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Retric on April 27, 2016 | parent | context | favorite | on: Why it's hard to write a dupefinder

Your math is off. You tried to calculate the odds that the next block would collide you want the odds that any of them would collide. https://en.wikipedia.org/wiki/Birthday_problem

X!/((X^n) * (X-n)!)

It only takes 23 people to have a 50/50 shot of sharing a birthday.

On the plus side MD5 is 128bits.

hvidgaard on April 28, 2016 [–]

No, I used a reasonable approximation. Sqrt(2^128) is roughly in the ballpark of what is needed to have 50% of a collision, in fact it is slightly lower than the actual number. Incidently sqrt(2^128) is 2^64, which is the number I used.

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