I might have been able to figure this out after crushing my college statistics final many years ago but am not even going to even attempt it now.
Lets say there are an unknown but finite quantity of different items. You have a random sampling of 600 of them. You are given another 50 of them of which that sample is also random. Going through them you find that you have 38 of them already and 12 are different from the original 600. How many total exist?
To put this in a beer can perspective and to keep it from getting moved to the "tricky math problems nobody cares about" category, think Rainier Jubilees. The 600 is the approximate size of my collection. If you gave me another 50 (please do), about a dozen would be new. I base that number on instances like this weekend where I sorted through a case of dumpers and 5 or 6 were new. I realize my case sample was not entirely random but nobody wants to get into higher level statistical analysis that just makes your head hurt. The 600 came from all sorts of different places and can be considered random or very close.
The 600 number are all production cans so oddball wind tunnel colors that were never produced do not figure into the equation. Half quarts and zips are also not included.
My guess is the answer is somewhere under 1,000 which would be considerably less than most estimates I've heard. Anybody up to the task?
Any statistics experts out there?
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Any statistics experts out there?
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Re: Any statistics experts out there?
Statistics make my head feel like and then . So I and where I do my best usually followed by and , and again . Now I need some , when I . I ponder what came first the or the egg. Instead of going out in public and I just quietly So I say lets just call it .And have a

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Re: Any statistics experts out there?
As it has been 40 years since my last stats class, I cannot easily help you. However, my experience in the real world has taught me that samples are seldom random. For instance, it is clear that not all cans were created in the same quantity. That would mean that you may have essentially all of the very common cans, but very few of the potentially unlimited extremely rare cans. There is also a way to look at every incremental 50 cans and estimate how many total cans there would be by looking at the decay rate of new cans. Still beyond what I can remember from long, long ago.
If you really need an answer, my soninlaw teaches stats at a community college; I can enlist his help if necessary.
Regardless of what you may come up with, the USBC Supplement is planning to build out the structure to cover all the permutations regardless of whether any have been found. Similar for the Esslinger cans.
Good luck in your search.
If you really need an answer, my soninlaw teaches stats at a community college; I can enlist his help if necessary.
Regardless of what you may come up with, the USBC Supplement is planning to build out the structure to cover all the permutations regardless of whether any have been found. Similar for the Esslinger cans.
Good luck in your search.
Jerry Cole

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Re: Any statistics experts out there?
You raise a good point about my collection maybe not being entirely random due to certain cans being less common than others. For the sake of this exercise though I think a number of assumptions are necessary and the randomness of my collection and the sample has to be one of those. I've sorted through enough case lots to know that's about the number I'll add versus this being a one time shot and the next time I might need 25.
This is not intended to answer once and for all a critical question that has been plaguing mankind since the early 1960's. I just thought it would be fun and interesting to take a run at it from a different angle rather than trying to figure out all the permutations. Premium Bill did that and arrived at 10,000+ which is ridiculously high.
If your son inlaw is willing to take a stab at it, that would be great. It's probably a simple problem for somebody with the expertise.
This is not intended to answer once and for all a critical question that has been plaguing mankind since the early 1960's. I just thought it would be fun and interesting to take a run at it from a different angle rather than trying to figure out all the permutations. Premium Bill did that and arrived at 10,000+ which is ridiculously high.
If your son inlaw is willing to take a stab at it, that would be great. It's probably a simple problem for somebody with the expertise.
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Re: Any statistics experts out there?
Let me take a stab at this one …
First of all, let me round your numbers off a bit Scott. Let's say you have exactly 600 different Rainier Jubilees. Let's also assume that all the Rainier Jubilees are equally likely to occurthat is, if you could randomly select a Rainier Jubilee at random from the universe of ALL Rainier Jubilees, every variation would have the same probability of being selected as every other one. You mentioned that in a "random case" of Jubilees, you might expect to get 5 or 6 that you do not already have. Let me round that number off a bit and assume that right now, your probability of selecting a Jubilee that you do not already have is 20% (which is close to 5 in 24). We can write this out this waythe number
(The number of Jubilees you do not already have) divided by (The total number of Jubilees)
is 20%, or 0.2. Now allow me to use just a tiny bit of algebraic symbolism (sorry!) … let N be the total number of Jubilees, and n be the number of Jubilees you do not already have. Then n + 600 would be equal to N, and what we get above is
0.2 = (n) / (N), or
0.2 = (n) / (n + 600).
