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

Jomo

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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

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

Jomo

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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

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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

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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

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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

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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

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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

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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??
 

Jomo

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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

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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

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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

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

Romsek

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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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