Quote from: tromp on June 15, 2018, 04:34:29 PM

Quote from: aliashraf on June 15, 2018, 01:52:56 PM

Quote from: tromp on June 15, 2018, 01:12:26 PM

Quote from: tromp on June 15, 2018, 11:27:45 AM

On further reflection, if you randomly accumulate shares of weight (fraction of required difficulty) >= 10^-4 until their sum weight exceeds 1, then the expected number of shares is 5000.

I calculated wrong.

n shares expect to accumulate about n*ln(n)*10^-4 in weight, so we expect
a little under 1400 shares to accumulate unit weight...

Interesting, appreciate it if you would share the logic beyond the formula.

Consider a uniformly random real x in the interval [10^-4,1]
Its expected inverse is the integral of 1/x dx from 10^-4 to 1, which equals ln 1 - ln (10^-4) = 4 ln 10.

Now if we scale this up by 10^4*T, where T is the target threshold, and assume that shares are not lucky enough to go below T, then the n hashes will be uniformly distributed in the interval [T, 10^4*T], and we get the formula above.

Solid. Will include this formula and the proof in the white paper, if you don't mind.