Quote from: zielar on December 27, 2019, 05:39:19 PM

This is not exactly the answer to my question - "Is there a number that creates a range that makes up 100% of the pool in which the prefix is located?"

The answer was here:

Quote from: arulbero on December 26, 2019, 11:18:37 AM

Search space size is not 173346595075428800, sometimes you have to generate more than 173346595075428800 addresses to get a match.
A group that creates 100% addresses where one of them will start with a given prefix has size ~~2^160 - 173346595075428800+ 1~~ (and I'm not considering the fact that there are 2^96 different private keys - means tries - for the same address).

in other terms: no, there is not such a number, or if you prefer there is but it is a very very large number, something like

2^256 - (2^160/difficulty)*2^96

where (2^160 / difficulty) is the size target (number of addresses with a given prefix, 2^96 is the average number of private keys to the same address).

But nobody knows exactly how many private keys generate addresses with a given prefix, it is only a estimate!

Quote from: zielar on December 27, 2019, 05:39:19 PM

What the case looks like in case of difficulties 10054102514374868992
What percentage does this number specify? (I can't find an effective calculation method myself)

difficulty = 10054102514374868992 = 2^(63.1)
number of keys we have to use (= number of addresses we have to generate to get on average 1 match)

target = 2^160 / 10054102514374868992 = 1.453637095147298e+29 = 2^(96.9)
number of addresses in target

difficulty * target = 2^160 (all addresses)!

The difficulty is how much the set of all addresses is bigger than set of the target addresses. In this case:

difficulty = set of all addresses / target set = 2^160 / 2^(96.9) = 2^(63.1)

Smaller is the target, bigger is the difficulty to hitting it and viceversa.

How to compute the probability?

The formula (probability to hit a element in the target set in n tries) is:

P(n) = 1-(1-p)^n

where p = 1/difficulty (probability to get a match each try).

In your case:
p = 9.946 * 10^(-20)

P(1) = 1-(1-p)^1 = p = 9.946 * 10^(-20)
P(100) = 1-(1-p)^100 = 9.94 * 10^(-18)
P(10054102514374868992) = 1-(1-p)^10054102514374868992 = 0.63 -> 63%
P(2*10054102514374868992) = 1-(1-p)^2*10054102514374868992 = 0.86 -> 86%
P(3*10054102514374868992) = 1-(1-p)^3*10054102514374868992 = 0.95 -> 95%

P(k*diff) = 1-(1-p)^k*diff = 1- e^-k (this way is more simple to compute)

If you want to know what is n to get P(n)=99%

--> 1-(1-p)^k*diff = 1 - e^-k = 0.99
--> e^-k = 0.01
--> k = -ln(0.01) = 4.6 --> n = 4.6 * 10054102514374868992

P(4.6*10054102514374868992) = 1-(1-p)^4.6*10054102514374868992 = 0.99 -> 99%

P(6.9*10054102514374868992) = 1-(1-p)^6.9*10054102514374868992 = 0.999 -> 99.9%

and so on.