Curious question, shouldnt it be 50% less so 00000000000000000000000000000000000000000000000000000000000000000000000000000172% because of the possibility of a collision (another key that works for the same hash, gives out the same hash?) How can we estimate the probability to find a colission?
Discussion
I don't this so, as this is the chance of 1(single) key. 🤔
If you were to reverse a PubKey to PrivKey, then indeed there is a reductionfdfrom 2^256 to 2^128 and that because Birthday Paradox(Misnomer)
You are also correct regarding collisions with addresses.
When calculating the Bitcoin address you at some point do RIPEMD160 and you can find colissions there.
This project focuses on this https://lbc.cryptoguru.org/about
The probability of collision is 100%(because of the pidgeon hole principle:p)
I think the question is how many steps it takes to find one, and the answer is sqrt(N).
cool thank you very much for this answer. I asked this question on calculating the probability on a possible collision on sha256 on stack overflow (crypto or math, dont recall) and no one was able to answer it.
i read up now on the pidgeon hole principle (never did hear about it)....