Quote from: drlukacs on May 01, 2013, 01:08:23 PM

Quote from: Meni Rosenfeld on May 01, 2013, 06:49:29 AM

It's not an error, though tricky and somewhat unclear.

He is assuming that \xi is constant while n and D are variable. For \xi=2, for example, he is considering a case that the hashes calculated are twice the average needed; he then considers what happens when D, and correspondingly n, go to infinity (continuous case). In this case the chance of not finding a block is indeed exp(-2).

So, is there an implicit assumption that n and D are linearly related?

You see, I have no problem with the observation that finding a block has a Poisson distribution. But the "proof" provided as it stands is simply wrong, unless D is a function of n, and \xi is the limit of their ration when n tends to infinity.

The assumption is explicit = \xi is defined as n / (2^32 D), meaning that D = n / (2^32 \xi).

And yes, in this calculation n and D both go to infinity at a specified ratio \xi. The assumption that \xi is fixed was not explicitly written but it follows from context.

D doesn't have to "go to" infinity in real life for the calculation to show that if D is sufficiently large, then for any n the probability of not finding a block is roughly exp ( - n / (2^32 D)).