A trading card series has 200 different cards in it, which are sold in 5-card packages.

Each package has a random sampling of the cards (assume that any card of the 200 has an equal chance of being in a package).

On the average, how many packages will need to be bought to collect the complete series if...

A: all the cards in a package will always be different

B: a package can have repeats

As others have pointed out, problem B is simpler than A. Since each package can have duplicate cards, a random series of packages can be treated the same as a random series of cards. Therefore, if we determine the average number of cards it takes to get the complete set, we simply divide it by 5 and round up.

There is a simple way to solve this problem – simple enough to do it in Excel in just a few seconds. Calculate how many cards you need to pull, on average, to get one more card for the set. If we calculate this number for all 200 cards and sum the results, that will give us the average number of cards it takes to acquire the whole set.

Obviously, when we have zero cards, the average number of "pulls" required to get a card we need is one. So what about the second card for the set? With one pull, the probability of getting one of the 199 cards we still need is 199 / 200. The average number of pulls to get that second card is the reciprocal of the probability: 1 / ( 199 / 200 ) = 200 / 199 = 1.005025...

This might be more intuitive if you look at the situation when you already have 199 cards and you just need the last one. The probability of getting it in one pull is 1/200, so the average number of cards you’ll have to pull to get it is 200.

Therefore, the formula is:

∑{i=0,199} 200/(200-i)
= 1175.606 cards (235.121 packs)

I suspect that the small difference between my answer and Charlie's (235.121 vs. 235.521) might be due to the fact that the probabilities in his formula are based on packs and mine are based on single cards. Just a hunch. Or maybe it's a bug in Excel.

I haven't figured out a way to apply this method to problem A.

Posted by Rick on 2003-02-05 12:06:52