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:
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 size2^160 - 173346595075428800+ 1 (and I'm not considering the fact that there are 2^96 different private keys - means tries - for the same address).
A group that creates 100% addresses where one of them will start with a given prefix has size
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!
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)
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.
