Post
Topic
Board Bitcoin Discussion
Re: How exactly would a 51% attack work?
by
sadpandatech
on 20/11/2011, 21:05:13 UTC

Ahh, I see. So 51% is the magic number because that's the point at which a person can make alternative blocks faster than the rest of the network combined, and then spring the alternative, longer blockchain on everybody all at once, later on, where it replaces the blocks everyone thought were already finalized and settled.

Thank you!
Exactly. Satoshi's original paper contains calculations for how many blocks the recipient has to wait to keep the chance of succeeding in double-spending at a given level (say 0.1%), as a function of the attacker's hashrate. At >50% hashrate the number of blocks if infinite - no matter how many blocks are waited, the attacker has 100% chance to eventually have the longer chain.

This attack isn't the only thing you can do with high hashrate, though.

  The key thing to note in Meni's explanation is 'eventually'. Due to variation it would be quite possible that the attacker might not find more blocks than the 49% for days or weeks even.. There is a HUGE amount of luck involved on the attackers end if they are to be successful. Plus the attacker would have to ensure he actually had complete control of the daemon that was keeping tabs on his attack blocks. A pool in itself over 50% means nothing unless it can be hacked and controlled completely or the OP was in on it. In the case of most pools they would lose more than they would stand to gain from what would certainly only be allowed to happen once.

  That being said, we are seeing more spread of hashing power these days and there are projects in the works that will help to maintain a high overall network hashrate, detering 'lone' 51% wolves from gaining enough to beat the entire network on their own.