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Number of combinations of dice having different number of faces that add to $10$
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A function $face_sum$ that takes a vector $f$ that represents the number of faces of each die, and a positive integer $s$ , and returns the probability that the sum of the top faces observed is $s$ .
For example, if $f=[3,4,5]$ and $s≤2$ or $s≥13$ , $face_sum$ returns $0$, and if $s=3$ or $s=12$ , it returns $frac13*frac14*frac15=frac160$
$face_sum([2,4,4,6], 10)$
I tried to solve this but I can't find the formula to calculate the number of combination of die faces that add to $10$, as the number of faces are different for each die.
Since max no of possible combination is $6*4*4*2=192$, all I need to find is the number of combination of die faces that add to $10$.
Thanks in advance.
combinatorics
edited Jul 18 at 20:02
GuySa
402313
asked Jul 18 at 19:10
Udolf
355
add a comment |Â
up vote
3
down vote
favorite
A function $face_sum$ that takes a vector $f$ that represents the number of faces of each die, and a positive integer $s$ , and returns the probability that the sum of the top faces observed is $s$ .
For example, if $f=[3,4,5]$ and $s≤2$ or $s≥13$ , $face_sum$ returns $0$, and if $s=3$ or $s=12$ , it returns $frac13*frac14*frac15=frac160$
$face_sum([2,4,4,6], 10)$
I tried to solve this but I can't find the formula to calculate the number of combination of die faces that add to $10$, as the number of faces are different for each die.
Since max no of possible combination is $6*4*4*2=192$, all I need to find is the number of combination of die faces that add to $10$.
Thanks in advance.
combinatorics
edited Jul 18 at 20:02
GuySa
402313
asked Jul 18 at 19:10
Udolf
355
The generating function for $f([3,4,5],n)$ is $$x^3fracx^3-1x-1fracx^4-1x-1fracx^5-1x-1= x^12+3x^11+6x^10+9x^9+11x^8+11x^7+9x^6+6x^5+3x^4+x^3$$. The generating function for $f([2,4,4,6],n)$ is $$scriptsize x^4fracx^2-1x-1fracx^4-1x-1fracx^4-1x-1fracx^6-1x-1 =x^16+4x^15+9x^14+16x^13+23x^12+28x^11+30x^10+28x^9+23x^8+16x^7+9x^6+4x^5+x^4$$.
– robjohn♦
Jul 18 at 19:20
don't dice 2 and 3 have the same number of faces?
– zhw.
Jul 18 at 19:48
yep they have the same number of faces.
– Udolf
Jul 19 at 3:33
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A function $face_sum$ that takes a vector $f$ that represents the number of faces of each die, and a positive integer $s$ , and returns the probability that the sum of the top faces observed is $s$ .
For example, if $f=[3,4,5]$ and $s≤2$ or $s≥13$ , $face_sum$ returns $0$, and if $s=3$ or $s=12$ , it returns $frac13*frac14*frac15=frac160$
$face_sum([2,4,4,6], 10)$
I tried to solve this but I can't find the formula to calculate the number of combination of die faces that add to $10$, as the number of faces are different for each die.
Since max no of possible combination is $6*4*4*2=192$, all I need to find is the number of combination of die faces that add to $10$.
Thanks in advance.
combinatorics
edited Jul 18 at 20:02
GuySa
402313
asked Jul 18 at 19:10
Udolf
355
A function $face_sum$ that takes a vector $f$ that represents the number of faces of each die, and a positive integer $s$ , and returns the probability that the sum of the top faces observed is $s$ .
For example, if $f=[3,4,5]$ and $s≤2$ or $s≥13$ , $face_sum$ returns $0$, and if $s=3$ or $s=12$ , it returns $frac13*frac14*frac15=frac160$
$face_sum([2,4,4,6], 10)$
I tried to solve this but I can't find the formula to calculate the number of combination of die faces that add to $10$, as the number of faces are different for each die.
Since max no of possible combination is $6*4*4*2=192$, all I need to find is the number of combination of die faces that add to $10$.
