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).