If you solve this equation for n, you get n = 150. This would mean that the total number of Jubilees is N = n + 600 = 750.
Incidentally, that last equation would work the same way given a different probability for getting a "new Jubilee". For example, if you assume your probability of getting a new one right now is 25% instead of 20%, it gives you a total number of Jubilees of 800 instead of 750. (You could make this into a formula, but I'll resist that given that it would have two variables in it and would probably scare the nonalgebraically inclined even more than they already are … )
Now, as Uncle Jackson has correctly pointed out, here's the rub … not all the Jubilees are equally likely!! It is very obvious that the 15 and 16 oz. cans are much harder than the 11 and 12 oz. cans. Even if you peel those off and only consider the 11/12 oz. cans, it is clear these cans are not equally likely either. From the Wind Tunnel find I got my first "flesh" colored cansbut only two of them. Do any others exist? If they do, they are not nearly as common as the other colors. I am reasonably sure there are combinations of Seattle/Spokane and exact brewery name that are more scarce than othersand I don't know enough about those to know what they are, because I don't save them that way.
You could probably construct a mathematical model of the Jubilees given some sort of a distribution of the expected scarcity of some of the colors, cities, and breweries that exist, and derive a reasonable estimate of the total number of Jubilees. But that would be a very difficult mathematical problem!! On the other hand, I have taught probability and statistics virtually every semester at the college level for the last 24 years, so if you wanted to spell out the problem that way, I'd take a stab at it, just because it's an interesting problem ...
First of all, let me round your numbers off a bit Scott. Let's say you have exactly 600 different Rainier Jubilees. Let's also assume that all the Rainier Jubilees are equally likely to occurthat is, if you could randomly select a Rainier Jubilee at random from the universe of ALL Rainier Jubilees, every variation would have the same probability of being selected as every other one. You mentioned that in a "random case" of Jubilees, you might expect to get 5 or 6 that you do not already have. Let me round that number off a bit and assume that right now, your probability of selecting a Jubilee that you do not already have is 20% (which is close to 5 in 24). We can write this out this waythe number
(The number of Jubilees you do not already have) divided by (The total number of Jubilees)
is 20%, or 0.2. Now allow me to use just a tiny bit of algebraic symbolism (sorry!) … let N be the total number of Jubilees, and n be the number of Jubilees you do not already have. Then n + 600 would be equal to N, and what we get above is
0.2 = (n) / (N), or
0.2 = (n) / (n + 600).
If you solve this equation for n, you get n = 150. This would mean that the total number of Jubilees is N = n + 600 = 750.
Incidentally, that last equation would work the same way given a different probability for getting a "new Jubilee". For example, if you assume your probability of getting a new one right now is 25% instead of 20%, it gives you a total number of Jubilees of 800 instead of 750. (You could make this into a formula, but I'll resist that given that it would have two variables in it and would probably scare the nonalgebraically inclined even more than they already are … )
Now, as Uncle Jackson has correctly pointed out, here's the rub … not all the Jubilees are equally likely!! It is very obvious that the 15 and 16 oz. cans are much harder than the 11 and 12 oz. cans. Even if you peel those off and only consider the 11/12 oz. cans, it is clear these cans are not equally likely either. From the Wind Tunnel find I got my first "flesh" colored cansbut only two of them. Do any others exist? If they do, they are not nearly as common as the other colors. I am reasonably sure there are combinations of Seattle/Spokane and exact brewery name that are more scarce than othersand I don't know enough about those to know what they are, because I don't save them that way.
You could probably construct a mathematical model of the Jubilees given some sort of a distribution of the expected scarcity of some of the colors, cities, and breweries that exist, and derive a reasonable estimate of the total number of Jubilees. But that would be a very difficult mathematical problem!! On the other hand, I have taught probability and statistics virtually every semester at the college level for the last 24 years, so if you wanted to spell out the problem that way, I'd take a stab at it, just because it's an interesting problem ...

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Re: Any statistics experts out there?
Didn't want to make one post too overwhelming so I've split it in half … I'll take off my mathematician's hat now and put on my beer can collector's hat instead.
