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Posted by Brian Callahan
on October 29, 2009, 7:08 pm
99.55.214.245
My son's math (calculus HS) teacher had a problem of the day, that got extra credit if the student could solve it, or at least work out part of it. I gave her last year a Problem based on the Flintstones variations, which she liked, but did not actually give me a solution. My older son worked out some probablities at the time which made sense, but I have now actually sorted through the Flintstones glasses that I have, so I have what actually resulted.
Your results may vary.
First the problem:
FLINTSTONES WELCHS GLASSES Problem of the Day
the 14 scenes with 8 different colors can produce a possible 112 glasses without regard to BOTTOMs...but with the bottoms considered in the equation, that number swells to 688 possible glasses...1962 glasses had 6 titles x 8 colors x 5 bottoms (dino and Bamm-bamm not made in 1962)...the 1963 and 1964 glasses totaled 8 titles x 8 colors x 7 bottoms...240 + 448=688...now, we are not sure that all colors or bottoms exist for each title, but it's certainly fun to search..
There are two groups of glasses in the collection.
POPULATION A has 262 glasses, all of which are different.
POPULATION B has 134 glasses, it is unknown whether they are different from one another,
And it is unknown if they are different than the glasses in Population A.
1.What are the chances of any one glass in Population B being different than every glass
In Population A? In other words the chance of any one Pop B glass NOT being a duplicate to the glasses in Population A?
2.How many glasses in Population B would be expected to be NOT duplicates, if you
Assume that Population B has glasses that are all different from each other?
3.How many glasses in Population B would be expected to be NOT duplicates, if you
Assume that Population B has glasses that may or may not be duplicates to each other?
ANSWER based on my Flintstones glasses collection:
1. 57/134 or 0.4253 - I got 57 different NEW glasses to add to the 262 different glasses that I already had sorted through. I got 77 glasses that were duplicates - about a dozen were multiple duplicates.
2. unknown - the second group that I sorted through probably (and in fact did) have some duplicates.
3. 57 - found 57 new glasses.
Once again, I do not claim this is the actual mathematical solution to the problem, just the results I obtained with my glasses. For better or worse I now have over 300 Flintstones glasses -
NOT EVEN HALF WAY to a complete (?) set. But a heck of a lot of 'stones in anybody's book.
NOTE: thanks to Deacon for the explanation of the existing variations, and thanks to Matt for the Where's Freddie checklist I used to sort them out.
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