I am a non-statistician testing a randomising algorthm.
If I have a sample of size 100 (numbered balls) with replacement and that 50 such samples are independently drawn. To my understanding, the binomial distribution will describe very well the chance that any one particular numbered ball has not been selected in any of the draws (0.605).
if after tallying up all the numbers drawn, what is the probability that, say 40 OR MORE of the numbered balls (ANY 40) were not chosen in any of the 50 samples (i.e. any 40 balls have a tally/count of 0)?
If possible, could the answer be devoid of R (or other statistical software) arguments, as I wish to understand the logic, rather than just achieve a result.
Thank you.
Edit
Thanks to whuber, this question has been altered from its earlier erroneous use of the word "Permutations".In response to the second comment (which will probably not fit as a comment), the exact use of this is as part of a monte-carlo type simulation of a randomising algorithm. I am writing a randomising algorithm in C# which is quite complex, with variable block sizes (as twins of the same gender need to be randomised to the same arm), 2 different interventions and 1 control arm, and a different ratio of the 2 intervention arms. I am running tests on the algorithm before rolling out to trial sites. If a block size is for example 8, with 4 receiving treatment a, 2 receiving treatment b, and 2 control, there will be $$\frac{8!}{4!2!2!}$$ permutations of allocaton order for this block. I wanted to check if any possible permutations were not generated after thousands of runs, and if not, what the probability of having n unused permutations was.