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Re: comments on client auth
> I can't quite be bothered to solve this for n=2^1024 and P=.99 but for the
> sake of illustration, lets take k=2^100 (=10^33, roughly, or 2*10^23
> certificates per person in the world). Then P=1-2^200/2^1025=1-2^(-825).
> In other words, the probability of a collision is roughly 1 in 10^275.
> Vanishingly small, I would say.
What's the saying - "better to stay silent and let them think you're a
fool than open your mouth and prove it".
I'll just shut up now and start packing for Montreal.
Thanks for the numbers,
dpk