631 sats \ 0 replies \ @PlebeiusG 10 Sep 2023 \ on: deleted by author meta
This paradox also helps illustrate the bitcoin difficulty.
If enough hash attempts are in the room, one will show up that is above the difficulty. Even though there’s 2^256 possibilities (an impossibly high amount) the birthday paradox explains that only a relatively small amount of total hashes are needed before we find “collisions.” And in this case, because the difficulty is a threshold, there are many possible hash collisions.