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

1. Originally Posted by Jomo
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 $n$ uniform[0,1] addends to sum to greater than 1 is given by

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

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.

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

$E[k]=e$

2. Originally Posted by Romsek
after toying with this for a while I think i've got it.

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

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

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.

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

$E[k]=e$
Yep, you got the correct result.

3. Originally Posted by Jomo
Hi,
Suppose people randomly pick a number between 0 and 1.
This is meaningless until you specify a probability distribution.

4. Originally Posted by HallsofIvy
This is meaningless until you specify a probability distribution.
Come on Halls, you're better than that. Someone says randomly between zero and one and nothing else it's uniform[0,1].

If they come back and say, that's not what they meant then you can rail on them.

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