> >
> > 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.
> 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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