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

Jomo

Elite Member
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

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Denis

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

Jomo

Elite Member
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

Senior Member
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

Jomo

Elite Member
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.

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Denis

Senior Member
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"

Jomo

Elite Member
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

Elite Member
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

Full Member
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(X[SUB]1[/SUB]) = 0.5
E(X[SUB]2[/SUB]) = 0.5
E(X[SUB]3[/SUB]) = 0.5

So, since they are independent, E(X[SUB]1[/SUB]+X[SUB]2[/SUB]+X[SUB]3[/SUB]) = 1.5 >1, E(X[SUB]1[/SUB] +X[SUB]2[/SUB]) is not >1.

Is that too simple??

Jomo

Elite Member
I would think 3.

E(X[SUB]1[/SUB]) = 0.5
E(X[SUB]2[/SUB]) = 0.5
E(X[SUB]3[/SUB]) = 0.5

So, since they are independent, E(X[SUB]1[/SUB]+X[SUB]2[/SUB]+X[SUB]3[/SUB]) = 1.5 >1, E(X[SUB]1[/SUB] +X[SUB]2[/SUB]) is not >1.

Is that too simple??
The answer is some number between 2 and 3.

Romsek

Junior Member
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.

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

$$\displaystyle E[k]=e$$

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Jomo

Elite Member
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

Elite Member
Hi,
Suppose people randomly pick a number between 0 and 1.
This is meaningless until you specify a probability distribution.

Romsek

Junior Member
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.