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

[Steven G]

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

 

[Denis]

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

 

[Steven G]

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:shock:. Didn't you learn anything from this forum?

 

[Denis]

Of course not, there are more numbers between 1 and 10, than 0 and 1:shock:. Didn't you learn anything from this forum?

Hokay: how many numbers are there between 0 and 1? Zillions?
.1, .11, .111, ......, .111111111111111111111111111111111111111111

 

[Steven G]

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.

 

[Denis]

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"

 

[Steven G]

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?

 

[pka]

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 $\displaystyle N_1~\&~N_2$ choices then if $\displaystyle \{N_1,N_2\}\subset[0.5,1]$ then $\displaystyle N_1+N_2\ge1$ and $\displaystyle \mathcal{P}(N_1,N_2)=.25$.
BUT if $\displaystyle \{N_1,N_2\}\subset[0,0.5)$ then $\displaystyle N_1+N_2<1$, while still $\displaystyle \mathcal{P}(N_1,N_2)=.25$.

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

 

[Harry_the_cat]

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??

 

[Steven G]

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.

 

[Romsek]

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 $\displaystyle n$ uniform[0,1] addends to sum to greater than 1 is given by

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

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.

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

$\displaystyle E[k]=e$

 

[Steven G]

after toying with this for a while I think i've got it.

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

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

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.

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

$\displaystyle E[k]=e$

Yep, you got the correct result.

 

[HallsofIvy]

Hi,
Suppose people randomly pick a number between 0 and 1.

This is meaningless until you specify a probability distribution.

 

[Romsek]

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.