Quote from: joan on June 29, 2011, 09:17:04 PM

Quote from: EricJ2190 on June 28, 2011, 10:12:09 PM

Quote from: TiagoTiago on June 28, 2011, 09:14:06 PM

If mining hardware was instead dedicated to generating new addresses; how long do you think would it take till someone stumbled on an existing address that had more BTC stored than what the person would have earned by mining?

(…)

Their are 2^160 possible addresses. Lets say 2^32 (4 billion) people use Bitcoin and each generate 2^16 (65 thousand) address. That gives us 2^48 total addresses out of 2^160 possible. The probability of a generated address matching one of these is 1/(2^112).

The probability for two addresses to match is much higher than 1/(2^112) though. It's more complex than doing 2^160/2^48. Check the Birthday attack.
Also, TiagoTiago mentionned using current mining resources to generate addresses, so I guess it's fair to say that the 2^16 figure is underestimated, that would be more like… 2^16 per second. (But there is not 4 billion miners, yet)

While we are on probabilities, and I'm by no means an expert in these, someone mentioned in a previous page of this thread that to find the "1Kahlahan…" vanity address (8 fixed chars) would take roughly 1.28e14 attempts.
I think this is underestimated. (It seems to come from the intuitive 58^8).
My own computation gives me 2.99e+15, which is an order of magnitude higher. (I decode the base58 and do the maths on the hash160).

The 2^16 figure is just the for the number of addresses actually being used by each normal user, not what you can generate. My point in that part was that there are no more than 2^48 bitcoin addresses that would be of any interest to find a private key for.

For the Birthday Attack, this just means it would be easier to find some two private keys that result in a common but unspecified address. However, we are not looking for any collision here. We specifically are looking for a key with an address in the set of previously used addresses. This is the same as PoW hashing, which I am comparing it with, where we are looking for value with a hash in the set of acceptable hashes.