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Probability of throwing exactly $V$ distinct sides on $N$ sided dice by $K$ rolls
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I'm struggling with finding a non-recursive formula for calculating the following probability:
Probability of throwing exactly $V$ distinct sides (numbers) on $N$ sided dice by $K$ rolls
Example for a normal dice where number of sides $N=6$, rolls $K=6$ and required distinct sides $V=2$ can be found here: Six throws, only two distinct numbers: coincidence?
The formula in the example is not general and only solution I can find is recursive one which I cannot use.
probability dice
asked Jul 20 at 14:38
L D
82
add a comment |Â
up vote
1
down vote
favorite
I'm struggling with finding a non-recursive formula for calculating the following probability:
Probability of throwing exactly $V$ distinct sides (numbers) on $N$ sided dice by $K$ rolls
Example for a normal dice where number of sides $N=6$, rolls $K=6$ and required distinct sides $V=2$ can be found here: Six throws, only two distinct numbers: coincidence?
The formula in the example is not general and only solution I can find is recursive one which I cannot use.
probability dice
asked Jul 20 at 14:38
L D
82
It would surprise me if the fact that you can only find recursive solutions is coincidental. I suspect there is no closed-form general solution for this.
– Vera
Jul 20 at 14:44
I hope it's just because I'm quite lame at closed formulas rather than the closed formula doesn't exist :-)
– L D
Jul 20 at 14:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm struggling with finding a non-recursive formula for calculating the following probability:
Probability of throwing exactly $V$ distinct sides (numbers) on $N$ sided dice by $K$ rolls
Example for a normal dice where number of sides $N=6$, rolls $K=6$ and required distinct sides $V=2$ can be found here: Six throws, only two distinct numbers: coincidence?
The formula in the example is not general and only solution I can find is recursive one which I cannot use.
probability dice
asked Jul 20 at 14:38
L D
82
I'm struggling with finding a non-recursive formula for calculating the following probability:
Probability of throwing exactly $V$ distinct sides (numbers) on $N$ sided dice by $K$ rolls
Example for a normal dice where number of sides $N=6$, rolls $K=6$ and required distinct sides $V=2$ can be found here: Six throws, only two distinct numbers: coincidence?
The formula in the example is not general and only solution I can find is recursive one which I cannot use.
probability dice
asked Jul 20 at 14:38
L D
82
asked Jul 20 at 14:38
L D
82
asked Jul 20 at 14:38
L D
82
asked Jul 20 at 14:38
L D
82
82
It would surprise me if the fact that you can only find recursive solutions is coincidental. I suspect there is no closed-form general solution for this.
– Vera
Jul 20 at 14:44
I hope it's just because I'm quite lame at closed formulas rather than the closed formula doesn't exist :-)
– L D
Jul 20 at 14:56
add a comment |Â
It would surprise me if the fact that you can only find recursive solutions is coincidental. I suspect there is no closed-form general solution for this.
– Vera
Jul 20 at 14:44
I hope it's just because I'm quite lame at closed formulas rather than the closed formula doesn't exist :-)
– L D
Jul 20 at 14:56
It would surprise me if the fact that you can only find recursive solutions is coincidental. I suspect there is no closed-form general solution for this.
– Vera
Jul 20 at 14:44
It would surprise me if the fact that you can only find recursive solutions is coincidental. I suspect there is no closed-form general solution for this.
– Vera
Jul 20 at 14:44
I hope it's just because I'm quite lame at closed formulas rather than the closed formula doesn't exist :-)
– L D
Jul 20 at 14:56
I hope it's just because I'm quite lame at closed formulas rather than the closed formula doesn't exist :-)
– L D
Jul 20 at 14:56
add a comment |Â
1 Answer
1
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0
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This would appear to be using Stirling numbers of the second kind:
$$frac1N^K times Nchoose V times V! times Kbrace V.$$
These probabilities sum to one ($Kbrace 0 = 0$ when $Kge 1$):
$$frac1N^K sum_V=0^N ·Nchoose V V!
K! [z^K] frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V V!
frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V
(exp(z)-1)^V = frac1N^K K! [z^K] exp(Nz)
= frac1N^K N^K = 1.$$
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
Sir, thank you! It works! For anyone wanting R code:pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )
– L D
Jul 20 at 15:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This would appear to be using Stirling numbers of the second kind:
$$frac1N^K times Nchoose V times V! times Kbrace V.$$
These probabilities sum to one ($Kbrace 0 = 0$ when $Kge 1$):
$$frac1N^K sum_V=0^N ·Nchoose V V!
K! [z^K] frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V V!
frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V
(exp(z)-1)^V = frac1N^K K! [z^K] exp(Nz)
= frac1N^K N^K = 1.$$
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
Sir, thank you! It works! For anyone wanting R code:pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )
– L D
Jul 20 at 15:40
add a comment |Â
up vote
0
down vote
accepted
This would appear to be using Stirling numbers of the second kind:
$$frac1N^K times Nchoose V times V! times Kbrace V.$$
These probabilities sum to one ($Kbrace 0 = 0$ when $Kge 1$):
$$frac1N^K sum_V=0^N ·Nchoose V V!
K! [z^K] frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V V!
frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V
(exp(z)-1)^V = frac1N^K K! [z^K] exp(Nz)
= frac1N^K N^K = 1.$$
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
Sir, thank you! It works! For anyone wanting R code:pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )
– L D
Jul 20 at 15:40
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This would appear to be using Stirling numbers of the second kind:
$$frac1N^K times Nchoose V times V! times Kbrace V.$$
These probabilities sum to one ($Kbrace 0 = 0$ when $Kge 1$):
$$frac1N^K sum_V=0^N ·Nchoose V V!
K! [z^K] frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V V!
frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V
(exp(z)-1)^V = frac1N^K K! [z^K] exp(Nz)
= frac1N^K N^K = 1.$$
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
This would appear to be using Stirling numbers of the second kind:
$$frac1N^K times Nchoose V times V! times Kbrace V.$$
These probabilities sum to one ($Kbrace 0 = 0$ when $Kge 1$):
$$frac1N^K sum_V=0^N ·Nchoose V V!
K! [z^K] frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V V!
frac(exp(z)-1)^VV!
\ = frac1N^K K! [z^K] sum_V=0^N ·Nchoose V
(exp(z)-1)^V = frac1N^K K! [z^K] exp(Nz)
= frac1N^K N^K = 1.$$
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
edited Jul 21 at 21:04
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
answered Jul 20 at 15:18
Marko Riedel
36.5k333107
36.5k333107
Sir, thank you! It works! For anyone wanting R code:pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )
– L D
Jul 20 at 15:40
add a comment |Â
Sir, thank you! It works! For anyone wanting R code:pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )
– L D
Jul 20 at 15:40
Sir, thank you! It works! For anyone wanting R code:
pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )– L D
Jul 20 at 15:40
Sir, thank you! It works! For anyone wanting R code:
pform <- function(K, N, V) ( 1/N**K ) * choose(N, V) * factorial(V) * as.numeric( Stirling2(K, V) )– L D
Jul 20 at 15:40
add a comment |Â
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It would surprise me if the fact that you can only find recursive solutions is coincidental. I suspect there is no closed-form general solution for this.
– Vera
Jul 20 at 14:44
I hope it's just because I'm quite lame at closed formulas rather than the closed formula doesn't exist :-)
– L D
Jul 20 at 14:56