Mixing
Drawing Balls of Different Colors from an Urn
$begingroup$
Let consider an urn model containing $n$ balls such that $m_g$ balls are green, $m_r$ are red and $m_b$ are blue and we have $n = m_g + m_r +m_b$.
Especially this means that the $m_g$ green balls are indistinguishable with respect to each other (and the same game for red and blue ones)
We draw $l le n$ times without placing the balls back.
My question how many posibilities there exist for different $l$-draws taking the order and the indistinguishability of balls of same color into account.
More generally: the urn contains $m_1, m_2, ..., m_k$ balls with $sum _i ^k m_i =n$ such that the $m_j$ are undistiguishable and one make $l$ draws without placing the balls back.
What is the cardinality of the resulting probability space?
Could anybody explain if there exist an intuitive way to describe and the count all cases?
probability combinatorics stochastic-analysis
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
$endgroup$
add a comment |
$begingroup$
Let consider an urn model containing $n$ balls such that $m_g$ balls are green, $m_r$ are red and $m_b$ are blue and we have $n = m_g + m_r +m_b$.
Especially this means that the $m_g$ green balls are indistinguishable with respect to each other (and the same game for red and blue ones)
We draw $l le n$ times without placing the balls back.
My question how many posibilities there exist for different $l$-draws taking the order and the indistinguishability of balls of same color into account.
More generally: the urn contains $m_1, m_2, ..., m_k$ balls with $sum _i ^k m_i =n$ such that the $m_j$ are undistiguishable and one make $l$ draws without placing the balls back.
What is the cardinality of the resulting probability space?
Could anybody explain if there exist an intuitive way to describe and the count all cases?
probability combinatorics stochastic-analysis
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
$endgroup$
$begingroup$
Are you saying the order of the balls matters? That is, GBB is considered different from BBG?
$endgroup$
– Mike Earnest
Jan 28 at 18:11
$begingroup$
@MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right?
$endgroup$
– KarlPeter
Jan 28 at 18:13
add a comment |
$begingroup$
Let consider an urn model containing $n$ balls such that $m_g$ balls are green, $m_r$ are red and $m_b$ are blue and we have $n = m_g + m_r +m_b$.
Especially this means that the $m_g$ green balls are indistinguishable with respect to each other (and the same game for red and blue ones)
We draw $l le n$ times without placing the balls back.
My question how many posibilities there exist for different $l$-draws taking the order and the indistinguishability of balls of same color into account.
More generally: the urn contains $m_1, m_2, ..., m_k$ balls with $sum _i ^k m_i =n$ such that the $m_j$ are undistiguishable and one make $l$ draws without placing the balls back.
What is the cardinality of the resulting probability space?
Could anybody explain if there exist an intuitive way to describe and the count all cases?
probability combinatorics stochastic-analysis
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
$endgroup$
Let consider an urn model containing $n$ balls such that $m_g$ balls are green, $m_r$ are red and $m_b$ are blue and we have $n = m_g + m_r +m_b$.
Especially this means that the $m_g$ green balls are indistinguishable with respect to each other (and the same game for red and blue ones)
We draw $l le n$ times without placing the balls back.
My question how many posibilities there exist for different $l$-draws taking the order and the indistinguishability of balls of same color into account.
More generally: the urn contains $m_1, m_2, ..., m_k$ balls with $sum _i ^k m_i =n$ such that the $m_j$ are undistiguishable and one make $l$ draws without placing the balls back.
What is the cardinality of the resulting probability space?
Could anybody explain if there exist an intuitive way to describe and the count all cases?
probability combinatorics stochastic-analysis
probability combinatorics stochastic-analysis
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
edited Jan 28 at 0:30
KarlPeter
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
asked Jan 28 at 0:21
KarlPeterKarlPeter
5841316
5841316
$begingroup$
Are you saying the order of the balls matters? That is, GBB is considered different from BBG?
$endgroup$
– Mike Earnest
Jan 28 at 18:11
$begingroup$
@MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right?
$endgroup$
– KarlPeter
Jan 28 at 18:13
add a comment |
$begingroup$
Are you saying the order of the balls matters? That is, GBB is considered different from BBG?
$endgroup$
– Mike Earnest
Jan 28 at 18:11
$begingroup$
@MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right?
$endgroup$
– KarlPeter
Jan 28 at 18:13
$begingroup$
Are you saying the order of the balls matters? That is, GBB is considered different from BBG?
