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MD has probability of collision 1 / 2^64 under birthday paradox but does this apply to any input given to MD5 ? Not quite. In the general case, the probability of a collision depends on the number of messages. The $2^{64}$ comes into play because we need $\approx 2^{64}$ messages for a $50$% probability that any two of those messages collide under MD5. ...

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MD5 was designed with the goal that any change in the input uniformly affects all the bits of the output. It's not perfect, but it's pretty good. If you're choosing the input "randomly enough" (e.g., by appending random bits before hashing) then your question approaches this one: Given two randomly generated 8-bit strings, what is the probability that ...

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