Calculating probability of intersecting sets (Dice rolls arranged in a grid)
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I've recently come across a stat rolling method used by some tabletop gamers that involves rolling 36 stat scores, placing them in a 6x6 grid, and selecting a row, column or diagonal of 6 values to use as your 6 scores. Lets call this method B.
I want to compare this system to the standard method of simply rolling 6 scores and using just those. As this is the original, we'll call it method A. With both methods each individual score is achieved by rolling 4 six-sided dice and adding the three highest values. I want to be able to compare these two methods with similar graphs/metrics to this anydice article.
The standard method obviously returns just one array of stats, so we need just one array to compare from the grid method.
I want to know 2 things:
- How do the bell curves/at-leasts for the SUM of all scores compare between the two methods
- For this the standard method's results should be compared to the array from the grid with the highest sum of scores; eg. Chance Method A yields at least x for the total sum of scores, compared to chance Method B yields at least one array with at least x for the total sum of scores.
- How do the bell curves/at-leasts for the number of 18s you're likely to get compare between the two methods.
- For this the standard method's results should be compared to the array that yields the highest number of 18s; eg. Chance Method A yields x scores of 18 compared to chance Method B yields at least one array with x scores of 18.
How do I calculate these probabilities?
I understand that for each roll, the probability of any specific result is the number of ways that result can be achieved divided by the number of possible results, so getting a 6 on a 6 sided dice is 1 in 6 (16.67%), getting an 18 on 3 6 sided dice (3d6) is 1 in 216 etc (0.4%), and the actual way of getting a single result, 4d6 drop lowest, can return a score of 18 by rolling 6 on any three dice, and anything on the other; so 21 in 1296 (1.62%).
What is getting me, though, is how I work out the number of possibilities and number of desired outcomes when 36 results are randomly entered into a 6x6 grid, and only certain sets (14) of 6 of these outcomes can be combined.
I'm not sure how to reliably test any hypothesis I may have, and even if I do formulate an idea and check it against some test tries I'd really want to get the actual principles of it.
probability dice
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
$endgroup$
|
show 1 more comment
$begingroup$
I've recently come across a stat rolling method used by some tabletop gamers that involves rolling 36 stat scores, placing them in a 6x6 grid, and selecting a row, column or diagonal of 6 values to use as your 6 scores. Lets call this method B.
I want to compare this system to the standard method of simply rolling 6 scores and using just those. As this is the original, we'll call it method A. With both methods each individual score is achieved by rolling 4 six-sided dice and adding the three highest values. I want to be able to compare these two methods with similar graphs/metrics to this anydice article.
The standard method obviously returns just one array of stats, so we need just one array to compare from the grid method.
I want to know 2 things:
- How do the bell curves/at-leasts for the SUM of all scores compare between the two methods
- For this the standard method's results should be compared to the array from the grid with the highest sum of scores; eg. Chance Method A yields at least x for the total sum of scores, compared to chance Method B yields at least one array with at least x for the total sum of scores.
- How do the bell curves/at-leasts for the number of 18s you're likely to get compare between the two methods.
- For this the standard method's results should be compared to the array that yields the highest number of 18s; eg. Chance Method A yields x scores of 18 compared to chance Method B yields at least one array with x scores of 18.
How do I calculate these probabilities?
I understand that for each roll, the probability of any specific result is the number of ways that result can be achieved divided by the number of possible results, so getting a 6 on a 6 sided dice is 1 in 6 (16.67%), getting an 18 on 3 6 sided dice (3d6) is 1 in 216 etc (0.4%), and the actual way of getting a single result, 4d6 drop lowest, can return a score of 18 by rolling 6 on any three dice, and anything on the other; so 21 in 1296 (1.62%).
What is getting me, though, is how I work out the number of possibilities and number of desired outcomes when 36 results are randomly entered into a 6x6 grid, and only certain sets (14) of 6 of these outcomes can be combined.
I'm not sure how to reliably test any hypothesis I may have, and even if I do formulate an idea and check it against some test tries I'd really want to get the actual principles of it.
probability dice
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
$endgroup$
1
$begingroup$
Seems like it would be easy enough to code up a simulation and run it a few million times.
