Mixing
Find overlapping outcomes in independent draws
$begingroup$
I have an urn containing 100 numbered balls. I randomly draw 5 balls (5 %) in 5 individual draws with replacement.
How many unique balls do I get? I.e., what is the number of unique values?
I've written some R code to approximate it, but I have no idea how to do this in a more 'mathematical' way. E.g., a formula of some sort.
draw <- function(size){
n <- 1:size
sa <- lapply(1:5,function(x) sample(n,size*0.05))
length(unique(unlist(sa)))/size
}
mean(unlist(lapply(1:1000,function(x) draw(100)))) # ~0.226
probability combinatorics probability-theory problem-solving
asked Jan 27 at 17:19
lezeleze
134
$endgroup$
add a comment |
$begingroup$
I have an urn containing 100 numbered balls. I randomly draw 5 balls (5 %) in 5 individual draws with replacement.
How many unique balls do I get? I.e., what is the number of unique values?
I've written some R code to approximate it, but I have no idea how to do this in a more 'mathematical' way. E.g., a formula of some sort.
draw <- function(size){
n <- 1:size
sa <- lapply(1:5,function(x) sample(n,size*0.05))
length(unique(unlist(sa)))/size
}
mean(unlist(lapply(1:1000,function(x) draw(100)))) # ~0.226
probability combinatorics probability-theory problem-solving
asked Jan 27 at 17:19
lezeleze
134
$endgroup$
$begingroup$
Are you after the expected number of unique values? If so, use indicator variables and linearity.
$endgroup$
– lulu
Jan 27 at 17:20
$begingroup$
Yes (edited question to clarify). Can you elaborate on your suggestion? I am not familiar with the terms.
$endgroup$
– leze
Jan 27 at 17:56
add a comment |
$begingroup$
I have an urn containing 100 numbered balls. I randomly draw 5 balls (5 %) in 5 individual draws with replacement.
How many unique balls do I get? I.e., what is the number of unique values?
I've written some R code to approximate it, but I have no idea how to do this in a more 'mathematical' way. E.g., a formula of some sort.
draw <- function(size){
n <- 1:size
sa <- lapply(1:5,function(x) sample(n,size*0.05))
length(unique(unlist(sa)))/size
}
mean(unlist(lapply(1:1000,function(x) draw(100)))) # ~0.226
probability combinatorics probability-theory problem-solving
asked Jan 27 at 17:19
lezeleze
134
$endgroup$
I have an urn containing 100 numbered balls. I randomly draw 5 balls (5 %) in 5 individual draws with replacement.
How many unique balls do I get? I.e., what is the number of unique values?
I've written some R code to approximate it, but I have no idea how to do this in a more 'mathematical' way. E.g., a formula of some sort.
draw <- function(size){
n <- 1:size
sa <- lapply(1:5,function(x) sample(n,size*0.05))
length(unique(unlist(sa)))/size
}
mean(unlist(lapply(1:1000,function(x) draw(100)))) # ~0.226
probability combinatorics probability-theory problem-solving
probability combinatorics probability-theory problem-solving
asked Jan 27 at 17:19
lezeleze
134
asked Jan 27 at 17:19
lezeleze
134
edited Jan 27 at 20:03
leze
asked Jan 27 at 17:19
lezeleze
134
asked Jan 27 at 17:19
lezeleze
134
asked Jan 27 at 17:19
lezeleze
134
134
$begingroup$
Are you after the expected number of unique values? If so, use indicator variables and linearity.
$endgroup$
– lulu
Jan 27 at 17:20
$begingroup$
Yes (edited question to clarify). Can you elaborate on your suggestion? I am not familiar with the terms.
$endgroup$
– leze
Jan 27 at 17:56
add a comment |
$begingroup$
Are you after the expected number of unique values? If so, use indicator variables and linearity.
$endgroup$
– lulu
Jan 27 at 17:20
$begingroup$
Yes (edited question to clarify). Can you elaborate on your suggestion? I am not familiar with the terms.
$endgroup$
– leze
Jan 27 at 17:56
$begingroup$
Are you after the expected number of unique values? If so, use indicator variables and linearity.
$endgroup$
– lulu
Jan 27 at 17:20
$begingroup$
Are you after the expected number of unique values? If so, use indicator variables and linearity.
$endgroup$
– lulu
Jan 27 at 17:20
$begingroup$
Yes (edited question to clarify). Can you elaborate on your suggestion? I am not familiar with the terms.
$endgroup$
– leze
Jan 27 at 17:56
$begingroup$
Yes (edited question to clarify). Can you elaborate on your suggestion? I am not familiar with the terms.
$endgroup$
– leze
Jan 27 at 17:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We will use Linearity of Expectation.
