Quote from: Meni Rosenfeld on May 08, 2014, 09:17:36 AM

Quote from: jl2012 on May 07, 2014, 04:21:12 PM

How about another scenario:

Hash rate A:B = 1:H . A is attacker, B is honest

A wants to outpace the latest X year's blocks. What is the probability for him to succeed in Y years? (A only wants to outpace the latest blocks. For example, with X=1, on 1 Jan 2014, A wants to mine a single block to outpace all the blocks since 1 Jan 2013; on 15 Jan 2014, the target will be the blocks since 15 Jan 2013). I think it's even easier, right?

Yes. The integral is then over the constant function 1/(HX) which gives a result of 1 - Exp (-Y/(HX)).

e.g., for X=1, Y=1, H=4 this gives 22.1%.

As a reminder, these calculations are for the case that everyone's hashrate is fixed. If it grows over time, the numbers will be different (in the attacker's favor).

Let say I have 1% of total hashrate (H=99), and want to outpace 1 day of blocks in 30 days (X=1, Y=30).

The probability is: 1-Exp(-30/99)=26.14%

and 97.5% in 1 year

The cost of this attack is about 1080BTC for 30 days or 13140BTC for 1 year, plus depreciation and electricity. It's expensive but still much cheaper than I thought.