Formula for calculating probability on customized dice.
$begingroup$
Given the following:
Player may choose from among the following:
1-10 White dice [0,0,0,1,1,2]
0-40 Blue dice [0,0,1,1,1,2]
0-10 Red dice [0,1,1,1,1,2]
and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.
I know this can be brute forced for each combination, but is there a simple formula to do this?
probability dice
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
$endgroup$
add a comment |
$begingroup$
Given the following:
Player may choose from among the following:
1-10 White dice [0,0,0,1,1,2]
0-40 Blue dice [0,0,1,1,1,2]
0-10 Red dice [0,1,1,1,1,2]
and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.
I know this can be brute forced for each combination, but is there a simple formula to do this?
probability dice
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
$endgroup$
1
$begingroup$
What does1-10mean?
$endgroup$
– Henry
Dec 4 '18 at 20:51
$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26
$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07
add a comment |
$begingroup$
Given the following:
Player may choose from among the following:
1-10 White dice [0,0,0,1,1,2]
0-40 Blue dice [0,0,1,1,1,2]
0-10 Red dice [0,1,1,1,1,2]
and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.
I know this can be brute forced for each combination, but is there a simple formula to do this?
probability dice
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
$endgroup$
Given the following:
Player may choose from among the following:
1-10 White dice [0,0,0,1,1,2]
0-40 Blue dice [0,0,1,1,1,2]
0-10 Red dice [0,1,1,1,1,2]
and rolls them. What is the formula to calculate the probability that xWhite + yBlue + zRed ≥ D where D is an arbitrary whole number, i.e. 8.
I know this can be brute forced for each combination, but is there a simple formula to do this?
probability dice
probability dice
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
edited Dec 4 '18 at 23:15
Ross Millikan
293k23197371
293k23197371
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
asked Dec 4 '18 at 19:42
Jared TritschJared Tritsch
1085
1085
1
$begingroup$
What does1-10mean?
$endgroup$
– Henry
Dec 4 '18 at 20:51
$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26
$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07
add a comment |
1
$begingroup$
What does1-10mean?
$endgroup$
– Henry
Dec 4 '18 at 20:51
$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26
$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07
1
1
$begingroup$
What does
1-10 mean?$endgroup$
– Henry
Dec 4 '18 at 20:51
$begingroup$
What does
1-10 mean?$endgroup$
– Henry
Dec 4 '18 at 20:51
$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26
$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26
$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07
$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
Worked it out using Generating functions.
These dice can be expressed as:
(1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R
where W, B, and R are the quantities of each dice.
Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
$endgroup$
For reasonably large numbers of dice you can use the normal approximation. Compute the mean and variance for each kind of die. For example, a blue die has a mean of $frac 16(0+0+1+1+1+2)=frac 56$ and a variance of $frac 16(0^2+0^2+1^2+1^2+2^2)-(frac 56)^2=frac {17}{36}.$ The sum will be (approximately) normally distributed with mean the sum of the means and variance the sum of the variances. Compute how many standard deviations high or low you are compared to the mean and use a z-score table. I would guess $10$ or $12$ dice would get you reasonably close, but am just guessing. For smaller numbers you can just do it in a spreadsheet, computing the probability of each sum for a pair of dice, then adding in another, and so on.
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
edited Dec 4 '18 at 23:12
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
answered Dec 4 '18 at 19:51
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
$begingroup$
Worked it out using Generating functions.
These dice can be expressed as:
(1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R
where W, B, and R are the quantities of each dice.
Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
$endgroup$
add a comment |
$begingroup$
Worked it out using Generating functions.
These dice can be expressed as:
(1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R
where W, B, and R are the quantities of each dice.
Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
$endgroup$
add a comment |
$begingroup$
Worked it out using Generating functions.
These dice can be expressed as:
(1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R
where W, B, and R are the quantities of each dice.
Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
$endgroup$
Worked it out using Generating functions.
These dice can be expressed as:
(1/2 + x/3 + x^2/6)^W (1/3 + x/2 + x^2/6)^B (1/6 + (2 x)/3 + x^2/6)^R
where W, B, and R are the quantities of each dice.
Expand out into polynomials, then sum up the coefficients starting with the Dth Term (D being the target value). Set that to 1-result and you have your probability of D or higher.
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
answered Dec 5 '18 at 0:39
Jared TritschJared Tritsch
1085
1085
add a comment |
add a comment |
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1
$begingroup$
What does
1-10mean?$endgroup$
– Henry
Dec 4 '18 at 20:51
$begingroup$
@Henry: I believe it means the player can choose how many white dice to roll within the range one to ten. That is $x$
$endgroup$
– Ross Millikan
Dec 4 '18 at 21:26
$begingroup$
@ross, this is what I meant. Players can choose an arbitrary amount of dice within that range.
$endgroup$
– Jared Tritsch
Dec 4 '18 at 23:07