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  1. #1

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    Resolved [RESOLVED] Portions

    I need another brain on this one to generalize a specific case, and I'm having some difficulty putting it into words.

    We mark fish in a variety of ways. Normally, there is one or two marks, but there could be more. The quality of each type of mark is independent of any other type of mark, but can be categorized as good or bad. Therefore, if you have 10,000 fish marked with marks A and B, then it might be that 90% of the fish have mark A (90% good, 10% bad for that mark), while 99% of the fish have mark B (99% good, 1% bad for that mark). I believe this means that of the 10,000, there will be 8910 with both A and B, 910 will have just B, 90 will have just A, and 90 will have neither. I think that's right, anyways.

    I need to categorize these as good, partial, or bad, so I would make it to be 8910 good, 1000 partial, and 90 bad.

    What I want to do is generalize this for N marks. So, given different rates of good and bad for each mark, what is the number that has all the marks, and what is the number that has none of the marks. The rest ends up as partial. Is there any calculation that simplifies this? I don't believe that N can go above 5, and probably won't go above 3, so working it out for each would be possible, but I'd rather generalize for all N.
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  2. #2
    Super Moderator RobDog888's Avatar
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    Re: Portions

    So the fish will have A and B, but never A or B only correct? Like if its marked as A it could also have a B or if bad B it will only have an A

    90% of 10,000 fish with mark A = 9,000 (90% or 8,910 of this is good marks and 10% or 90 bad or no mark)

    99% of 10,000 fish with mark B = 9,900 (99% or 9,900 of this is good marks and 1% or 100 bad or no mark)

    I dont think you can say 190 will have bad or no mark as the 90 bad or no mark from A could be the same 90 bad or no mark that make up B's 100? Correct?

    I believe this means that of the 10,000, there will be 8910 with both A and B, 910 will have just B, 90 will have just A, and 90 will have neither. I think that's right, anyways.
    Think your logic is off there?
    8910 with both A and B I believe this is correct. but to break down the remainder is not correct as there is no separate percentage for bad vs unmarked, correct?


    Trying to get the analysis down here first before the forumlas as need a control to check against to determine of the formula is correct outputs
    Last edited by RobDog888; Jun 8th, 2017 at 06:08 PM.
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  3. #3

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    Re: Portions

    Actually, the fish could have any number of marks, so it is entirely possible for some fish to just have A and some fish to just have B. They might also have C, or A, B and C, or many other combinations. However, when checking mark retention, it will always be for A, or B, or C, etc. The marks are independent of each other, so retention is independent as well.

    You are right that it would be possible for the 100 fish that shed mark B to be entirely part of the 900 that shed A, though that is unlikely. The chance of shedding A is independent of the chance of shedding B, so the odds that one is a subset of the other is pretty slim.

    So, I was saying that all 10,000 were supposed to get both A and B. If all had been perfect, then there would be 10,000 with both A and B. However, 10% of the A shed the tag, which means that only 9,000 of the 10,000 had A. Also, 1% of the B shed the tag, so of the 10,000, 9,900 had B. I also felt it was correct that to say that of any subset of the fish, the same percentages applied (this may not be true, but we wouldn't know, so we would have to assume that it was true). Therefore, of the 1000 that shed the A tag, they all should have had the B tag, but 1% of them did not. This means that of the 1000 that shed the A tag, 990 had the B tag and 10 did not have the B tag. So, it looks like I did screw up the math.

    Of the 10,000 fish, 9,000 had the A tag. Of those 1% would have shed the B tag, so there would be 8910 with both A and B, while 90 had shed the B tag and were A only.

    Of the 10,000 fish, 9900 had the B tag, but of that, 10% would have shed the A tag, so there would be 8910 with both A and B, while 990 would have shed the A tag and would be B only.
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    Super Moderator RobDog888's Avatar
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    Re: Portions

    Sounds correct enough for me. But honestly,... too many damn fish!!! Time for a fish fry of epic proportions!!
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    Re: Portions

    When they get those marks, they're pretty small. We generally tried to mark them when they were around 100 fish per pound, so it would take a goodly number to make a meal, even if you ate them whole.

    Anyways, that math may be right, but I'm not so clear on generalizing the pattern out for more than two marks. It does seem like I kind of laid out a pattern as to how to do it, and that may be good enough, but I was hoping for better.
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  6. #6

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    Re: Portions

    Actually, after thinking over what I want to do with this, I greatly simplified it. The only thing I need to know is the number that (probably) had all the marks. The partial and bad groups aren't really all that useful. Therefore, the calculation becomes a whole lot easier, because the percent good for the whole set of marks is the product of the percent good for each of the individual marks.
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