Birthday problem variation with balls and bins
$begingroup$
Balls are thrown randomly and uniformly into $n$ bins up until one bin has 3 balls.
Let there be $T=T(n)$ be the number of throws we made until the
occasion occurs. I am to assume $n$ is large and requested to find
$f(n)$ so that the probability:
$$ P(T > 0.1f(n))$$ Is close to 1 (for example bigger than 0.9999)
and the probability: $$ P(T > 10f(n))$$ Is close to 0 (for example
smaller than 0.001)
Our hint was to investigate the random variable, when throwing $m$
balls $X(m)$ is the number of triplets $[i,j,k] subset [1,2,...,m]$
so that the $i,j,k$ balls thrown fall in the same bin.
I have tried doing something similar to the birthday problem by defining a uniform random variable $X_i$ over $i in [1,...,n]$ so its basically if there is a ball in the $i$ bin. then I made another variable $$Y_i,_j,_k =begin{cases} 1, & text{$X_i = X_j = X_k$} \0, & text{otherwise}end{cases}$$
And then went on to calculate the expected value.
Overall I received the result ${m choose 3}*frac{1}{n^2}$ but the results don't match the definition of $X(m)$.
I am really kind of lost as this was the main way I felt might work, where am I wrong in this, whether its my way of thought or just my math, all help would be appreciated.
p.s. this is a homework question but the due date already passed
probability balls-in-bins
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
$endgroup$
add a comment |
$begingroup$
Balls are thrown randomly and uniformly into $n$ bins up until one bin has 3 balls.
Let there be $T=T(n)$ be the number of throws we made until the
occasion occurs. I am to assume $n$ is large and requested to find
$f(n)$ so that the probability:
$$ P(T > 0.1f(n))$$ Is close to 1 (for example bigger than 0.9999)
and the probability: $$ P(T > 10f(n))$$ Is close to 0 (for example
smaller than 0.001)
Our hint was to investigate the random variable, when throwing $m$
balls $X(m)$ is the number of triplets $[i,j,k] subset [1,2,...,m]$
so that the $i,j,k$ balls thrown fall in the same bin.
I have tried doing something similar to the birthday problem by defining a uniform random variable $X_i$ over $i in [1,...,n]$ so its basically if there is a ball in the $i$ bin. then I made another variable $$Y_i,_j,_k =begin{cases} 1, & text{$X_i = X_j = X_k$} \0, & text{otherwise}end{cases}$$
And then went on to calculate the expected value.
Overall I received the result ${m choose 3}*frac{1}{n^2}$ but the results don't match the definition of $X(m)$.
I am really kind of lost as this was the main way I felt might work, where am I wrong in this, whether its my way of thought or just my math, all help would be appreciated.
p.s. this is a homework question but the due date already passed
probability balls-in-bins
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
$endgroup$
$begingroup$
"one bin > has $3$ balls" looks like a typo.
$endgroup$
– saulspatz
Dec 21 '18 at 16:29
$begingroup$
Hey, thanks i edited
$endgroup$
– LonelyStudent
Dec 21 '18 at 16:43
add a comment |
$begingroup$
Balls are thrown randomly and uniformly into $n$ bins up until one bin has 3 balls.
Let there be $T=T(n)$ be the number of throws we made until the
occasion occurs. I am to assume $n$ is large and requested to find
$f(n)$ so that the probability:
$$ P(T > 0.1f(n))$$ Is close to 1 (for example bigger than 0.9999)
and the probability: $$ P(T > 10f(n))$$ Is close to 0 (for example
smaller than 0.001)
Our hint was to investigate the random variable, when throwing $m$
balls $X(m)$ is the number of triplets $[i,j,k] subset [1,2,...,m]$
so that the $i,j,k$ balls thrown fall in the same bin.
I have tried doing something similar to the birthday problem by defining a uniform random variable $X_i$ over $i in [1,...,n]$ so its basically if there is a ball in the $i$ bin. then I made another variable $$Y_i,_j,_k =begin{cases} 1, & text{$X_i = X_j = X_k$} \0, & text{otherwise}end{cases}$$
And then went on to calculate the expected value.
Overall I received the result ${m choose 3}*frac{1}{n^2}$ but the results don't match the definition of $X(m)$.
I am really kind of lost as this was the main way I felt might work, where am I wrong in this, whether its my way of thought or just my math, all help would be appreciated.
p.s. this is a homework question but the due date already passed
probability balls-in-bins
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
$endgroup$
Balls are thrown randomly and uniformly into $n$ bins up until one bin has 3 balls.
