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The problem: My daughter wants to have a collection of certain Kinder Surprise toys. There're 10 different toys in this series. Assuming the chances of getting each toy are equal, how many Kinder Surprises should I buy to get her a full collection?

What I did was 1/1+1/0.9+1/0.8+1/0.7+1/0.6+1/0.5+1/0.4+1/0.3+1/0.2+1/0.1 and it gave me the answer of 29.28968

Question 1: is it correct? Is there a better way to do it?

Question 2: I want to know how dispersed the hypothetical data of buying Kinder Surprises until you get all 10 would be . If it is applicable in this situation, standard deviation would answer my question. If I knew the standard deviation, I would know that ~95% of values lie within 2 standard deviations, so I could decide on how many Kinder Surprises to buy to get a full collection for sure.

Edit: if you have an answer which you can explain both mathematically and as an R idiom [software], including the latter would be very helpful.

Th334
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    Enough. You should buy enough. Sadly, not much more can be said than that, for sure. – Cameron Buie Nov 01 '13 at 03:54
  • For a closely related question, see here. – Cameron Buie Nov 01 '13 at 03:56
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    You have rediscovered the classic solution to the coupon collector's problem. Excellent work thinking about it. Unfortunately, the expected value is not so hard to find, but the higher confidence levels seem to be very difficult unless you run a simulation. – Ross Millikan Nov 01 '13 at 03:59
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    You are aware that children have died from choking on those toys? – bof Nov 01 '13 at 03:59
  • @bof, Yes, I stumbled upon this information while trying to find a solution to my problem floating around. Although it was an actual conversation about Kinder Surprises that stirred up my curiosity, I don't have kids and I made up the daughter example just to avoid unimportant background details and make it simpler to answer :) I will now surely consider this issue when I do have a daughter. – Th334 Nov 01 '13 at 04:21
  • @CameronBuie, I understand the solution to the problem you linked me to, but I can't quite understand how to relate it to my problem. I tried using binomial distribution before even posting this questions, but I couldn't make it work. We have a series of Bernoulli trials with changing p (from 1 to 0.1). So is it really a closely related question and there is a way to get our 29.28968 using binomial distribution? [how?] – Th334 Nov 01 '13 at 04:30
  • It relates not because of the distribution, but rather because it seeks a level at which there will be certainty...when there can be none. – Cameron Buie Nov 01 '13 at 04:41
  • @RossMillikan, thank you! Seems like you nailed it, I'll look through the link more carefully now. Regarding the simulation, would you by any chance know how to make it in R? Just in case you do :) – Th334 Nov 01 '13 at 04:49
  • @CameronBuie, hm, why not? Looking back at our example, when I buy 30,40,50,100,500 Kinder Surprises, at some point the chances of me having a full collection will be so high, that it's virtually certain. Estimating this point is what I was asking about, not the point when it will be literally 100%.

    As to your coin example, if we flip it twice we'll get heads at least once in 75% cases, isn't this certainty?

     – Th334 Nov 01 '13 at 05:11
  • Ah, I see what you're asking, now. As for the coin flip, no, not at all. It is quite possible (though unlikely) to flip $10,000$ tails in a row, for example. That's $5,000$ tests, and in none of them was heads flipped at least once, so the empirical probability can be zero, even though the theoretical probability is $75%$. Indeed, in the other post, the OP's simulations have garnered no successes, while the expected number of successes is almost $4$. – Cameron Buie Nov 01 '13 at 14:17
  • I simply wanted to make the distinction between certainty (when you have actually bought enough to have the full collection) and confidence (when you are theoretically likely to have bought enough). – Cameron Buie Nov 01 '13 at 14:18
  • Does this answer your question? Coupon collector problem for collecting set k times. – user1147844 Apr 07 '23 at 08:06

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