$endgroup$
– amd
Jan 8 at 2:06
$begingroup$
"With both methods each individual score is achieved by ..." If I understand the setting correctly, this means the outcomes are independent. Thus it doesn't matter how many "total rolls" you make and how you arrange them. It's all the same as the "standard method". The entire paragraph of "What is getting me, though..." applies only to situations where there are constraints (the 36 results are correlated). I don't see any constraints here.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:46
$begingroup$
If for example the scores in the 6x6 grid cannot repeat, then that's a different story. Here these 36 scores are just independent.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:53
$begingroup$
@LeeDavidChungLin Perhaps the problem isn't clear: I want to compare the chances of getting certain results within set groups of 6 scores. The individual scores are independent, but there's overlap in the groups. One potential wrinkle this produces is that scores that land in the diagonals are more ... "important" than the others, as they contribute to 3 potential groups, while the others contribute only to two. I'm not sure how/if that affects the probabilities though...
$endgroup$
– Isaac Reefman
Feb 5 at 2:11
$begingroup$
Then the key is how you model the selection of the "group". There are 6 rows, 6 columns, and 2 diagonals. For example, is it uniform among these 14 choices? Or is it a hierarchy uniform one-third (row, column , diagonal) then uniform within (6,6,2 respectively)? The results are different. I don't see this procedure specified.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:12
|
show 1 more comment
$begingroup$
I've recently come across a stat rolling method used by some tabletop gamers that involves rolling 36 stat scores, placing them in a 6x6 grid, and selecting a row, column or diagonal of 6 values to use as your 6 scores. Lets call this method B.
I want to compare this system to the standard method of simply rolling 6 scores and using just those. As this is the original, we'll call it method A. With both methods each individual score is achieved by rolling 4 six-sided dice and adding the three highest values. I want to be able to compare these two methods with similar graphs/metrics to this anydice article.
The standard method obviously returns just one array of stats, so we need just one array to compare from the grid method.
I want to know 2 things:
- How do the bell curves/at-leasts for the SUM of all scores compare between the two methods
- For this the standard method's results should be compared to the array from the grid with the highest sum of scores; eg. Chance Method A yields at least x for the total sum of scores, compared to chance Method B yields at least one array with at least x for the total sum of scores.
- How do the bell curves/at-leasts for the number of 18s you're likely to get compare between the two methods.
- For this the standard method's results should be compared to the array that yields the highest number of 18s; eg. Chance Method A yields x scores of 18 compared to chance Method B yields at least one array with x scores of 18.
How do I calculate these probabilities?
I understand that for each roll, the probability of any specific result is the number of ways that result can be achieved divided by the number of possible results, so getting a 6 on a 6 sided dice is 1 in 6 (16.67%), getting an 18 on 3 6 sided dice (3d6) is 1 in 216 etc (0.4%), and the actual way of getting a single result, 4d6 drop lowest, can return a score of 18 by rolling 6 on any three dice, and anything on the other; so 21 in 1296 (1.62%).
What is getting me, though, is how I work out the number of possibilities and number of desired outcomes when 36 results are randomly entered into a 6x6 grid, and only certain sets (14) of 6 of these outcomes can be combined.
I'm not sure how to reliably test any hypothesis I may have, and even if I do formulate an idea and check it against some test tries I'd really want to get the actual principles of it.
probability dice
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
$endgroup$
I've recently come across a stat rolling method used by some tabletop gamers that involves rolling 36 stat scores, placing them in a 6x6 grid, and selecting a row, column or diagonal of 6 values to use as your 6 scores. Lets call this method B.
I want to compare this system to the standard method of simply rolling 6 scores and using just those. As this is the original, we'll call it method A. With both methods each individual score is achieved by rolling 4 six-sided dice and adding the three highest values. I want to be able to compare these two methods with similar graphs/metrics to this anydice article.
The standard method obviously returns just one array of stats, so we need just one array to compare from the grid method.
I want to know 2 things:
- How do the bell curves/at-leasts for the SUM of all scores compare between the two methods
- For this the standard method's results should be compared to the array from the grid with the highest sum of scores; eg. Chance Method A yields at least x for the total sum of scores, compared to chance Method B yields at least one array with at least x for the total sum of scores.
- How do the bell curves/at-leasts for the number of 18s you're likely to get compare between the two methods.
