## Re: Does Poisson hate me?

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• ... in 90 ... extremely ... unlucky. ... numbers in ... or 3 ... was. ... taken ... triple of ... So the probability of a pair yielding a triplet is 1 in 240
Message 1 of 3 , Jul 9, 2008
--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:
>
> Ken Davis wrote:
> > These are contructed such that they are BLS provable for c =
> > 1,5,7,11,13.
> > I sieved to a level where I expected around 1 in 360 to be PRPs
> > So far I have tested 17,221,365 numbers (of the c=7 type)
> > Finding 49949 prps (1 in 344 tests. This is good)
> > For each of these 49949 I prp'ed the other 4 types, expecting 1
in 90
> > (360/4) to be a prp
> > This has yielded 557 pairs (1 in 89.6 stil going to plan).
> > Based on the same logic as above I would have expected 1 in 120
> > (360/3)
> > of these pairs to have yielded a triple of some sort
> > However as yet I have had no luck.
> > Is there something wrong with my logic, or have I just been
extremely
> > unlucky?
>
> There is something wrong *and* you are unlucky, but not extremely
unlucky.
>
> Suppose that after getting a prp for c=7, you test the four other
numbers in
> some fixed order. If a second prime is found, there can be 0, 1, 2
or 3
> candidates left for a third prime, depending on where the second
was.
> Your calculation corresponds to saying it's always 3.
>
> The problem is that "the second prime" and "the third prime" are
taken
> from the same pool of 4 candidates. Here is another way of looking
at it:
> For each of the 49949 initial prp's, there are 6 ways to get a
triple of
> some sort, because there are 6 ways to choose 2 out of 4 numbers.
> Each way has probability 1 in 360^2, so the expected number of
> triples of some sort is 49949*6 / 360^2 = 2.31.
> The probability of having no triples of any sort is around
> 1/exp(2.31) = 10%.
So the probability of a pair yielding a triplet is 1 in 240 instead
of 1 in 120.
I still think poisson has it in for me :-(
> --
> Jens Kruse Andersen
>
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