Regardless of the type of avg.many.n.birthdays' output, it should have n elements, right? n probabilities, for 1, 2, 3, ....n
Yes. The function should return a vector of n averages.
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Regardless of the type of avg.many.n.birthdays' output, it should have n elements, right? n probabilities, for 1, 2, 3, ....n