Quote from: theymos on March 26, 2011, 09:37:32 PM

Quote from: stonetz on March 26, 2011, 09:22:34 PM

The probability of any specific address is 1 in 2^160 or 1.4x10^48 right?

It's ~6.84x10^-49. Did you put "1 in 2^160" in Wolfram Alpha? Wink

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So the more addresses you have out there the higher the chance of a collision right (birthday problem?)

Yes. It is a birthday problem.

That depends. I've lost track of the specific problem being addressed, so pardon me if I'm adding extra detail that isn't necessary.

If you are just trying to find out if any two keys match, then it's the birthday problem, i.e. "do any two people in my class have the same birthday?" If you are trying to match a specific key, then it's a much huger problem to solve - you're now asking "does anyone else in my class have the same birthday as me?"

And the whole problem still has another layer of complexity, because you aren't trying to find just any old block of 160 bits. You're trying to find a 160 bit hash of a ~~much larger~~ public key, for which you also need the private key. So to create a collision, you have to generate a key pair, hash the public key and see if it matches. That's assuming you're trying to deliberately create a duplicate, as opposed to stumbling upon one.

Edit: I realized the public key isn't necessarily larger, it's just different.