Birthday problem with indistinguishable clones. Suppose we have created an army of n clones which are completely identical(except they may have different birthdays). The cloning happened at different times such that all 365(disregarding the 366th day) birthdays are equally likely.
What is the probability of at least 2 people sharing a birthday in this indistinguishable setting.
In the original birthday problem the solution is $$1-\frac{365Pn}{365^n}$$
But here the solution assumes the distinguishability of the people.
Also let us note that for the indistinguishable case the solution $$1-\frac{365 \choose n}{365+n-1 \choose n}$$ is incorrect because it fails to regard the probability weighting of each outcome as they are not equally likely(for example probability of two people having two Sep 1s is less that probability of having Sep 1 and Sep 2 as it can happen in two cases).
 A: Since we are not given whether or not the clone's birthdays are independent or not, there is not enough information to determine the probability that no two share a birthday.
Suppose you were given that the clone's birthdays were independent. Let us number the clones in an arbitrary order. The definition of independence implies that for every sequence $(b_1,b_2,\dots,b_n)$, where $b_i$ is a day of year for each $i\in \{1,\dots,365\}$, the probability of that sequence occurring (meaning the $i^\text{th}$ clone has birthday $b_i$ for all $i$) is $(1/365)^n$. In particular, all of these sequences are equally likely. Therefore,we can find the probability of all birthdays being different by counting the number of sequences where all entries are different, and dividing by $365^n$. The result is
$$
365\cdot 364\cdots\cdot (365-n+1) \over 365^n
$$
which is the same result as the distinguishable person case.
The takeaway message here is that independence implies that the underlying sample space is most conveniently thought of as distinguishable, at least for the purpose of counting cases.
