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Re: Does Poisson hate me?

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  • Ken Davis
    ... 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
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      --- 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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