# 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 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"
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?

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

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



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
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??

10. 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.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•