medians of the candidates jump discontinuously from 100 to
zero and from 0 to 100, respectively, near t = 1/2.

The XA scores vary continuously in the same time interval.


On Mon, Oct 31, 2016 at 4:27 PM, Forest Simmons <fsimmons at pcc.edu> wrote:

>
> And any method that fails Participation also fails Consistency.
>
> However, this example is not as bad as it looks:
>
> (1) It shows that Majority Judgment and other forms of Bucklin fail
> Participation in exactly the same way.
>
> (2) Obviously, the voters were not aware of minimum strategy, i.e. to give
> max support to Favorite, and no support to Worst;  No two-candidate
> election can fail Participation if the ballots are normalized.
>
> (3)  If the purpose of the ballots is to estimate the "worth" of the
> respective candidates on a scale of zero to 100, then we don't talk about
> "winner" or "loser."
>
> Before the additional participants joined in, A's estimated worth (50%)
> was within one standard deviation (about 11 points) of B's estimated worth
> of 40%.
>
> After the additional ballots are added, the estimation intervals overlap
> even more.
>
> So the results are not inconsistent with the expected errors of estimation.
>
> On Fri, Oct 28, 2016 at 8:56 AM, Toby Pereira <tdp201b at yahoo.co.uk> wrote:
>
>> I think Chiastic Approval would fail participation, assuming I've done
>> this right. Take the following ballots with scores out of 100:
>>
>> 2 voters: A=50, B=40
>> 1 voter: A=50, B=60
>>
>> A would have a score of 50. B would have a score of 40. Everyone gives B
>> a score of at least 40, and only a third give B a score higher. Now imagine
>> there are two extra voters and we have these ballots:
>>
>> 2 voters: A=50, B=40
>> 1 voter: A=50, B=60
>> 2 voters: A=100, B=60
>>
>> A still has a score of 50, but B now has a score of 60. So these two
>> ballots cause B to overtake A despite them both preferring A to B.
>>
>> ------------------------------
>> *From:* Forest Simmons <fsimmons at pcc.edu>
>>
>>
>> Does it satisfy Participation?
>>
>>
>>
>
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