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Calculate no of sequences with sum k, out of die toss n times, with die having m outcomes
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I am trying to calculate total no of sequences possible with given sum k, out of a die (with m outcomes), tossed n times.
I could go after each combination, but $m^n$ grows very large quickly, so want a faster way to generate this. For coin flip, I simply grabbed binomial coefficients of respective row from Pascal's triangle, but here I am stuck on how to proceed. Can you kindly help?
Big picture:I am trying to understand how underlying nature of combinations evolve into normal distribution both theoretically and statistically.
Below is when I throw dice with 3 outcomes (0,1,2), 2 times. I get $[1,2,3,2,1]$ as the frequency of sequence as you see on LHs of the diagram.
Below is when I throw dice with 3 outcomes (0,1,2), 3 times. I get $[1,3,6,7,6,3,1]$ as frequency of sequence.
I have generated these, by directly generating all possible combinations and counting. When n is large, computing time is exponentially large. So I want a formula shortcut to come up with these frequency values pro grammatically. This is why above question.
probability probability-distributions normal-distribution dice
asked Aug 2 at 11:47
Paari Vendhan
355
 |Â
show 14 more comments
up vote
0
down vote
favorite
I am trying to calculate total no of sequences possible with given sum k, out of a die (with m outcomes), tossed n times.
I could go after each combination, but $m^n$ grows very large quickly, so want a faster way to generate this. For coin flip, I simply grabbed binomial coefficients of respective row from Pascal's triangle, but here I am stuck on how to proceed. Can you kindly help?
Big picture:I am trying to understand how underlying nature of combinations evolve into normal distribution both theoretically and statistically.
Below is when I throw dice with 3 outcomes (0,1,2), 2 times. I get $[1,2,3,2,1]$ as the frequency of sequence as you see on LHs of the diagram.
Below is when I throw dice with 3 outcomes (0,1,2), 3 times. I get $[1,3,6,7,6,3,1]$ as frequency of sequence.
I have generated these, by directly generating all possible combinations and counting. When n is large, computing time is exponentially large. So I want a formula shortcut to come up with these frequency values pro grammatically. This is why above question.
probability probability-distributions normal-distribution dice
asked Aug 2 at 11:47
Paari Vendhan
355
here is a discussion of a related problem.
– lulu
Aug 2 at 11:52
checked it, could not relate directly. can you please transform that to this problem?
– Paari Vendhan
Aug 2 at 12:28
$a_i$ is the value of the $i^th$ die (or one less than that). Your problem is the case where all the $r_i$ are equal. Specifically, if you are dealing with ordinary dice, all of your $r_i$ are $6$. The fact that your rolls are all at least $1$ is not a significant difference, as you can just subtract $n$ from the total and remove that constraint.
– lulu
Aug 2 at 12:31
Worth noting: the first answer specifically addresses the case where all the $r_i$ are equal.
– lulu
Aug 2 at 12:33
Please check if my inference is correct. Suppose I have a die with 3 outcomes (say 0,1,2). I flip it 2 times. Let X be sum of outcomes. I want to calculate $X_2$, that is, how many outcomes where sum is 2. So, $a_1 + a_2 + a_3 = 2$. If so, we need to solve $(1+X+X^2+X^3)^2 = dfrac (1-X^4)^2(1-X)^2$?
– Paari Vendhan
Aug 2 at 13:49
 |Â
show 14 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to calculate total no of sequences possible with given sum k, out of a die (with m outcomes), tossed n times.
I could go after each combination, but $m^n$ grows very large quickly, so want a faster way to generate this. For coin flip, I simply grabbed binomial coefficients of respective row from Pascal's triangle, but here I am stuck on how to proceed. Can you kindly help?
Big picture:I am trying to understand how underlying nature of combinations evolve into normal distribution both theoretically and statistically.
Below is when I throw dice with 3 outcomes (0,1,2), 2 times. I get $[1,2,3,2,1]$ as the frequency of sequence as you see on LHs of the diagram.
Below is when I throw dice with 3 outcomes (0,1,2), 3 times. I get $[1,3,6,7,6,3,1]$ as frequency of sequence.
I have generated these, by directly generating all possible combinations and counting. When n is large, computing time is exponentially large. So I want a formula shortcut to come up with these frequency values pro grammatically. This is why above question.
probability probability-distributions normal-distribution dice
asked Aug 2 at 11:47
Paari Vendhan
355
I am trying to calculate total no of sequences possible with given sum k, out of a die (with m outcomes), tossed n times.
I could go after each combination, but $m^n$ grows very large quickly, so want a faster way to generate this. For coin flip, I simply grabbed binomial coefficients of respective row from Pascal's triangle, but here I am stuck on how to proceed. Can you kindly help?
Big picture:I am trying to understand how underlying nature of combinations evolve into normal distribution both theoretically and statistically.
Below is when I throw dice with 3 outcomes (0,1,2), 2 times. I get $[1,2,3,2,1]$ as the frequency of sequence as you see on LHs of the diagram.
Below is when I throw dice with 3 outcomes (0,1,2), 3 times. I get $[1,3,6,7,6,3,1]$ as frequency of sequence.
I have generated these, by directly generating all possible combinations and counting. When n is large, computing time is exponentially large. So I want a formula shortcut to come up with these frequency values pro grammatically. This is why above question.
probability probability-distributions normal-distribution dice
asked Aug 2 at 11:47
Paari Vendhan
355
edited Aug 2 at 14:31
asked Aug 2 at 11:47
Paari Vendhan
355
asked Aug 2 at 11:47
Paari Vendhan
355
asked Aug 2 at 11:47
Paari Vendhan
355
355
here is a discussion of a related problem.
