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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.

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  • $\begingroup$ By definition, a "permutation" is a rearrangement of the set. Therefore every permutation selected (out of the set of all possible permutations) will include every particular number. If you really meant to refer to sample of size $100$ with replacement and that $50$ such samples are independently drawn, then please edit the question to make that as clear as possible. Otherwise, what do you mean? $\endgroup$
    – whuber
    Commented Jul 15, 2014 at 0:11
  • $\begingroup$ Thank you for the clarification. Is there a reason for the complicated description of this circumstance as a collection of samples rather than a single sample? That is, what is the difference between what you are doing and just taking a sample of size $5000 = 50\times 100$? It seems to me that the event "$40$ or more of the balls were not chosen" is the same in each case and will have the same probability. $\endgroup$
    – whuber
    Commented Jul 15, 2014 at 2:32

1 Answer 1

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You have 50 draws. Each of the draws have size 100 and are drawn with replacement. If we want to mark one ball and find the probabibility that this specific ball is not drawn in one specific draw we can note that this fits into the binomial set-up as you mentioned yourself. Every time we pick a ball there's 1/100 probability of success (drawing the marked ball). As we replace the drawn balls this remains true. Thus, we want to find the probability of getting no successes in 100 Bernoulli draws. This probability is described by the binomial distribution as the probability of getting zero successes when the $n = 100$ and $p = 1/100$. Alternatively, we could argue that the only way of not drawing this specific ball would be to draw one of the 99 others every time. As the draws are independent we can find the probability as $\left( \frac{99}{100}\right)^{100}$.

As whuber points out, the above readily generalizes to the situation with more independent draws. It doesn't make a difference if you divide the 5000 single draws into subgroups of draws. This is of course a result of the replacement strategy: we replace drawn balls, thus, the urn of balls is the same at every draw.

If you want to consider the probability of not drawing any of, e.g., 40 balls you can do the exact same thing. Each draw has a success probability of $40/100$ and you can use the binomial distribution with $n = 5000$ and $p = 0.4$ to find the probability of zero successes. It's miniscule, by the way.

EDIT: the question has been edited and the above is an answer of the original question as I understood it. It concerns 40 specific balls that have been chosen in advance.

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