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  1. #1
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    Expected value: Suppose people randomly pick a number between 0 and 1....

    Hi,
    Suppose people randomly pick a number between 0 and 1. What would be the expected number of people you would have to ask for numbers if you want the sum to be greater than 1 for the 1st time?
    Steve
    Last edited by Jomo; 12-05-2018 at 11:32 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

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    Would 1 to 10 and ending with > 10 be the same?
    I'm a man of few words...but I use 'em often!!

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    Quote Originally Posted by Denis View Post
    Would 1 to 10 and ending with > 10 be the same?
    Of course not, there are more numbers between 1 and 10, than 0 and 1. Didn't you learn anything from this forum?
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

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    Quote Originally Posted by Jomo View Post
    Of course not, there are more numbers between 1 and 10, than 0 and 1. Didn't you learn anything from this forum?
    Hokay: how many numbers are there between 0 and 1? Zillions?
    .1, .11, .111, ......, .111111111111111111111111111111111111111111
    I'm a man of few words...but I use 'em often!!

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    Quote Originally Posted by Denis View Post
    Hokay: how many numbers are there between 0 and 1? Zillions?
    .1, .11, .111, ......, .111111111111111111111111111111111111111111
    Between 0 and 1, I'd say about 100. Just like in the supermarkets. A zillion is out of my range.
    Last edited by Jomo; 12-05-2018 at 11:33 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

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    Well, your post is too mysterious for li'l ole me...
    going back to learning the "times 3" table; much tuffer than the "times 1"
    I'm a man of few words...but I use 'em often!!

  7. #7
    Quote Originally Posted by Jomo View Post
    Hi,
    Suppose people randomly pick a number between 0 and 1. What would be the expected number of people you would have to ask for numbers if you want the sum to be greater than 1 for the 1st time?
    Steve
    I would think 3.

    E(X1) = 0.5
    E(X2) = 0.5
    E(X3) = 0.5

    So, since they are independent, E(X1+X2+X3) = 1.5 >1, E(X1 +X2) is not >1.

    Is that too simple??
    Heavens to Murgatroyd!!

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    Quote Originally Posted by Harry_the_cat View Post
    I would think 3.

    E(X1) = 0.5
    E(X2) = 0.5
    E(X3) = 0.5

    So, since they are independent, E(X1+X2+X3) = 1.5 >1, E(X1 +X2) is not >1.

    Is that too simple??
    The answer is some number between 2 and 3.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  9. #9
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    Quote Originally Posted by Jomo View Post
    Hi,
    Suppose people randomly pick a number between 0 and 1. What would be the expected number of people you would have to ask for numbers if you want the sum to be greater than 1 for the 1st time?
    Steve
    after toying with this for a while I think i've got it.

    the probability it takes [tex]n[/tex] uniform[0,1] addends to sum to greater than 1 is given by

    [tex]\Large \displaystyle \int_0^1 \int_0^{1-x_1}\int_0^{1-x_1-x_2} \dots \int_0^{1-\sum \limits_{k=1}^{n-2}x_k}\int_{1-\sum \limits_{k=1}^{n-1}~x_k}^1~1~dx_n~dx_{n-1}~\dots dx_2~dx_1[/tex]

    This generates numbers in agreement with what I saw in the sim.

    This is the integral over the probability mass where the first (n-1) variables sum to less than one but the sum over all n sums to greater than 1.

    [tex]p[2]=\dfrac 1 2\\p[3]=\dfrac 1 3\\p[4] = \dfrac 1 8\\p[5]=\dfrac{1}{144} \\ p[k] = \dfrac{1}{k!+(k-1)!}[/tex]

    [tex]E[k]=e[/tex]
    Last edited by Romsek; 12-07-2018 at 04:10 AM.

  10. #10
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    Quote Originally Posted by Romsek View Post
    after toying with this for a while I think i've got it.

    the probability it takes [tex]n[/tex] uniform[0,1] addends to sum to greater than 1 is given by

    [tex]\Large \displaystyle \int_0^1 \int_0^{1-x_1}\int_0^{1-x_1-x_2} \dots \int_0^{1-\sum \limits_{k=1}^{n-2}x_k}\int_{1-\sum \limits_{k=1}^{n-1}~x_k}^1~1~dx_n~dx_{n-1}~\dots dx_2~dx_1[/tex]

    This generates numbers in agreement with what I saw in the sim.

    This is the integral over the probability mass where the first (n-1) variables sum to less than one but the sum over all n sums to greater than 1.

    [tex]p[2]=\dfrac 1 2\\p[3]=\dfrac 1 3\\p[4] = \dfrac 1 8\\p[5]=\dfrac{1}{144} \\ p[k] = \dfrac{1}{k!+(k-1)!}[/tex]

    [tex]E[k]=e[/tex]
    Yep, you got the correct result.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

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