Let there be $T=T(n)$ be the number of throws we made until the
occasion occurs. I am to assume $n$ is large and requested to find
$f(n)$ so that the probability:
$$ P(T > 0.1f(n))$$ Is close to 1 (for example bigger than 0.9999)
and the probability: $$ P(T > 10f(n))$$ Is close to 0 (for example
smaller than 0.001)
Our hint was to investigate the random variable, when throwing $m$
balls $X(m)$ is the number of triplets $[i,j,k] subset [1,2,...,m]$
so that the $i,j,k$ balls thrown fall in the same bin.
I have tried doing something similar to the birthday problem by defining a uniform random variable $X_i$ over $i in [1,...,n]$ so its basically if there is a ball in the $i$ bin. then I made another variable $$Y_i,_j,_k =begin{cases} 1, & text{$X_i = X_j = X_k$} \0, & text{otherwise}end{cases}$$
And then went on to calculate the expected value.
Overall I received the result ${m choose 3}*frac{1}{n^2}$ but the results don't match the definition of $X(m)$.
I am really kind of lost as this was the main way I felt might work, where am I wrong in this, whether its my way of thought or just my math, all help would be appreciated.
p.s. this is a homework question but the due date already passed
probability balls-in-bins
probability balls-in-bins
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
edited Dec 21 '18 at 16:43
LonelyStudent
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
asked Dec 21 '18 at 16:27
LonelyStudentLonelyStudent
523
523
$begingroup$
"one bin > has $3$ balls" looks like a typo.
$endgroup$
– saulspatz
Dec 21 '18 at 16:29
$begingroup$
Hey, thanks i edited
$endgroup$
– LonelyStudent
Dec 21 '18 at 16:43
add a comment |
$begingroup$
"one bin > has $3$ balls" looks like a typo.
$endgroup$
– saulspatz
Dec 21 '18 at 16:29
$begingroup$
Hey, thanks i edited
$endgroup$
– LonelyStudent
Dec 21 '18 at 16:43
$begingroup$
"one bin > has $3$ balls" looks like a typo.
$endgroup$
– saulspatz
Dec 21 '18 at 16:29
$begingroup$
"one bin > has $3$ balls" looks like a typo.
$endgroup$
– saulspatz
Dec 21 '18 at 16:29
$begingroup$
Hey, thanks i edited
$endgroup$
– LonelyStudent
Dec 21 '18 at 16:43
$begingroup$
Hey, thanks i edited
$endgroup$
– LonelyStudent
Dec 21 '18 at 16:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are asked to get the number of balls right within a factor of $10$ so we can be rather rough. If $n$ is fairly large you will have a Poisson distribution of the number of balls in each bin. If we throw $k$ balls the parameter in the Poisson distribution is $lambda=frac kn$. We want to choose this so there is a reasonable chance that at least one bin has three balls. The chance a given bin has three balls is $frac {lambda^3e^{-lambda}}{3!}$. Since there are $n$ bins, we want this to be about $frac 1n$ So (using $=$ instead of $approx$) we have
$$frac {lambda^3e^{-lambda}}{3!}=frac 1n\
frac {(frac kn)^3e^{-frac kn}}{3!}=frac 1n\
e^{-frac kn}=frac {6n^2}{k^3}$$
I did an approximate solution of this for $n$ from $5$ to $65$ in steps of $5$ in a spreadsheet. The graph is below.
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
$endgroup$
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
2
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
1
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
|
show 3 more comments
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are asked to get the number of balls right within a factor of $10$ so we can be rather rough. If $n$ is fairly large you will have a Poisson distribution of the number of balls in each bin. If we throw $k$ balls the parameter in the Poisson distribution is $lambda=frac kn$. We want to choose this so there is a reasonable chance that at least one bin has three balls. The chance a given bin has three balls is $frac {lambda^3e^{-lambda}}{3!}$. Since there are $n$ bins, we want this to be about $frac 1n$ So (using $=$ instead of $approx$) we have
$$frac {lambda^3e^{-lambda}}{3!}=frac 1n\
frac {(frac kn)^3e^{-frac kn}}{3!}=frac 1n\
e^{-frac kn}=frac {6n^2}{k^3}$$
I did an approximate solution of this for $n$ from $5$ to $65$ in steps of $5$ in a spreadsheet. The graph is below.