– lulu
Aug 2 at 11:52
checked it, could not relate directly. can you please transform that to this problem?
– Paari Vendhan
Aug 2 at 12:28
$a_i$ is the value of the $i^th$ die (or one less than that). Your problem is the case where all the $r_i$ are equal. Specifically, if you are dealing with ordinary dice, all of your $r_i$ are $6$. The fact that your rolls are all at least $1$ is not a significant difference, as you can just subtract $n$ from the total and remove that constraint.
– lulu
Aug 2 at 12:31
Worth noting: the first answer specifically addresses the case where all the $r_i$ are equal.
– lulu
Aug 2 at 12:33
Please check if my inference is correct. Suppose I have a die with 3 outcomes (say 0,1,2). I flip it 2 times. Let X be sum of outcomes. I want to calculate $X_2$, that is, how many outcomes where sum is 2. So, $a_1 + a_2 + a_3 = 2$. If so, we need to solve $(1+X+X^2+X^3)^2 = dfrac (1-X^4)^2(1-X)^2$?
– Paari Vendhan
Aug 2 at 13:49
 |Â
show 14 more comments
here is a discussion of a related problem.
– lulu
Aug 2 at 11:52
checked it, could not relate directly. can you please transform that to this problem?
– Paari Vendhan
Aug 2 at 12:28
$a_i$ is the value of the $i^th$ die (or one less than that). Your problem is the case where all the $r_i$ are equal. Specifically, if you are dealing with ordinary dice, all of your $r_i$ are $6$. The fact that your rolls are all at least $1$ is not a significant difference, as you can just subtract $n$ from the total and remove that constraint.
– lulu
Aug 2 at 12:31
Worth noting: the first answer specifically addresses the case where all the $r_i$ are equal.
– lulu
Aug 2 at 12:33
Please check if my inference is correct. Suppose I have a die with 3 outcomes (say 0,1,2). I flip it 2 times. Let X be sum of outcomes. I want to calculate $X_2$, that is, how many outcomes where sum is 2. So, $a_1 + a_2 + a_3 = 2$. If so, we need to solve $(1+X+X^2+X^3)^2 = dfrac (1-X^4)^2(1-X)^2$?
– Paari Vendhan
Aug 2 at 13:49
here is a discussion of a related problem.
– lulu
Aug 2 at 11:52
here is a discussion of a related problem.
– lulu
Aug 2 at 11:52
checked it, could not relate directly. can you please transform that to this problem?
– Paari Vendhan
Aug 2 at 12:28
checked it, could not relate directly. can you please transform that to this problem?
– Paari Vendhan
Aug 2 at 12:28
$a_i$ is the value of the $i^th$ die (or one less than that). Your problem is the case where all the $r_i$ are equal. Specifically, if you are dealing with ordinary dice, all of your $r_i$ are $6$. The fact that your rolls are all at least $1$ is not a significant difference, as you can just subtract $n$ from the total and remove that constraint.
– lulu
Aug 2 at 12:31
$a_i$ is the value of the $i^th$ die (or one less than that). Your problem is the case where all the $r_i$ are equal. Specifically, if you are dealing with ordinary dice, all of your $r_i$ are $6$. The fact that your rolls are all at least $1$ is not a significant difference, as you can just subtract $n$ from the total and remove that constraint.
– lulu
Aug 2 at 12:31
Worth noting: the first answer specifically addresses the case where all the $r_i$ are equal.
– lulu
Aug 2 at 12:33
Worth noting: the first answer specifically addresses the case where all the $r_i$ are equal.
– lulu
Aug 2 at 12:33
Please check if my inference is correct. Suppose I have a die with 3 outcomes (say 0,1,2). I flip it 2 times. Let X be sum of outcomes. I want to calculate $X_2$, that is, how many outcomes where sum is 2. So, $a_1 + a_2 + a_3 = 2$. If so, we need to solve $(1+X+X^2+X^3)^2 = dfrac (1-X^4)^2(1-X)^2$?
– Paari Vendhan
Aug 2 at 13:49
Please check if my inference is correct. Suppose I have a die with 3 outcomes (say 0,1,2). I flip it 2 times. Let X be sum of outcomes. I want to calculate $X_2$, that is, how many outcomes where sum is 2. So, $a_1 + a_2 + a_3 = 2$. If so, we need to solve $(1+X+X^2+X^3)^2 = dfrac (1-X^4)^2(1-X)^2$?
– Paari Vendhan
Aug 2 at 13:49
 |Â
show 14 more comments
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here is a discussion of a related problem.
– lulu
Aug 2 at 11:52
checked it, could not relate directly. can you please transform that to this problem?
– Paari Vendhan
Aug 2 at 12:28
$a_i$ is the value of the $i^th$ die (or one less than that). Your problem is the case where all the $r_i$ are equal. Specifically, if you are dealing with ordinary dice, all of your $r_i$ are $6$. The fact that your rolls are all at least $1$ is not a significant difference, as you can just subtract $n$ from the total and remove that constraint.
– lulu
Aug 2 at 12:31
Worth noting: the first answer specifically addresses the case where all the $r_i$ are equal.
– lulu
Aug 2 at 12:33
Please check if my inference is correct. Suppose I have a die with 3 outcomes (say 0,1,2). I flip it 2 times. Let X be sum of outcomes. I want to calculate $X_2$, that is, how many outcomes where sum is 2. So, $a_1 + a_2 + a_3 = 2$. If so, we need to solve $(1+X+X^2+X^3)^2 = dfrac (1-X^4)^2(1-X)^2$?
– Paari Vendhan
Aug 2 at 13:49