yOn Tue, 17 Sep 2019, Riccardo (Jack) Lucchetti wrote:
Yes. But doubles are 8 byte large. But the probability of extracting
k
different numbers out of n choices without repetition is
$\frac{(n-1)!}{(n-k)! n^{k-1}}$ (if I recall correctly), so when n = 2^32 I
guess we're pretty safe for any reasonable value of k.
Whoops, sorry, it's 2^64.
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Riccardo (Jack) Lucchetti
Dipartimento di Scienze Economiche e Sociali (DiSES)
Università Politecnica delle Marche
(formerly known as Università di Ancona)
r.lucchetti(a)univpm.it
http://www2.econ.univpm.it/servizi/hpp/lucchetti
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