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There are $c$ different types of coupon, and each coupon obtained is equally likely to be any one of the $c$ types. Find the probability that the first $n$ coupons which you collect do not form a complete set.

It is clear that the probability of getting a new coupon on the first draw is 1, and that the probability of getting a new coupon on the second is $(c-1)/c$, or $(c-i)/c$ for the $i$-th new coupon after the initial draw (or $(c-(i-1))=c$ if you include the first draw as a new $i$-th coupon) I don't know where to go from here.

StubbornAtom
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Housefire
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2 Answers2

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Just because you accepted a wrong answer.

There are in total $c^n$ possible collections among the first $n$ collected coupons. Among them there are $$ c!{n \brace c} $$ combinations with each coupon collected at least once. Here ${n \brace c}$ is the Stirling number of second kind.

Therefore the probability in question is: $$ 1-\frac{c!}{c^n}{n \brace c}. $$

user
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Here is another answer: The probability that none of the n coupons are of type 1 is ${(\frac{c-1}{c})}^n$. Likewise for all the other types. The sum of these is an overcount of the total probability. We can use the inclusion exclusion principle. Subtract the probability that two types are excluded, then add the probability that three types are, and so forth until all possibilities are covered.

RobertTheTutor
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  • I am very familiar with the stars and bars argument. So I can see how this would create the number of combinations for the c types of coupons, but how do I turn this into a probability that the first n coupons do not form a complete set? – Housefire Feb 07 '21 at 19:37
  • The solution is wrong. – user Feb 08 '21 at 00:16
  • Well I think this is still a valuable answer and it could be helpful to see why it's incorrect vs. the correct answer. – Jared Feb 08 '21 at 08:51
  • The problem with this approach is that not all the arrangements you have counted are equally likely. – awkward Feb 08 '21 at 15:32
  • As you can see, not being allowed to delete this Answer doesn't prevent you from editing it. I recommend editing it to be as full and correct as possible, rather than dwelling on what is incorrect. Future Readers will come looking for good information, and there is something to said in the way of a simple or comprehensive solution. The edit history is there for anyone what wants to see exactly how the Answer was improved. – hardmath Feb 10 '21 at 17:06