Toward that end, let $X_i$ denote the indicator variable for the $i^{th}$ ball. That is, $X_i=1$ if the $i^{th}$ ball is in your selection and $=0$ otherwise. The Linearity tell us that the answer we want, $E$, is $$E=Eleft[ sum X_iright]=sum E[X_i]=100times E[X_1]$$
Where, in the last equation, we used symmetry to conclude that all the $X_i$ have the same expected value.
To compute $E[X_1]$: note that this is simply $p$, the probability that we do in fact select ball $1$. The probability that we don't select it is $left(frac {99}{100}right)^5$ so $E[X_1]=p=1-left(frac {99}{100}right)^5approx .049$
It follows that our answer is $$Eapprox 4.9$$
answered Jan 27 at 18:11
lulululu
43.2k25080
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
We will use Linearity of Expectation.
Toward that end, let $X_i$ denote the indicator variable for the $i^{th}$ ball. That is, $X_i=1$ if the $i^{th}$ ball is in your selection and $=0$ otherwise. The Linearity tell us that the answer we want, $E$, is $$E=Eleft[ sum X_iright]=sum E[X_i]=100times E[X_1]$$
Where, in the last equation, we used symmetry to conclude that all the $X_i$ have the same expected value.
To compute $E[X_1]$: note that this is simply $p$, the probability that we do in fact select ball $1$. The probability that we don't select it is $left(frac {99}{100}right)^5$ so $E[X_1]=p=1-left(frac {99}{100}right)^5approx .049$
It follows that our answer is $$Eapprox 4.9$$
answered Jan 27 at 18:11
lulululu
43.2k25080
$endgroup$
add a comment |
$begingroup$
We will use Linearity of Expectation.
Toward that end, let $X_i$ denote the indicator variable for the $i^{th}$ ball. That is, $X_i=1$ if the $i^{th}$ ball is in your selection and $=0$ otherwise. The Linearity tell us that the answer we want, $E$, is $$E=Eleft[ sum X_iright]=sum E[X_i]=100times E[X_1]$$
Where, in the last equation, we used symmetry to conclude that all the $X_i$ have the same expected value.
To compute $E[X_1]$: note that this is simply $p$, the probability that we do in fact select ball $1$. The probability that we don't select it is $left(frac {99}{100}right)^5$ so $E[X_1]=p=1-left(frac {99}{100}right)^5approx .049$
It follows that our answer is $$Eapprox 4.9$$
answered Jan 27 at 18:11
lulululu
43.2k25080
$endgroup$
add a comment |
$begingroup$
We will use Linearity of Expectation.
Toward that end, let $X_i$ denote the indicator variable for the $i^{th}$ ball. That is, $X_i=1$ if the $i^{th}$ ball is in your selection and $=0$ otherwise. The Linearity tell us that the answer we want, $E$, is $$E=Eleft[ sum X_iright]=sum E[X_i]=100times E[X_1]$$
Where, in the last equation, we used symmetry to conclude that all the $X_i$ have the same expected value.
To compute $E[X_1]$: note that this is simply $p$, the probability that we do in fact select ball $1$. The probability that we don't select it is $left(frac {99}{100}right)^5$ so $E[X_1]=p=1-left(frac {99}{100}right)^5approx .049$
It follows that our answer is $$Eapprox 4.9$$
answered Jan 27 at 18:11
lulululu
43.2k25080
$endgroup$
We will use Linearity of Expectation.
Toward that end, let $X_i$ denote the indicator variable for the $i^{th}$ ball. That is, $X_i=1$ if the $i^{th}$ ball is in your selection and $=0$ otherwise. The Linearity tell us that the answer we want, $E$, is $$E=Eleft[ sum X_iright]=sum E[X_i]=100times E[X_1]$$
Where, in the last equation, we used symmetry to conclude that all the $X_i$ have the same expected value.
To compute $E[X_1]$: note that this is simply $p$, the probability that we do in fact select ball $1$. The probability that we don't select it is $left(frac {99}{100}right)^5$ so $E[X_1]=p=1-left(frac {99}{100}right)^5approx .049$
It follows that our answer is $$Eapprox 4.9$$
answered Jan 27 at 18:11
lulululu
43.2k25080
edited Jan 27 at 18:32
answered Jan 27 at 18:11
lulululu
43.2k25080
answered Jan 27 at 18:11
lulululu
43.2k25080
answered Jan 27 at 18:11
lulululu
43.2k25080
43.2k25080
add a comment |
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$begingroup$
Are you after the expected number of unique values? If so, use indicator variables and linearity.
$endgroup$
– lulu
Jan 27 at 17:20
$begingroup$
Yes (edited question to clarify). Can you elaborate on your suggestion? I am not familiar with the terms.
$endgroup$
– leze
Jan 27 at 17:56