Thanks in advance.
combinatorics
edited Jul 18 at 20:02
GuySa
402313
asked Jul 18 at 19:10
Udolf
355
edited Jul 18 at 20:02
GuySa
402313
edited Jul 18 at 20:02
GuySa
402313
edited Jul 18 at 20:02
GuySa
402313
402313
asked Jul 18 at 19:10
Udolf
355
asked Jul 18 at 19:10
Udolf
355
asked Jul 18 at 19:10
Udolf
355
355
The generating function for $f([3,4,5],n)$ is $$x^3fracx^3-1x-1fracx^4-1x-1fracx^5-1x-1= x^12+3x^11+6x^10+9x^9+11x^8+11x^7+9x^6+6x^5+3x^4+x^3$$. The generating function for $f([2,4,4,6],n)$ is $$scriptsize x^4fracx^2-1x-1fracx^4-1x-1fracx^4-1x-1fracx^6-1x-1 =x^16+4x^15+9x^14+16x^13+23x^12+28x^11+30x^10+28x^9+23x^8+16x^7+9x^6+4x^5+x^4$$.
– robjohn♦
Jul 18 at 19:20
don't dice 2 and 3 have the same number of faces?
– zhw.
Jul 18 at 19:48
yep they have the same number of faces.
– Udolf
Jul 19 at 3:33
add a comment |Â
The generating function for $f([3,4,5],n)$ is $$x^3fracx^3-1x-1fracx^4-1x-1fracx^5-1x-1= x^12+3x^11+6x^10+9x^9+11x^8+11x^7+9x^6+6x^5+3x^4+x^3$$. The generating function for $f([2,4,4,6],n)$ is $$scriptsize x^4fracx^2-1x-1fracx^4-1x-1fracx^4-1x-1fracx^6-1x-1 =x^16+4x^15+9x^14+16x^13+23x^12+28x^11+30x^10+28x^9+23x^8+16x^7+9x^6+4x^5+x^4$$.
– robjohn♦
Jul 18 at 19:20
don't dice 2 and 3 have the same number of faces?
– zhw.
Jul 18 at 19:48
yep they have the same number of faces.
– Udolf
Jul 19 at 3:33
The generating function for $f([3,4,5],n)$ is $$x^3fracx^3-1x-1fracx^4-1x-1fracx^5-1x-1= x^12+3x^11+6x^10+9x^9+11x^8+11x^7+9x^6+6x^5+3x^4+x^3$$. The generating function for $f([2,4,4,6],n)$ is $$scriptsize x^4fracx^2-1x-1fracx^4-1x-1fracx^4-1x-1fracx^6-1x-1 =x^16+4x^15+9x^14+16x^13+23x^12+28x^11+30x^10+28x^9+23x^8+16x^7+9x^6+4x^5+x^4$$.
– robjohn♦
Jul 18 at 19:20
The generating function for $f([3,4,5],n)$ is $$x^3fracx^3-1x-1fracx^4-1x-1fracx^5-1x-1= x^12+3x^11+6x^10+9x^9+11x^8+11x^7+9x^6+6x^5+3x^4+x^3$$. The generating function for $f([2,4,4,6],n)$ is $$scriptsize x^4fracx^2-1x-1fracx^4-1x-1fracx^4-1x-1fracx^6-1x-1 =x^16+4x^15+9x^14+16x^13+23x^12+28x^11+30x^10+28x^9+23x^8+16x^7+9x^6+4x^5+x^4$$.
– robjohn♦
Jul 18 at 19:20
don't dice 2 and 3 have the same number of faces?
– zhw.
Jul 18 at 19:48
don't dice 2 and 3 have the same number of faces?
– zhw.
Jul 18 at 19:48
yep they have the same number of faces.
– Udolf
Jul 19 at 3:33
yep they have the same number of faces.
– Udolf
Jul 19 at 3:33
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The formula is a bit messy. Suppose there are $n$ dice, and the list is $f=[f_1,f_2,dots,f_n]$. You want to count the number of integer solutions to the constrained equation
$$
beginalign
x_1+x_2+dots+x_n &= s\ 1le x_i&le f_ihspace2cm text for i=1,2,dots,n
endalign
$$
To solve this, first, count the number of solutions without the constraint $x_ile f_i$. This can be shown to be $binoms-1n-1$ (imagine a line of $s$ dots, and you place dividers in $n-1$ of the $s-1$ spaces between the dots). Now, use inclusion exclusion to subtract out the "bad" solutions where some $x_i > f_i$. The result is this:
$$
boxedBbb P(n text dice with f_1,dots,f_n text faces sum to s) = frac1f_1timesdotstimes f_nsum_Ssubseteq 1,2,dots,n(-1)^binoms-1-sum_iin Sf_in-1.
$$
There are $2^n$ summands, one for each subset of $1,2,dots,n$ (i.e. subset of the available dice).