I've been collecting Jubilees for close to 40 years now myself. I do save the three major sets and color variations of each. I collect the ziptabs as well as flats. I also collect the gold band variations (blank or not). I would save the Christmas cans, but have never managed to get one … What I do not collect is the different brewery (Sicks' Rainier vs. Sicks' Spokane vs. Sicks' Seattle) or city (Seattle vs. Spokane) variations, nor do I save both 11 and 12 oz., if the cans are otherwise the same.
I have managed to amass a total of 477 different over the years. Of these, 33 are 12 oz. zips and 10 are 16 oz. zips, and 58 of the others are either 15 or 16 oz. In order to guesstimate how many 11/12 oz. flattop variations actually exist, here's what I've got so far:
55 from the Cartoon set
89 from the Party set with blank gold band
151 from the Party set with "Rainier for Life"
24 from the Brewery set with blank gold band (this set is a challenge!)
57 from the Brewery set wtih "Rainier for Life"
So that's a total of 376 11/12 oz. cans without accounting for city or brewery variations. I'm assuming Scott does save these given his interests.
As I've accumulated more of the Wind Tunnel cans, almost all of them being Party/"Rainier for Life" cans, I've noticed that there are 15 colors that are well represented in my collection: Black, deep purple, "pinkish purple" (definitely different from "pink"), violet, red, redorange, yellow, enamel yellow green, metallic green, enamel green, turquoise, metallic blue, enamel blue, pink, and gray. I have about 80% of the color/design combinations in these 15 colors, and I'm guessing that all of them exist. I have just a couple other colors (metallic blue green, brown, and "flesh") with a grand total of 4 cans. I am guessing these probably only come in a couple colors each. I can't document a few other colors listed in the Wolpe Catalog for this set, like orange, metallic yellow green, enamel blue green, and metallic bronzethough these colors definitely exist in other sets.
If I were going to make an educated guess how many 11/12 oz., flat top Jubilees exist without accounting for brewery or city differences, I'd guesstimate probably about 550600 or so as maybe an upper estimate. I'll leave it to Scott or to others who know a lot more than I do how many exist if you take all the variations into account. If Scott's estimate of under 1000 is too low, I'd be pretty confident it's under 1500.
I've been collecting Jubilees for close to 40 years now myself. I do save the three major sets and color variations of each. I collect the ziptabs as well as flats. I also collect the gold band variations (blank or not). I would save the Christmas cans, but have never managed to get one … What I do not collect is the different brewery (Sicks' Rainier vs. Sicks' Spokane vs. Sicks' Seattle) or city (Seattle vs. Spokane) variations, nor do I save both 11 and 12 oz., if the cans are otherwise the same.
I have managed to amass a total of 477 different over the years. Of these, 33 are 12 oz. zips and 10 are 16 oz. zips, and 58 of the others are either 15 or 16 oz. In order to guesstimate how many 11/12 oz. flattop variations actually exist, here's what I've got so far:
55 from the Cartoon set
89 from the Party set with blank gold band
151 from the Party set with "Rainier for Life"
24 from the Brewery set with blank gold band (this set is a challenge!)
57 from the Brewery set wtih "Rainier for Life"
So that's a total of 376 11/12 oz. cans without accounting for city or brewery variations. I'm assuming Scott does save these given his interests.
As I've accumulated more of the Wind Tunnel cans, almost all of them being Party/"Rainier for Life" cans, I've noticed that there are 15 colors that are well represented in my collection: Black, deep purple, "pinkish purple" (definitely different from "pink"), violet, red, redorange, yellow, enamel yellow green, metallic green, enamel green, turquoise, metallic blue, enamel blue, pink, and gray. I have about 80% of the color/design combinations in these 15 colors, and I'm guessing that all of them exist. I have just a couple other colors (metallic blue green, brown, and "flesh") with a grand total of 4 cans. I am guessing these probably only come in a couple colors each. I can't document a few other colors listed in the Wolpe Catalog for this set, like orange, metallic yellow green, enamel blue green, and metallic bronzethough these colors definitely exist in other sets.
If I were going to make an educated guess how many 11/12 oz., flat top Jubilees exist without accounting for brewery or city differences, I'd guesstimate probably about 550600 or so as maybe an upper estimate. I'll leave it to Scott or to others who know a lot more than I do how many exist if you take all the variations into account. If Scott's estimate of under 1000 is too low, I'd be pretty confident it's under 1500.