Chirayu Patel wrote:
>
> > >
> > > Assuming that the pool size is "n", where n = 2^40.
> > >
> > > As per your formula the probability of choosing two unique numbers is (n-
> > 1)/n,
> > > and of three unique numbers is ((n-1)/n)*((n-1)/n).
> > >
> > > As per Geoff, the probability of choosing two unique numbers is (n-1)/n,
> > and
> > > of three unique numbers is ((n-1)/n)*((n-2)/n).
> > >
> > > Since the space from which you can choose a unique number diminishes by
> > one
> > > with each draw. I think Geoff's formula is correct in this regards.
> >
> > No. The space does not diminish at all in the local assignment method, since
> > every draw is totally independent of every other draw.
>
> If n = 5, and the set of numbers is {1, 2, 3, 4, 5}
>
> Assume that the first number chosen is "3".
> The probability that the second number is unique is (5-1)/5 = 4/5, because
> four unique numbers are available - {1, 2, 4, 5}
>
> Assume that the second number chosen is "5".
> The probability that the second number is also unique is (4/5)*(3/5) because
> three unique numbers are available - (1, 2, 4}.
>
> The total number space (which is 5) does not diminish with every draw.
> However, the space from which *unique* (from all the previous draws, and not
> only the first one) numbers are chosen *does* diminish by one with each draw.
OK, if the property of interest is the probability that two (or more) members
of a set sites have chosen the same number, I guess that is correct.
For n = 2**40, how large is that set to give one chance in a million of
a collision?
Brian
>
> > The space diminishes by one in the central assignment method, but it
> > doesn't matter since collisions are excluded by construct in that case.
>
> Agree.
>
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