Mixing
Probability of having m groups of at least 2 empty consecutive cells by putting k balls in n bins
Suppose k balls are placed uniformly at random in n cells, each one with probability 1/n. Let's call "r-block" a group of at least r consecutive cells.
I would like to know if it's possible to compute the probability of having m empty 2-block.
I know the probability of having exactly t empty cells is $$P(n,k,t)=dfrac{binom{n}{t}S(k,n-t)(n-t)!}{n^k}$$ where $S(a,b)$ represents the Stirling's number of the second kind, but in the case of interest it seems to me too difficult to compute the number of possible configurations, as I did to derive the previous formula.
I've also found that $$(n-r)binom{n-r-1}{t-r}+binom{n-r}{t-r}$$ expresses the ways of choosing t out of n numbers with at least r consecutive numbers. If my cells are numbered ${1,2,dots,n}$ and a list of numbers represents the empty ones, using that formula I obtain the total occurrencies of r-blocks that are empty when the total number of empty cells is t.
The problem is in this way I only have the total number of r-blocks for a given t, which could be used in principle to obtain the probability of finding at least 1 empty r-block, but not exactly m empty r-block.
probability combinatorics balls-in-bins
asked Nov 19 '18 at 23:39
ARWarrior
64
add a comment |
Suppose k balls are placed uniformly at random in n cells, each one with probability 1/n. Let's call "r-block" a group of at least r consecutive cells.
I would like to know if it's possible to compute the probability of having m empty 2-block.
I know the probability of having exactly t empty cells is $$P(n,k,t)=dfrac{binom{n}{t}S(k,n-t)(n-t)!}{n^k}$$ where $S(a,b)$ represents the Stirling's number of the second kind, but in the case of interest it seems to me too difficult to compute the number of possible configurations, as I did to derive the previous formula.
I've also found that $$(n-r)binom{n-r-1}{t-r}+binom{n-r}{t-r}$$ expresses the ways of choosing t out of n numbers with at least r consecutive numbers. If my cells are numbered ${1,2,dots,n}$ and a list of numbers represents the empty ones, using that formula I obtain the total occurrencies of r-blocks that are empty when the total number of empty cells is t.
The problem is in this way I only have the total number of r-blocks for a given t, which could be used in principle to obtain the probability of finding at least 1 empty r-block, but not exactly m empty r-block.
probability combinatorics balls-in-bins
asked Nov 19 '18 at 23:39
ARWarrior
64
add a comment |
Suppose k balls are placed uniformly at random in n cells, each one with probability 1/n. Let's call "r-block" a group of at least r consecutive cells.
I would like to know if it's possible to compute the probability of having m empty 2-block.
I know the probability of having exactly t empty cells is $$P(n,k,t)=dfrac{binom{n}{t}S(k,n-t)(n-t)!}{n^k}$$ where $S(a,b)$ represents the Stirling's number of the second kind, but in the case of interest it seems to me too difficult to compute the number of possible configurations, as I did to derive the previous formula.
I've also found that $$(n-r)binom{n-r-1}{t-r}+binom{n-r}{t-r}$$ expresses the ways of choosing t out of n numbers with at least r consecutive numbers. If my cells are numbered ${1,2,dots,n}$ and a list of numbers represents the empty ones, using that formula I obtain the total occurrencies of r-blocks that are empty when the total number of empty cells is t.
The problem is in this way I only have the total number of r-blocks for a given t, which could be used in principle to obtain the probability of finding at least 1 empty r-block, but not exactly m empty r-block.
probability combinatorics balls-in-bins
asked Nov 19 '18 at 23:39
ARWarrior
64
Suppose k balls are placed uniformly at random in n cells, each one with probability 1/n. Let's call "r-block" a group of at least r consecutive cells.
I would like to know if it's possible to compute the probability of having m empty 2-block.
I know the probability of having exactly t empty cells is $$P(n,k,t)=dfrac{binom{n}{t}S(k,n-t)(n-t)!}{n^k}$$ where $S(a,b)$ represents the Stirling's number of the second kind, but in the case of interest it seems to me too difficult to compute the number of possible configurations, as I did to derive the previous formula.
I've also found that $$(n-r)binom{n-r-1}{t-r}+binom{n-r}{t-r}$$ expresses the ways of choosing t out of n numbers with at least r consecutive numbers. If my cells are numbered ${1,2,dots,n}$ and a list of numbers represents the empty ones, using that formula I obtain the total occurrencies of r-blocks that are empty when the total number of empty cells is t.
The problem is in this way I only have the total number of r-blocks for a given t, which could be used in principle to obtain the probability of finding at least 1 empty r-block, but not exactly m empty r-block.
probability combinatorics balls-in-bins
probability combinatorics balls-in-bins
asked Nov 19 '18 at 23:39
ARWarrior
64
asked Nov 19 '18 at 23:39
ARWarrior
64
edited Nov 21 '18 at 19:44
asked Nov 19 '18 at 23:39
ARWarrior
64
asked Nov 19 '18 at 23:39
ARWarrior
64
asked Nov 19 '18 at 23:39
ARWarrior
64
64
add a comment |
add a comment |
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