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

  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!!

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    Quote Originally Posted by Denis View Post
    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"
    Asked 1st person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 (which is impossible at this stage) stop. Otherwise,
    Asked 2nd person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 stop. Otherwise,
    Asked 3rd person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 stop. Otherwise,
    .
    .
    .
    What is the expected number of people you would have to ask?
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  8. #8
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    Quote Originally Posted by Jomo View Post
    Asked 1st person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 (which is impossible at this stage) stop. Otherwise,
    Asked 2nd person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 stop. Otherwise,
    Asked 3rd person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 stop. Otherwise,
    .
    .
    .
    What is the expected number of people you would have to ask?
    Suppose that [tex]N_1~\&~N_2[/tex] choices then if [tex]\{N_1,N_2\}\subset[0.5,1][/tex] then [tex]N_1+N_2\ge1[/tex] and [tex]\mathcal{P}(N_1,N_2)=.25[/tex].
    BUT if [tex]\{N_1,N_2\}\subset[0,0.5)[/tex] then [tex]N_1+N_2<1[/tex], while still [tex]\mathcal{P}(N_1,N_2)=.25[/tex].

    However it gets even more messy. What if [tex]N_1=0.2~\&~N_2=0.9[/tex] with the same probability, [tex](N_1+N_2)>1[/tex].
    What if [tex]N_1=0.2~\&~N_2=0.7[/tex] with the same probability, [tex](N_1+N_2)<1[/tex]
    I am saying the the modeling here stumps me. I'll sleep on it.

    [tex][/tex][tex][/tex][tex][/tex][tex][/tex][tex][/tex][tex][/tex]
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  9. #9
    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

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