24 Variants in blind packaging

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djtrepp

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Math problem: Albertsons is offering 24 different Micro Pops in blind packaging. Assuming all produced in equal number, how many blind packages do I need to collect to get all 24?
 
Sorry, I dont realky know where to start..would it be: 24/24 + 23/24 + 22/24...?
 
djtrepp said:
Math problem: Albertsons is offering 24 different Micro Pops in blind packaging. Assuming all produced in equal number, how many blind packages do I need to collect to get all 24?
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If this is a problem you were given, I don't think you've quoted it fully. If it is your own invention, you'll need to clarify it.

Does "blind packaging" mean that a package contains one MicroPop (whatever that is -- it sounds like something small that would come more than one to a package), but the package doesn't indicate what flavor/kind it is? Or are there more than one?

Are you looking for the number you need to buy to be certain that you will get all 24 types, or perhaps for the expected number you need to buy? The wording seems to imply certainty, but that would be impossible as I read it.
 
It is a real world problem. 24 variants. Assume equal production and distribution. One variant per blind package. I want to know the minimum number of packages that I would need to buy to ensure to some level of confidence that I get all 24.
 
If I understand your question then the question can not be answered.

Suppose that 1 of the 24 pops is extremely rare. Suppose there are 1 million pops and there is only 1 of this rare one. Then you would have to buy 1 million pops to be assured that you get the rare one (you need this one before you have a chance on having all of them). On the other hand if all the pops are evenly distributed it will take less than 1 million purchases.
 
djtrepp said:
It is a real world problem. 24 variants. Assume equal production and distribution. One variant per blind package. I want to know the minimum number of packages that I would need to buy to ensure to some level of confidence that I get all 24.
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You said that the variants are equally distributed (equally likely), so Jomo's concern is moot.

Your question is related to the Coupon Collector's Problem. Unfortunately, that asks for the expected number you need in order to collect all, rather than what you are asking, the number to collect so that the probability of getting all is greater than some threshold. I'm not sure how to find that. But the link may be of use, depending on your specific need.
 
Dr.Peterson said:
You said that the variants are equally distributed (equally likely), so Jomo's concern is moot.

Your question is related to the Coupon Collector's Problem. Unfortunately, that asks for the expected number you need in order to collect all, rather than what you are asking, the number to collect so that the probability of getting all is greater than some threshold. I'm not sure how to find that. But the link may be of use, depending on your specific need.
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This is exactly the answer I was looking for!
 
I think that the answer depends on how many pops are there. For example, if there are exactly one of each then you will need to pick 24.
Suppose there are two of each. The worst case scenario would be to get 2 of the 1st kind, then 2 of the 2nd kind,..., then 2 of the 23 kind and then 1 of the 24th kind. This takes 47 picks.
 
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Jomo said:
I think that the answer depends on how many pops are there. For example, if there are exactly one of each then you will need to pick 24.
Suppose there are two of each. The worst case scenario would be to get 2 of the 1st kind, then 2 of the 2nd kind,..., then 2 of the 23 kind and then 1 of the 24th kind. This takes 47 picks.
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I believe the assumption (based on the fact that these are products sold in a store, not a set held in a single jar, is that there are perhaps millions of each type, not one or two. And that is the assumption in the problem I referred to.
 
Dr.Peterson said:
I believe the assumption (based on the fact that these are products sold in a store, not a set held in a single jar, is that there are perhaps millions of each type, not one or two. And that is the assumption in the problem I referred to.
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Yes, I t0o believe that there are millions. I just showed, using small numbers, that the answer depends on how many there are. Are you suggesting that for some number N and beyond that the answer will be constant?
 
It is standard to consider a very large population as effectively infinite, so that individual probabilities can be thought of as independent. That is, there is a point beyond which you can ignore effects imposed by the population, because the errors introduced are far smaller than other possible sources of error (especially the approximate nature of the formula in the page I referred to). Emphasizing the effects of small numbers doesn't particularly help in solving the actual problem.
 
djtrepp said:
So for n=24 variants, t = 91 blind packages
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Correct, whether you got that from the graph they show or applying the formula (and rounding up).

Of course, this doesn't tell us how likely it is that you will actually need more; but in the long run, you will on average have all 24 when you get to 91 packages.
 
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