3 ms·

See my reply to the sibiling comment. I don't believe we've seen an accidential collicion yet.

• Retric 11y ago
  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 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 11y ago
    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.