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

1. 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

2. Would 1 to 10 and ending with > 10 be the same?

3. Originally Posted by Denis
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?

4. Originally Posted by Jomo
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

5. Originally Posted by Denis
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.

6. 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"

7. 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
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??

8. Originally Posted by Harry_the_cat
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.

9. 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.

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

$E[k]=e$

10. 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.

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