- For this the standard method's results should be compared to the array that yields the highest number of 18s; eg. Chance Method A yields x scores of 18 compared to chance Method B yields at least one array with x scores of 18.
How do I calculate these probabilities?
I understand that for each roll, the probability of any specific result is the number of ways that result can be achieved divided by the number of possible results, so getting a 6 on a 6 sided dice is 1 in 6 (16.67%), getting an 18 on 3 6 sided dice (3d6) is 1 in 216 etc (0.4%), and the actual way of getting a single result, 4d6 drop lowest, can return a score of 18 by rolling 6 on any three dice, and anything on the other; so 21 in 1296 (1.62%).
What is getting me, though, is how I work out the number of possibilities and number of desired outcomes when 36 results are randomly entered into a 6x6 grid, and only certain sets (14) of 6 of these outcomes can be combined.
I'm not sure how to reliably test any hypothesis I may have, and even if I do formulate an idea and check it against some test tries I'd really want to get the actual principles of it.
probability dice
probability dice
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
edited Feb 6 at 22:25
Isaac Reefman
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
asked Jan 7 at 2:12
Isaac ReefmanIsaac Reefman
1236
1236
1
$begingroup$
Seems like it would be easy enough to code up a simulation and run it a few million times.
$endgroup$
– amd
Jan 8 at 2:06
$begingroup$
"With both methods each individual score is achieved by ..." If I understand the setting correctly, this means the outcomes are independent. Thus it doesn't matter how many "total rolls" you make and how you arrange them. It's all the same as the "standard method". The entire paragraph of "What is getting me, though..." applies only to situations where there are constraints (the 36 results are correlated). I don't see any constraints here.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:46
$begingroup$
If for example the scores in the 6x6 grid cannot repeat, then that's a different story. Here these 36 scores are just independent.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:53
$begingroup$
@LeeDavidChungLin Perhaps the problem isn't clear: I want to compare the chances of getting certain results within set groups of 6 scores. The individual scores are independent, but there's overlap in the groups. One potential wrinkle this produces is that scores that land in the diagonals are more ... "important" than the others, as they contribute to 3 potential groups, while the others contribute only to two. I'm not sure how/if that affects the probabilities though...
$endgroup$
– Isaac Reefman
Feb 5 at 2:11
$begingroup$
Then the key is how you model the selection of the "group". There are 6 rows, 6 columns, and 2 diagonals. For example, is it uniform among these 14 choices? Or is it a hierarchy uniform one-third (row, column , diagonal) then uniform within (6,6,2 respectively)? The results are different. I don't see this procedure specified.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:12
|
show 1 more comment
1
$begingroup$
Seems like it would be easy enough to code up a simulation and run it a few million times.
$endgroup$
– amd
Jan 8 at 2:06
$begingroup$
"With both methods each individual score is achieved by ..." If I understand the setting correctly, this means the outcomes are independent. Thus it doesn't matter how many "total rolls" you make and how you arrange them. It's all the same as the "standard method". The entire paragraph of "What is getting me, though..." applies only to situations where there are constraints (the 36 results are correlated). I don't see any constraints here.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:46
$begingroup$
If for example the scores in the 6x6 grid cannot repeat, then that's a different story. Here these 36 scores are just independent.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:53
$begingroup$
@LeeDavidChungLin Perhaps the problem isn't clear: I want to compare the chances of getting certain results within set groups of 6 scores. The individual scores are independent, but there's overlap in the groups. One potential wrinkle this produces is that scores that land in the diagonals are more ... "important" than the others, as they contribute to 3 potential groups, while the others contribute only to two. I'm not sure how/if that affects the probabilities though...
$endgroup$
– Isaac Reefman
Feb 5 at 2:11
$begingroup$
Then the key is how you model the selection of the "group". There are 6 rows, 6 columns, and 2 diagonals. For example, is it uniform among these 14 choices? Or is it a hierarchy uniform one-third (row, column , diagonal) then uniform within (6,6,2 respectively)? The results are different. I don't see this procedure specified.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:12
1
1
$begingroup$
Seems like it would be easy enough to code up a simulation and run it a few million times.
$endgroup$
– amd
Jan 8 at 2:06
$begingroup$
Seems like it would be easy enough to code up a simulation and run it a few million times.
