**[holger--rebol--com]**<</Holger>> Your second formula is incorrect for any n except for (n = 2 ** 160). Especially it is incorrect for (n = 2), where it yields twice as much as it should. OTOH for higher n, (2 < n) and (n < (2 ** 160)), the probability of a collision may be much higher than your formula yields. The correct formula should be: m: 2 ** 160 q: 1 * (1 - (1 / m)) * (1 - (2 / m)) * ... * (1 - (n - 1 / m)) p: 1 - q , where Q is the probability that no collision occurs.

## [REBOL] Re: Question about checksum/secure

### From: lmecir:mbox:vol:cz at: 19-Nov-2001 17:38

Hi Holger, <<Brett>>> I hope this isn't a silly question. > > Can the result of checksum/secure be used like a GUID, is it guaranteed to > be unique?<</Brett>> <<Holger>> No, hashing is a many-to-one operation, mapping from an infinite space (arbitrarily long string series) to a finite space (string series of length 20), so there always has to be the chance of collisions. The chance of a collision is very small though, especially if you deal with a small number of values, say, less than a billion billion billion :-). For any two arbitrary strings to have the same checksum the chance is 1/(2^160). For at least one collision among n strings the chance is n/(2^160) (for n <= 2^160, obviously). -- Holger Kruse