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Math Help - Group porobability

  1. #1
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    Group porobability

    We choose two groups where group A can be of size X where X ranges from 1 to n-1. Group B is the remaining (n-X).
    All values of X are equally likely and all group sizes are equally likely

    If we choose one item from 1 to n where all choices are equally likely, what is the average size of the group containing the item?

    3. The attempt at a solution

    Originally I tried the following:

    Average size of group A is simply n/2 as it is uniformly distributed and therefore, since the item can be in either group, the size must also be n/2

    However, given that the item is more likely to be in the larger of the two groups, we must skew the average to greater than n/2 and indeed, when I run an Excel simulation, that proves correct. Any thoughts on this?

    The solution can be obtained form considering the first 1 to n-1 samples which must all be equally likely to be chosen

    This yields

    1/n(n-1) * [sum from 1 to (n-1) (x^2 + (n-x)^2)]

    and has been verified. Can be simplified further. There must be another way to do this though? A more elegant method perhaps?
    Last edited by Gekko; September 10th 2010 at 11:07 PM.
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  2. #2
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    Quote Originally Posted by Gekko View Post
    We choose two groups where group A can be of size X where X ranges from 1 to n-1. Group B is the remaining (n-X).
    All values of X are equally likely and all group sizes are equally likely

    If we choose one item from 1 to n where all choices are equally likely, what is the average size of the group containing the item?

    3. The attempt at a solution

    Originally I tried the following:

    Average size of group A is simply n/2 as it is uniformly distributed and therefore, since the item can be in either group, the size must also be n/2

    However, given that the item is more likely to be in the larger of the two groups, we must skew the average to greater than n/2 and indeed, when I run an Excel simulation, that proves correct. Any thoughts on this?
    Hmm I think I've got it although maybe in a roundabout way

    \displaystyle \sum_{k=1}^{n-1} \frac{2k}{n-1}\cdot\frac{\binom{n-1}{k-1}}{\binom{n}{k}}

    and although I haven't proven it, based on the numbers this simplifies to 1+\frac{2}{3}(n-2).

    The way I arrived was a bit complicated, involved using Bayes' theorem, if you want I can explain but maybe someone will post something simpler.
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