3

Given a collection of $n$ coupons, you can purchase a package of $k$ unique coupons. What is the probability that you will collect all $n$ with $T$ packages?

For the case where $k=n$, you always get the full collection after 1 package.

For the case $k=n-1$, after one draw it is the probability of not drawing the last remaining card, so it should be similar to a geometric distribution.

For the case $k=1$, the answer reduces to the usual coupon collector's problem, found here Probability distribution in the coupon collector's problem

What is the general formula?

I just realized this is a duplicate of a question on MO, so the answers there may also be helpful: https://mathoverflow.net/questions/229060/batched-coupon-collector-problem

Frank Vel
  • 5,339
  • Does "with $T$ packages" mean "with at most $T$ packages" or "with exactly $T$ packages". I ask because the given link to"usual coupon collector's problem" gives the second kind of probability. – user Oct 24 '22 at 21:05
  • @user I had the second one in my head when asking the question, but since $P(x) = F(x) - F(x-1)$, it's trivial to go from one to the other. – Frank Vel Oct 24 '22 at 21:15
  • That's true. But obviously the answer below was given for the first version. By the way why don't you accept it? – user Oct 24 '22 at 21:16
  • @user I wanted to verify that the answer was correct. I wasn't too sure of the differences between that answer and the one on MO. It's a pretty neat answer. – Frank Vel Oct 25 '22 at 16:17

1 Answers1

3

The Coupon Collector's Problem with the coupons arriving in packages of unique coupons can be solved by inclusion-exclusion, much like the standard Coupon Collector's Problem.

Let's say a collection of $T$ packages of $k$ coupons has "Property $i$" if none of the packages include coupon number $i$, for $1 \le i \le n$, and define $S_j$ to be the totals of the probabilites of all the collections with $j$ of the properties, for $1 \le j \le n$. Then $$S_j = \binom{n}{j} \left( \frac{\binom{n-j}{k}}{\binom{n}{k}} \right) ^T$$ By inclusion-exclusion, the probability that a collection of $T$ packages has none of the properties (in other words, contains a complete set of coupons) is $$1+\sum_{j=1}^n (-1)^j S_j$$

awkward
  • 14,736
  • 1
    Thanks for your answer, I just found a duplicate of my question on MO, which appears to agree with your answer https://mathoverflow.net/questions/229060/batched-coupon-collector-problem – Frank Vel Oct 24 '22 at 15:05
  • Is there any advantage not to include $S_0=1$ and write $\sum_{j=0}^n (-1)^j S_j$? – user Oct 25 '22 at 06:46
  • @user I think either form is fine. – awkward Oct 25 '22 at 12:54