For example, when $f=[2,4,4,6]$ and $s=10$, you get
$$
1over 2cdot 4cdot 4cdot 6left[
beginarraylbinom10-14-1\
-binom10-2-14-1-binom10-4-14-1
-binom10-4-14-1-binom10-6-14-1\
+binom10-2-4-14-1+binom10-2-4-14-1
endarrayright].
$$
Terms with $sum_iin Sf_i>s-n$ have been omitted, since in these cases the binomial coefficient is zero.
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
add a comment |Â
up vote
0
down vote
Do you know how to find the number of solutions for an equation like the following?
$$
x_1 + x_2 + ... + x_n = k
$$
given $forall i in [n]: x_i in S_i$ for some set $S_i$.
answered Jul 18 at 19:25
GuySa
402313
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The formula is a bit messy. Suppose there are $n$ dice, and the list is $f=[f_1,f_2,dots,f_n]$. You want to count the number of integer solutions to the constrained equation
$$
beginalign
x_1+x_2+dots+x_n &= s\ 1le x_i&le f_ihspace2cm text for i=1,2,dots,n
endalign
$$
To solve this, first, count the number of solutions without the constraint $x_ile f_i$. This can be shown to be $binoms-1n-1$ (imagine a line of $s$ dots, and you place dividers in $n-1$ of the $s-1$ spaces between the dots). Now, use inclusion exclusion to subtract out the "bad" solutions where some $x_i > f_i$. The result is this:
$$
boxedBbb P(n text dice with f_1,dots,f_n text faces sum to s) = frac1f_1timesdotstimes f_nsum_Ssubseteq 1,2,dots,n(-1)^binoms-1-sum_iin Sf_in-1.
$$
There are $2^n$ summands, one for each subset of $1,2,dots,n$ (i.e. subset of the available dice).
For example, when $f=[2,4,4,6]$ and $s=10$, you get
$$
1over 2cdot 4cdot 4cdot 6left[
beginarraylbinom10-14-1\
-binom10-2-14-1-binom10-4-14-1
-binom10-4-14-1-binom10-6-14-1\
+binom10-2-4-14-1+binom10-2-4-14-1
endarrayright].
$$
Terms with $sum_iin Sf_i>s-n$ have been omitted, since in these cases the binomial coefficient is zero.
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
add a comment |Â
up vote
1
down vote
accepted
The formula is a bit messy. Suppose there are $n$ dice, and the list is $f=[f_1,f_2,dots,f_n]$. You want to count the number of integer solutions to the constrained equation
$$
beginalign
x_1+x_2+dots+x_n &= s\ 1le x_i&le f_ihspace2cm text for i=1,2,dots,n
endalign
$$
To solve this, first, count the number of solutions without the constraint $x_ile f_i$. This can be shown to be $binoms-1n-1$ (imagine a line of $s$ dots, and you place dividers in $n-1$ of the $s-1$ spaces between the dots). Now, use inclusion exclusion to subtract out the "bad" solutions where some $x_i > f_i$. The result is this:
$$
boxedBbb P(n text dice with f_1,dots,f_n text faces sum to s) = frac1f_1timesdotstimes f_nsum_Ssubseteq 1,2,dots,n(-1)^binoms-1-sum_iin Sf_in-1.
$$
There are $2^n$ summands, one for each subset of $1,2,dots,n$ (i.e. subset of the available dice).
For example, when $f=[2,4,4,6]$ and $s=10$, you get
$$
1over 2cdot 4cdot 4cdot 6left[
beginarraylbinom10-14-1\
-binom10-2-14-1-binom10-4-14-1
-binom10-4-14-1-binom10-6-14-1\
+binom10-2-4-14-1+binom10-2-4-14-1
endarrayright].
$$
Terms with $sum_iin Sf_i>s-n$ have been omitted, since in these cases the binomial coefficient is zero.
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The formula is a bit messy. Suppose there are $n$ dice, and the list is $f=[f_1,f_2,dots,f_n]$. You want to count the number of integer solutions to the constrained equation
$$
beginalign
x_1+x_2+dots+x_n &= s\ 1le x_i&le f_ihspace2cm text for i=1,2,dots,n
endalign
$$
To solve this, first, count the number of solutions without the constraint $x_ile f_i$. This can be shown to be $binoms-1n-1$ (imagine a line of $s$ dots, and you place dividers in $n-1$ of the $s-1$ spaces between the dots). Now, use inclusion exclusion to subtract out the "bad" solutions where some $x_i > f_i$. The result is this:
$$
boxedBbb P(n text dice with f_1,dots,f_n text faces sum to s) = frac1f_1timesdotstimes f_nsum_Ssubseteq 1,2,dots,n(-1)^binoms-1-sum_iin Sf_in-1.
$$
There are $2^n$ summands, one for each subset of $1,2,dots,n$ (i.e. subset of the available dice).
