This is what I don't understand too well (sorry to bother you).
Let's assume, as you wrote, that your device has 1GH/s of computing power.
So what T do you like to choose?
If you choose T=1 sec, then the probabilty to find one nonce in T is the mean, so is lambda. Lambda is 1/4.3=0.23. 23%.
The probability to find two nonces in T, according to Poisson distribution, is ((0.2325^2)*e^-0.2325)/2!=0.021. 2,1%.
Depends on what question you are trying to solve. The question you answered above is "What is the probability of finding k nonces in a period of 1s? And in a period of 2s?"
That was not the question I was answering. My question was "What is the probability of finding k nonces
at the same time?"
If you choose T=1/1E9 sec, AKA the clock tick duration of your device, then I calculate the probability to simultaneously find two or more nonces this way:
a) if your device processes hashes sequentially (one thread), of course there cannot be simultaneous nonces if we consider T=1/1E9 sec =1/(GPU hashes per timeframe);
b) if your device processes more than one hash simultaneously (more threads), there can be simultaneous nonces, but every thread uses just a part of the device computing power. Let's say we have five threads. Each thread is capable of 200MH/s, so it finds a nonce in about 21.47 sec (that is: 2^32 H 200MH/s). In 1/1E9 sec each thread finds a mean of 1/21.47G nonces.
The problem is independent of thread execution time, because that is relatively constant. They all end at mostly the same time in a wavefront. The problem is: when they all end, how many have nonces?
The probability that at least N of our five threads find a nonce in the same clock is 1/21.47G^N. We can say zero, and we don't need any Poisson distribution for it.
When n->inf and p->0 the Poisson follows the Binomial, so either gives the same results:
https://en.wikipedia.org/wiki/Poisson_limit_theoremI can rewrite everything in another way, using another example:
Assuming each wavefront composed of 256 threads with 2 vectors each:
lambda = 1/(2^32)*256*2
Poisson(k) = (lambda^k)*(exp^(-lambda))/(k!)
Probability of finding 1 nounce in a wavefront:
P = Poisson(1) = 1.1921e-07
P = Binomial(1, 512, 1/(2^32)) = 1.1921e-07
Probability of finding 2 nounces in a wavefront:
P = Poisson(2) = 7.1054e-15
P = Binomial(2, 512, 1/(2^32)) = 7.0915e-15
Either way we look at the problem, the answer is always the same: the probability of finding 2 nonces at the same time is 10^8 smaller than for finding 1 nonce! In another words, a ~0.1 Exahash/s hardware will find 2 nonces simultaneously at about the same rate a 1GH/s card finds 1 nonce right now.
