I have a question about a probability problem located here:


And quoted here:

“If you were trying to collect 6 baseball cards that came in packets of cheese puffs, assuming they are distributed evenly how many packets of cheese puffs would you expect to buy before you have all 6 cards?”

I was in agreement with the first part (Case I), in that one is guaranteed to get some card on the first packet.

However, they lost me at part II (Case II), with their odds calculation on that step. I would have assumed since we still needed 5 cards of the 6 cards total, with each being equally likely, and given that we already have one of the cards, the odds of getting a card we needed would have been 5/6. However, they say 4/5. (Also, they have 4/5 = 125/100, which sure looks fishy to me!)

I would have further guessed 4/6, then 3/6, then 2/6 and finally 1/6 for the last card. I also figured that the reciprocal of these summed is the number of cheese puff packets one would have to buy to collect all 6 cards, or 13.7 packets total

Can somebody point out where I am going astray with my method? Or, is this a rare case where I'm okay and the reference is wrong?

  • $\begingroup$ it lost me immediately! how many cards there are in one packet and how many there are in the whole array? $\endgroup$ – carlo Apr 3 '17 at 16:28
  • $\begingroup$ ok, I got it: the answeres to my questions are 1 and 6 $\endgroup$ – carlo Apr 3 '17 at 16:39
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    $\begingroup$ This is the Coupon Collector problem. $\endgroup$ – whuber Apr 3 '17 at 17:04

It's a dodgy site, I would not trust it.

The key to solving this problem is to realize that the expected number of packets needed to get the 2nd coupon (out of 6) is:

$E_2 = 5/6 * 1 + 1/6 * (1 + E_2)$

That is, with probability $5/6$ we will get it on the first go, and with probability $1/6$ we are back to where we started, but have now used up an extra turn.

Solving the formula for $E_2$, we can find that $E_2 = 6/5$

Similarly, you can find the expected number of packets you need to buy to get the rest of the coupons. This is a very useful method that comes up in a lot of these probability problems.

The answer should be $14.7$, or 15, but your reasoning seems correct.

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  • $\begingroup$ thank you, this was very helpful. I see I forgot to include the original packet in my count of 13.7, and I can see that 14.7 is the correct answer now. I think you're right about that site, which is a shame, but I have since found a couple other sites that look more trustworthy to replace that one with. $\endgroup$ – EthanT Apr 3 '17 at 16:56

the site first defines the problem with 6 cards, then provides the solution with 5 cards, here is the main problem.

also, you must be aware to the difference between probability and expectation: expected value is the mean of all possible outputs of some probability process, that is called random variable; so, if we collect six cards in the way the site says, we can call $X$ the number of packets we have to buy, one after the other; if we do this, we can call $E(X)$ the expected value of $X$, that is, the number of packet we expect to buy.

expected value can be computed multiplying all possible outputs with the probability we have to observe them, and this is where the very good answere of rinspy comes handy:

$$E_2= 5/6∗1 + 1/6∗(1+E_2)$$

here the two possible outputs are $1$ and $(1+E_2)$, and the two probabilities are, this is easy, $5/6$ and $1/6$.

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  • $\begingroup$ Thanks @carlo. Just out of curiosity, how would the formula for E(3) be defined? On second glance, I'm not quite sure I understand what the two possible "outputs" stand for in your formula for E(2), specifically the (1 + E2) term. $\endgroup$ – EthanT Apr 3 '17 at 17:03

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