For example, when $f=[2,4,4,6]$ and $s=10$, you get
$$
1over 2cdot 4cdot 4cdot 6left[
beginarraylbinom10-14-1\
-binom10-2-14-1-binom10-4-14-1
-binom10-4-14-1-binom10-6-14-1\
+binom10-2-4-14-1+binom10-2-4-14-1
endarrayright].
$$
Terms with $sum_iin Sf_i>s-n$ have been omitted, since in these cases the binomial coefficient is zero.
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
The formula is a bit messy. Suppose there are $n$ dice, and the list is $f=[f_1,f_2,dots,f_n]$. You want to count the number of integer solutions to the constrained equation
$$
beginalign
x_1+x_2+dots+x_n &= s\ 1le x_i&le f_ihspace2cm text for i=1,2,dots,n
endalign
$$
To solve this, first, count the number of solutions without the constraint $x_ile f_i$. This can be shown to be $binoms-1n-1$ (imagine a line of $s$ dots, and you place dividers in $n-1$ of the $s-1$ spaces between the dots). Now, use inclusion exclusion to subtract out the "bad" solutions where some $x_i > f_i$. The result is this:
$$
boxedBbb P(n text dice with f_1,dots,f_n text faces sum to s) = frac1f_1timesdotstimes f_nsum_Ssubseteq 1,2,dots,n(-1)^binoms-1-sum_iin Sf_in-1.
$$
There are $2^n$ summands, one for each subset of $1,2,dots,n$ (i.e. subset of the available dice).
For example, when $f=[2,4,4,6]$ and $s=10$, you get
$$
1over 2cdot 4cdot 4cdot 6left[
beginarraylbinom10-14-1\
-binom10-2-14-1-binom10-4-14-1
-binom10-4-14-1-binom10-6-14-1\
+binom10-2-4-14-1+binom10-2-4-14-1
endarrayright].
$$
Terms with $sum_iin Sf_i>s-n$ have been omitted, since in these cases the binomial coefficient is zero.
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
edited Jul 24 at 18:08
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
answered Jul 18 at 19:40
Mike Earnest
15.4k11644
15.4k11644
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
add a comment |Â
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
Thank you for this detailed answer.
– Udolf
Jul 19 at 3:30
add a comment |Â
up vote
0
down vote
Do you know how to find the number of solutions for an equation like the following?
$$
x_1 + x_2 + ... + x_n = k
$$
given $forall i in [n]: x_i in S_i$ for some set $S_i$.
answered Jul 18 at 19:25
GuySa
402313
add a comment |Â
up vote
0
down vote
Do you know how to find the number of solutions for an equation like the following?
$$
x_1 + x_2 + ... + x_n = k
$$
given $forall i in [n]: x_i in S_i$ for some set $S_i$.
answered Jul 18 at 19:25
GuySa
402313
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Do you know how to find the number of solutions for an equation like the following?
$$
x_1 + x_2 + ... + x_n = k
$$
given $forall i in [n]: x_i in S_i$ for some set $S_i$.
answered Jul 18 at 19:25
GuySa
402313
Do you know how to find the number of solutions for an equation like the following?
$$
x_1 + x_2 + ... + x_n = k
$$
given $forall i in [n]: x_i in S_i$ for some set $S_i$.
answered Jul 18 at 19:25
GuySa
402313
answered Jul 18 at 19:25
GuySa
402313
answered Jul 18 at 19:25
GuySa
402313
answered Jul 18 at 19:25
GuySa
402313
402313
add a comment |Â
add a comment |Â
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The generating function for $f([3,4,5],n)$ is $$x^3fracx^3-1x-1fracx^4-1x-1fracx^5-1x-1= x^12+3x^11+6x^10+9x^9+11x^8+11x^7+9x^6+6x^5+3x^4+x^3$$. The generating function for $f([2,4,4,6],n)$ is $$scriptsize x^4fracx^2-1x-1fracx^4-1x-1fracx^4-1x-1fracx^6-1x-1 =x^16+4x^15+9x^14+16x^13+23x^12+28x^11+30x^10+28x^9+23x^8+16x^7+9x^6+4x^5+x^4$$.
– robjohn♦
Jul 18 at 19:20
don't dice 2 and 3 have the same number of faces?
– zhw.
Jul 18 at 19:48
yep they have the same number of faces.
– Udolf
Jul 19 at 3:33