[EM] Truncation error in STV

[Stephane Rouillon]

Olli,

I agree that the accuracy cannot be increased because the level of
precision used gives a error of zero (0). What makes you think
I said anything different?
The formula I gave "1/(nb of votes *2^ (nb cand-2) ) so no misranking appears."
is the precision necessary to get a zero error.
"nb cand" is the number of starting candidate and it does not
vary after each transfer. Was that the problem?
As the number of candidates grows, the precision diminishes, so the number of
digits increases slowly. For a little more than every 3 candidates, an extra digit

is necessary to preserve an error of zero.
But for a given election, once you know the number of starting candidates,
the number of digits for the calculations "mantisse" is fixed.

Stéphane

Olli Salmi a écrit :

> Dear Steph,
>
> Thanks.
>
> The passage from Wichmann gave me the impression that accuracy cannot
> be increased.
>
> In the election of the Australian senate the transfer value is taken
> to eight decimals but only whole votes are transferred and the quota
> is an integer.
> http://www.aec.gov.au/_content/what/voting/count_senate.htm
> In the example on this site the value of the transfer is 1513870 x
> 0.65350505 = 989 321, which means it's rounded down from
> 989321.69004. I can't understand what complexities arise. To me this
> looks much better than the ERS rules which would transfer 1513870 x
> 0.65=984015.5 votes. It's the same with Northern Ireland rules, but
> they only transfer whole votes. That's more than 5000 votes wasted,
> isn't it? Now I do feel stupid. Have I overlooked something?
>
> I found nothing on accuracy in the relevant passage of the law.
> http://scaleplus.law.gov.au/html/pasteact/0/57/0/PA003460.htm
> It's boring, you wouldn't want to read it.
>
> On the other hand, eight decimals are not required in a small election.
>
> I didn't formulate my questions very clearly.
>
> Best regards,
> Olli
>
> >So, as I said,
> >it is not because you have tor transfer a total of 666 = 999x2/3
> >votes that there are not splitted
> >like this
> >1 second preference to D
> >998 second preference to C.
> >So you could get expected to transfer 666x1/999 to D and 666x998/999 to C.
> >
> >When you add this to the number of votes already got by D, you add
> >errors. (remember,
> >with multiple rounds, it would not be an integer usually). So each
> >time there is an addition
> >you needed a precision twice better before the addition.
> >So maybe the good formula should be:
> >
> >1/(nb of votes *2^ (nb cand-2) ) so no misranking appears.
> >
> >Sorry, I am not that good at this stuff. After getting an order of magnitude,
> >my engineering side kicks in and I am not much thorough...
> >
> >Do you think the result is good?
> >
> >Steph