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A comment on The birthday paradox in action: calculating the probability of a hash collision. Parent comment

Re:The birthday paradox in action: calculating the probabili... by jlgreco on Wednesday November 07, 2012 @01:05PM (#4754413)
If there are max(SHA-1) hashes, and a 1/max(SHA-1) chance any particular one will hash to itself, does that mean we would expect only a single instance of a hash hashing to itself?

I have no idea how you might go about finding that one expected instance.

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Re:The birthday paradox in action: calculating the probabili... by Dylan16807 on Wednesday November 07, 2012 @01:26PM (#4754504)
You would expect about 1 per unflawed hash function tested. But the law of averages is less powerful when you focus on a specific hash. The chance of having no fixed points in a particular unflawed hash is 1/e, so there is probably a fixed point, and possibly more than one.
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