$endgroup$
– amd
Jan 8 at 2:06
$begingroup$
"With both methods each individual score is achieved by ..." If I understand the setting correctly, this means the outcomes are independent. Thus it doesn't matter how many "total rolls" you make and how you arrange them. It's all the same as the "standard method". The entire paragraph of "What is getting me, though..." applies only to situations where there are constraints (the 36 results are correlated). I don't see any constraints here.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:46
$begingroup$
"With both methods each individual score is achieved by ..." If I understand the setting correctly, this means the outcomes are independent. Thus it doesn't matter how many "total rolls" you make and how you arrange them. It's all the same as the "standard method". The entire paragraph of "What is getting me, though..." applies only to situations where there are constraints (the 36 results are correlated). I don't see any constraints here.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:46
$begingroup$
If for example the scores in the 6x6 grid cannot repeat, then that's a different story. Here these 36 scores are just independent.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:53
$begingroup$
If for example the scores in the 6x6 grid cannot repeat, then that's a different story. Here these 36 scores are just independent.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:53
$begingroup$
@LeeDavidChungLin Perhaps the problem isn't clear: I want to compare the chances of getting certain results within set groups of 6 scores. The individual scores are independent, but there's overlap in the groups. One potential wrinkle this produces is that scores that land in the diagonals are more ... "important" than the others, as they contribute to 3 potential groups, while the others contribute only to two. I'm not sure how/if that affects the probabilities though...
$endgroup$
– Isaac Reefman
Feb 5 at 2:11
$begingroup$
@LeeDavidChungLin Perhaps the problem isn't clear: I want to compare the chances of getting certain results within set groups of 6 scores. The individual scores are independent, but there's overlap in the groups. One potential wrinkle this produces is that scores that land in the diagonals are more ... "important" than the others, as they contribute to 3 potential groups, while the others contribute only to two. I'm not sure how/if that affects the probabilities though...
$endgroup$
– Isaac Reefman
Feb 5 at 2:11
$begingroup$
Then the key is how you model the selection of the "group". There are 6 rows, 6 columns, and 2 diagonals. For example, is it uniform among these 14 choices? Or is it a hierarchy uniform one-third (row, column , diagonal) then uniform within (6,6,2 respectively)? The results are different. I don't see this procedure specified.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:12
$begingroup$
Then the key is how you model the selection of the "group". There are 6 rows, 6 columns, and 2 diagonals. For example, is it uniform among these 14 choices? Or is it a hierarchy uniform one-third (row, column , diagonal) then uniform within (6,6,2 respectively)? The results are different. I don't see this procedure specified.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:12
|
show 1 more comment
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$begingroup$
Seems like it would be easy enough to code up a simulation and run it a few million times.
$endgroup$
– amd
Jan 8 at 2:06
$begingroup$
"With both methods each individual score is achieved by ..." If I understand the setting correctly, this means the outcomes are independent. Thus it doesn't matter how many "total rolls" you make and how you arrange them. It's all the same as the "standard method". The entire paragraph of "What is getting me, though..." applies only to situations where there are constraints (the 36 results are correlated). I don't see any constraints here.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:46
$begingroup$
If for example the scores in the 6x6 grid cannot repeat, then that's a different story. Here these 36 scores are just independent.
$endgroup$
– Lee David Chung Lin
Jan 29 at 11:53
$begingroup$
@LeeDavidChungLin Perhaps the problem isn't clear: I want to compare the chances of getting certain results within set groups of 6 scores. The individual scores are independent, but there's overlap in the groups. One potential wrinkle this produces is that scores that land in the diagonals are more ... "important" than the others, as they contribute to 3 potential groups, while the others contribute only to two. I'm not sure how/if that affects the probabilities though...
$endgroup$
– Isaac Reefman
Feb 5 at 2:11
$begingroup$
Then the key is how you model the selection of the "group". There are 6 rows, 6 columns, and 2 diagonals. For example, is it uniform among these 14 choices? Or is it a hierarchy uniform one-third (row, column , diagonal) then uniform within (6,6,2 respectively)? The results are different. I don't see this procedure specified.
$endgroup$
– Lee David Chung Lin
Feb 5 at 8:12