4 ms·
> For an organization with 45% hashing power to maliciously double-spend cross 20 confirmations, it'd have to win the coin flip 20 times, or 0.00001% of the tim
by fragsworth 13y ago
> For an organization with 45% hashing power to maliciously double-spend cross 20 confirmations, it'd have to win the coin flip 20 times, or 0.00001% of the time.
That's not right. If they want to maliciously double-spend, they'd just have to get 20 blocks in less time than the rest of the network got it in, and publish their result. This is not the same as beating the network at 20 individual blocks.
For instance, you can double spend by getting your 20 blocks in 19 minutes, when the rest of the network took 21 minutes. This is not the same as beating the rest of the network at all 20 blocks - some of yours may have taken longer.
- makomk 13y agoYou're technically right - the correct calculation is in section 11 of Satoshi's original paper [1], though the probability of success still reduces exponentially with the number of confirmations in the same way that dragontamer's calculation does. [1] http://bitcoin.org/bitcoin.pdf http://bitcoin.org/bitcoin.pdf
- dragontamer 13y agoThanks for the link. Binomial walk makes sense. So it isn't 2^n, but a convolution summation over n-choose-r. Yeah... I don't feel like doing the "correct" math on this one. But the concept is still correct. It is exponentially harder for the malicious user to double-spend based. It is much harder to break ten "one minute" confirmation, than one "ten minute" confirmation.