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
$endgroup$
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
2
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
1
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
|
show 3 more comments
$begingroup$
You are asked to get the number of balls right within a factor of $10$ so we can be rather rough. If $n$ is fairly large you will have a Poisson distribution of the number of balls in each bin. If we throw $k$ balls the parameter in the Poisson distribution is $lambda=frac kn$. We want to choose this so there is a reasonable chance that at least one bin has three balls. The chance a given bin has three balls is $frac {lambda^3e^{-lambda}}{3!}$. Since there are $n$ bins, we want this to be about $frac 1n$ So (using $=$ instead of $approx$) we have
$$frac {lambda^3e^{-lambda}}{3!}=frac 1n\
frac {(frac kn)^3e^{-frac kn}}{3!}=frac 1n\
e^{-frac kn}=frac {6n^2}{k^3}$$
I did an approximate solution of this for $n$ from $5$ to $65$ in steps of $5$ in a spreadsheet. The graph is below.
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
$endgroup$
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
2
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
1
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
|
show 3 more comments
$begingroup$
You are asked to get the number of balls right within a factor of $10$ so we can be rather rough. If $n$ is fairly large you will have a Poisson distribution of the number of balls in each bin. If we throw $k$ balls the parameter in the Poisson distribution is $lambda=frac kn$. We want to choose this so there is a reasonable chance that at least one bin has three balls. The chance a given bin has three balls is $frac {lambda^3e^{-lambda}}{3!}$. Since there are $n$ bins, we want this to be about $frac 1n$ So (using $=$ instead of $approx$) we have
$$frac {lambda^3e^{-lambda}}{3!}=frac 1n\
frac {(frac kn)^3e^{-frac kn}}{3!}=frac 1n\
e^{-frac kn}=frac {6n^2}{k^3}$$
I did an approximate solution of this for $n$ from $5$ to $65$ in steps of $5$ in a spreadsheet. The graph is below.
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
$endgroup$
You are asked to get the number of balls right within a factor of $10$ so we can be rather rough. If $n$ is fairly large you will have a Poisson distribution of the number of balls in each bin. If we throw $k$ balls the parameter in the Poisson distribution is $lambda=frac kn$. We want to choose this so there is a reasonable chance that at least one bin has three balls. The chance a given bin has three balls is $frac {lambda^3e^{-lambda}}{3!}$. Since there are $n$ bins, we want this to be about $frac 1n$ So (using $=$ instead of $approx$) we have
$$frac {lambda^3e^{-lambda}}{3!}=frac 1n\
frac {(frac kn)^3e^{-frac kn}}{3!}=frac 1n\
e^{-frac kn}=frac {6n^2}{k^3}$$
I did an approximate solution of this for $n$ from $5$ to $65$ in steps of $5$ in a spreadsheet. The graph is below.
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
edited Dec 22 '18 at 2:19
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
answered Dec 21 '18 at 16:58
Ross MillikanRoss Millikan
298k24200374
298k24200374
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
2
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
1
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
|
show 3 more comments
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
2
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
1
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
Ok, thanks for your answer! But do you think i can progress somehow with my way?
$endgroup$
– LonelyStudent
Dec 21 '18 at 18:00
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
$begingroup$
I found that I had located the wrong root in my previous graph. It was throwing so many balls that one would expect only one bin to have as few as three balls. The new graph has much smaller numbers of balls and I believe is correct.
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:20
2
2
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
Your approach is not so different. You are saying that any given set of three balls has $frac 1{n^2} $ chance of hitting the same bin. There are about $frac {k^3}6$ sets of three balls. That gives the same combination as on my right side. I think your logic would then be to set that to $1$, while I use a number somewhat less than $1$, but not so much. You would then say $k=sqrt[3]{6n^2}$
$endgroup$
– Ross Millikan
Dec 22 '18 at 2:23
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
$begingroup$
First of all thank you, second I tried doing what you said and reducing the number to 1 by $k = sqrt[3]{(6n^2)}$ but I cant seem to relate it to a function like requested in the question, when I use for example $n=10$ while making the function $f(n) = sqrt[3]{(6n^2)}$ then the result $ P(T > 0.1f(n))$ isnt really like expected because the number of balls must be larger then about $0.1*8.4$
$endgroup$
– LonelyStudent
Dec 22 '18 at 15:52
1
1
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
$begingroup$
What you would have to do would be to make an argument that involves approximations, bound those approximations, and show that the product means the error is less than a factor of $10$. Maybe you have three approximations, each of which can be wrong by a factor $2$. The product is then good to a factor $8$, which is good enough. I don't see a way there.
$endgroup$
– Ross Millikan
Dec 23 '18 at 17:06
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$begingroup$
"one bin > has $3$ balls" looks like a typo.
$endgroup$
– saulspatz
Dec 21 '18 at 16:29
$begingroup$
Hey, thanks i edited
$endgroup$
– LonelyStudent
Dec 21 '18 at 16:43