$endgroup$
– Mike Earnest
Jan 28 at 18:11
$begingroup$
Are you saying the order of the balls matters? That is, GBB is considered different from BBG?
$endgroup$
– Mike Earnest
Jan 28 at 18:11
$begingroup$
@MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right?
$endgroup$
– KarlPeter
Jan 28 at 18:13
$begingroup$
@MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right?
$endgroup$
– KarlPeter
Jan 28 at 18:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The obvious answer is
$$
sum_{substack{a_1+a_2+dots+a_k=ell\0le a_ile m_i}}binom{ell}{a_1,a_2,dots,a_k}
$$
but this is an unwieldy summation.
A less obvious answer: Let $E_m(x)=sum_{i=0}^m frac{x^i}{i!}$ be the partial exponential series. The size of your sample space is
$$
ell![x^ell]prod_{j=1}^k E_{m_j}(x)
$$
Here, $[x^ell] f(x)$ denotes the coefficient of $x^ell$ in the power series $f(x)$. This at least lets you compute the size much quicker, since polynomials can be multiplied quickly using the Fast Fourier Transform.
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The obvious answer is
$$
sum_{substack{a_1+a_2+dots+a_k=ell\0le a_ile m_i}}binom{ell}{a_1,a_2,dots,a_k}
$$
but this is an unwieldy summation.
A less obvious answer: Let $E_m(x)=sum_{i=0}^m frac{x^i}{i!}$ be the partial exponential series. The size of your sample space is
$$
ell![x^ell]prod_{j=1}^k E_{m_j}(x)
$$
Here, $[x^ell] f(x)$ denotes the coefficient of $x^ell$ in the power series $f(x)$. This at least lets you compute the size much quicker, since polynomials can be multiplied quickly using the Fast Fourier Transform.
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
$endgroup$
add a comment |
$begingroup$
The obvious answer is
$$
sum_{substack{a_1+a_2+dots+a_k=ell\0le a_ile m_i}}binom{ell}{a_1,a_2,dots,a_k}
$$
but this is an unwieldy summation.
A less obvious answer: Let $E_m(x)=sum_{i=0}^m frac{x^i}{i!}$ be the partial exponential series. The size of your sample space is
$$
ell![x^ell]prod_{j=1}^k E_{m_j}(x)
$$
Here, $[x^ell] f(x)$ denotes the coefficient of $x^ell$ in the power series $f(x)$. This at least lets you compute the size much quicker, since polynomials can be multiplied quickly using the Fast Fourier Transform.
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
$endgroup$
add a comment |
$begingroup$
The obvious answer is
$$
sum_{substack{a_1+a_2+dots+a_k=ell\0le a_ile m_i}}binom{ell}{a_1,a_2,dots,a_k}
$$
but this is an unwieldy summation.
A less obvious answer: Let $E_m(x)=sum_{i=0}^m frac{x^i}{i!}$ be the partial exponential series. The size of your sample space is
$$
ell![x^ell]prod_{j=1}^k E_{m_j}(x)
$$
Here, $[x^ell] f(x)$ denotes the coefficient of $x^ell$ in the power series $f(x)$. This at least lets you compute the size much quicker, since polynomials can be multiplied quickly using the Fast Fourier Transform.
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
$endgroup$
The obvious answer is
$$
sum_{substack{a_1+a_2+dots+a_k=ell\0le a_ile m_i}}binom{ell}{a_1,a_2,dots,a_k}
$$
but this is an unwieldy summation.
A less obvious answer: Let $E_m(x)=sum_{i=0}^m frac{x^i}{i!}$ be the partial exponential series. The size of your sample space is
$$
ell![x^ell]prod_{j=1}^k E_{m_j}(x)
$$
Here, $[x^ell] f(x)$ denotes the coefficient of $x^ell$ in the power series $f(x)$. This at least lets you compute the size much quicker, since polynomials can be multiplied quickly using the Fast Fourier Transform.
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
answered Jan 29 at 0:54
Mike EarnestMike Earnest
25.9k22151
25.9k22151
add a comment |
add a comment |
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$begingroup$
Are you saying the order of the balls matters? That is, GBB is considered different from BBG?
$endgroup$
– Mike Earnest
Jan 28 at 18:11
$begingroup$
@MikeEarnest: Yes. If not then I guess that the problem would be solved by multinomial hypergeometric distribution, right?
$endgroup$
– KarlPeter
Jan 28 at 18:13