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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.

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    $\begingroup$ Why would someone vote this down?! $\endgroup$ – mxcl Sep 18 '13 at 2:32
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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