Take it one more step, Christian:
> You wrote:
>
> The probability of a duplicate assignment somewhere is
> N^2/2^38...
>
> which is not quite correct. ...
> ...
> Thus
> (1-P) ~= e^-(N.N-1/2.X) ~= e^-(N^2/2.X)
> P ~= 1 - e^-(N^2/2.X) if N^3 << X^2
~= 1 - ( 1 - (N^2/2X) ) = (N^2)/2X
awfully close, after all your approximations, to what was given.
There's no use quibbling over a factor of 2 in these very small
numbers when, as the next paragraph says, this is not at all an
interesting question, and the real question of interest yields a
probability far far smaller.
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