Each Chocolate Frog comes with one collectable illustrated wizard card (very cool and not dorky at all, honest). There are equal odds of each card being in a pack (i.e., they have all been produced and distributed evenly). How many packs must we buy in order to have an 80% chance of having obtained all 12 cards? How about 90%? Thanks.

  • 3
    $\begingroup$ Why would someone vote this down?! $\endgroup$ – mxcl Sep 18 '13 at 2:32
  • 6
    $\begingroup$ Maybe they're a muggle. $\endgroup$ – anon Sep 18 '13 at 2:33
  • $\begingroup$ Aside from the Potter jokes, how many cards are in one deck? You may want to edit your question and add that answer. $\endgroup$ – jim mcnamara Sep 18 '13 at 2:41
  • $\begingroup$ Thanks @jimmcnamara, I amended my question. $\endgroup$ – mxcl Sep 18 '13 at 2:45

This website on the coupon collector's problem under item 11a (adapted to 12 items from 6) states that the chance you complete your set on purchase $n$ is $\sum_{j=0}^{11} (-1)^j{11 \choose j}\left( \frac {11-j}{12} \right)^{n-1}$ but gives no derivation. You can add up starting with $n=12$ until you get to the desired success probability.

  • $\begingroup$ The expression comes from the inclusion-exclusion principle, and really gives the chance you have not completed your set by purchase $n$. So find the $n$ that reduces it to $0.1$ or $0.2$ and you are there. I would argue the lower limit should be $1$, but it doesn't matter much. $\endgroup$ – Ross Millikan Sep 18 '13 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.