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Thread: Probability that X restaurants will be visited if Y people choose independently

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    Probability that X restaurants will be visited if Y people choose independently




    I am trying to determine for Y people selecting from X restaurants (with an equal probability) what the expected value of the number of restaurants chosen would be. All decisions are independent.

    Ex: 2 people and 2 restaurants - E(X) = 1(.5) + 2(.5) = 1.5 restaurants will be visited on average
    EX: 3 people and 3 restaurants - E(X) = 1(1/9) + 2(6/9) + 3(2/9) = 2.11 restaurants will be visited on average

    This is simple enough for small values like these. However, I would like to be able to do this for any combination of X and Y and create a table of expected values. For larger numbers I am having trouble determining the probabilities to plug into the expected value formula. When calculating the probabilities, I know that the denominator will always be X^Y but I am really struggling to find a way to calculate the numerator without writing down every possible combination.

    My ultimate goal would be to create a table in excel that for any given number of restaurants and people, returns the expected value.

    I'm running into a major road block so any help would be appreciated.

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    Re: Probability that X restaurants will be visited if Y people choose independently

    Let
    n be the total number of restaurants.
    X be the number of restaurants visited by at least 1 person
    k be the total number of people
    B_i, i = 1, 2, \ldots, n be the indicators of the i-th restaurant being visited by at least 1 person

    If you are only interested in the expected value E[X], we can try to calculate it without actually calculating the pmf of X. The trick is to consider the following decomposition, having the same spirit as the one we do in calculating the expected value of hypergeometric distribution:

    X = \sum_{i=1}^n B_i

    Note those B_i are not independent; but this does not matter - all we need is the linearity of the expectation. Further note that

    E[B_i] = \Pr\{B_i = 1\} = 1 - \left(1 - \frac {1} {n}\right)^k

    and therefore

    E[X] = nE[B_1] = n\left[1 - \left(1 - \frac {1} {n}\right)^k \right]

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    Re: Probability that X restaurants will be visited if Y people choose independently


    That worked perfectly. Thank you! Glad I asked because I was not headed in the right direction.

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