Re: PGP

Tim Smith wrote:
In article <4YKdnbP4xoAE0bvVnZ2dnUVZ_o3inZ2d@xxxxxxxxxxxxx>,
David Eather <eather@xxxxxxxxxx> wrote:
--- In PGP, what is the probability that a user with N public keys
will have at least one duplicate key ID?
Sub Key ID is 32 bits long, so the answer is 2^16 or 65536.

Uhm...should the probability be somewhere in [0,1]? :-)


Doh! Yes of course.

If you have N keys, out of a total possible M keys, the chance of a match is

P = 1 - ((M-1)/M * (M-2)/M * ...(M-(N-1))/M)

which would simplify into something much prettier - maybe this (no warranty expressed or implied - going from memory with no sleep for 30hrs). On the plus side, if I am wrong someone will correct me.

M!
P = 1 - ----------------
(M-(N-1))! x M^N
.

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