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  1. #1
    Groinlonger
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    BG Math Contest [Not Someone's Homework]

    First one to answer the question correctly wins.

    An occurrence occurs at a rate r. How many occurrences must occur for the occurrence of the occurrences to occur within +/-S of r n% of the time.

  2. #2
    Bagel
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  3. #3

    YouTube - Willy Wonka Computer

    "I won't tell, that would be ...cheating."

  4. #4
    Groinlonger
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    Although I know your answer isn't serious, I might as well state that the answer will be in the form of an equation that relate r, S, n%, and some number of occurrences.

    Also, maybe to be more clear, the occurrence is either true or false.

  5. #5
    Campaign
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    r = (n%/S)/S

  6. #6
    Groinlonger
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    Not quite, keep on trying though~

  7. #7
    Black Belt
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    I like this joke better when it involved woodchucks

  8. #8
    Death by snoo snoo
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    r/0

  9. #9
    Relic Shield
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    Quote Originally Posted by Mojo View Post
    First one to answer the question correctly wins.

    An occurrence occurs at a rate r. How many occurrences must occur for the occurrence of the occurrences to occur within +/-S of r n% of the time.
    the way you've stated it there is nothing to solve

    the occurrence of the occurrences is already r, so it's already within +/-S of r, whatever S is, 100% of the time

    plz state the assumptions more clearly

  10. #10
    Relic Shield
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    ok I think I deciphered what you're asking

    "every time an experiment is performed a certain outcome occurs with probability r

    how many times must the experiment be performed in order to be n% sure that the actual observed ratio of occurrences/(# of experiments) will be within +/-S of r"

    this reeks of the binomial distribution:

    the probability of having exactly k occurrences in m experiments (where the event occurs with probability p in each experiment) is given by the formula

    P(exactly k occurrences in m experiments) =
    p^k*(1-p)^{m-k} * m!/[k!(m-k)!]

    but we don't want the answer to be exact, we wanted it to be within a specific range, so we need the cumulative distribution function. it gives the probability of having less than or equal to k occurrences in m trials. to get this just calculate the probability that it's exactly 1, exactly 2, ..., exactly k, and add them all together:

    P(occurrences <= k in m experiments) =
    sum from i = 1 to [[ (p^i * (1-p)^{n-i} * m!/[i!(m-i)!] ]]

    we can plug non-integers in for k, but then we just round down.
    to get (k-s <= # of occurences in m trials <= k + s),
    we just do P(occurences <= k+s) - P(occurences < k-s)
    (or if this trick is too fancy, just think of it as adding up the probability for occurrences exactly equal to each integer between k-s and k+s, rounding down from k+s and up from k-s if either of them are not integers)

    now we have the formulas, we just need to figure out what gets plugged in where.

    we want the P[occurrences in m trials within +/-m*S of m*r (which is the number of occurrences we "expect" in m trials) ] = n%

    so we plug all the variables in to our formula and solve for m:

    sum from j = ceil(mr-mS) to floor(mr+mS) of
    r^j * (1 - r)^{m-j} * m!/[j!(m-j)!] = n/100

    the problem is that this sucker is impossible to solve unless some specific values are given for some of the variables and those values happen to nicely simplify things. in actual practice it would probably be better to approximate with a normal distribution

    if you can somehow get a clean formula for the answer i'd like to see that...

  11. #11

    ▲▲

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    Inb4 w∞zie

  12. #12
    Ridill
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    Are you asking about the rate of occurrence as measured by someone in a lab who has no idea what the true rate is and can only extrapolate from the data he records, or about the true rate that it really is, regardless of observations?

  13. #13
    Sea Torques
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    42

  14. #14
    Cerberus
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    Quote Originally Posted by Ensam View Post
    42
    I'm gonna go with this person. As the computer on Hitchiker's Guide to the galaxy says this is the answer to Life, the universe, and everything. As does google. And you saying it's wrong would mean that the internet and a book/movie are wrong. And that never happens. So obviously Ensam wins.

  15. #15

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