Mixing
Dungeons and Dragons: Ability Score combinations
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
I am trying to figure out how likely it is to roll a certain combination of 6 ability scores in Dungeons and Dragons.
edit:
Dungeons and Dragons is a tabletop role-playing game. You create a character in a fantasy world which has 6 main statistics, Strength, Dexterity, Constitution, Intelligence, Wisdom, and Charisma. Those ability scores determine the powers that a certain character can use and how strong they are.
The ability scores are rolled using 4 6-sided dice (4d6) and then using the 3 highest scores, giving you values $in [3, 18]$ for the possible values.
My goal is to compute the likelihood of getting a combination like (8, 8, 8, 8, 8, 8) or (15, 14, 13, 12, 10, 8)./edit
I already worked out the probabilities for rolling a single stat and a question on here confirmed that I did the math right:
3: $frac11296$, 4: $frac41296$, 5: $frac101296$, 6: $frac211296$, 7: $frac381296$, 8: $frac621296$, 9: $frac911296$, 10: $frac1221296$, 11: $frac1481296$, 12: $frac1671296$, 13: $frac1721296$, 14: $frac1601296$, 15: $frac1311296$, 16: $frac941296$, 17: $frac541296$, 18: $frac211296$
Now using the same reasoning I managed to work out the probabilities of combinations based on the sum of ability scores, but only if the outcomes are equally likely. I used Python to compute the probabilities like this:
def probability_of_outcomes(outcomes, repetitions, subsum=slice(None, None, None)):
probability_dict =
sum_combinations = 0
for c in combinations_with_replacement(outcomes, repetitions):
for p in set(permutations(c, repetitions)):
s = sum(sorted(p)[subsum])
if s in probability_dict:
probability_dict[s] += 1
else:
probability_dict[s] = 1
sum_combinations += 1
return probability_dict, sum_combinations
if __name__ == "__main__":
all_comb, num_combinations_stats = probability_of_outcomes(range(1, 7) 4, subsum=slice(1, None, None))
in the same way I can compute:
abilities_comb, num_combinations_ability = probability_of_outcomes(range(3, 19) 6)
Now with the first computation I can already see that not every number is equally likely, but I am unsure where to add that probability in. Also the approach is not very elegant, as I simply try out every possible combination.
Any ideas?
probability combinatorics dice
edited 3 hours ago
Especially Lime
19.7k22353
asked 4 hours ago
meetaig
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
6
down vote
favorite
I am trying to figure out how likely it is to roll a certain combination of 6 ability scores in Dungeons and Dragons.
edit:
Dungeons and Dragons is a tabletop role-playing game. You create a character in a fantasy world which has 6 main statistics, Strength, Dexterity, Constitution, Intelligence, Wisdom, and Charisma. Those ability scores determine the powers that a certain character can use and how strong they are.
The ability scores are rolled using 4 6-sided dice (4d6) and then using the 3 highest scores, giving you values $in [3, 18]$ for the possible values.
My goal is to compute the likelihood of getting a combination like (8, 8, 8, 8, 8, 8) or (15, 14, 13, 12, 10, 8)./edit
I already worked out the probabilities for rolling a single stat and a question on here confirmed that I did the math right:
3: $frac11296$, 4: $frac41296$, 5: $frac101296$, 6: $frac211296$, 7: $frac381296$, 8: $frac621296$, 9: $frac911296$, 10: $frac1221296$, 11: $frac1481296$, 12: $frac1671296$, 13: $frac1721296$, 14: $frac1601296$, 15: $frac1311296$, 16: $frac941296$, 17: $frac541296$, 18: $frac211296$
Now using the same reasoning I managed to work out the probabilities of combinations based on the sum of ability scores, but only if the outcomes are equally likely. I used Python to compute the probabilities like this:
def probability_of_outcomes(outcomes, repetitions, subsum=slice(None, None, None)):
probability_dict =
sum_combinations = 0
for c in combinations_with_replacement(outcomes, repetitions):
for p in set(permutations(c, repetitions)):
s = sum(sorted(p)[subsum])
if s in probability_dict:
probability_dict[s] += 1
else:
probability_dict[s] = 1
sum_combinations += 1
return probability_dict, sum_combinations
if __name__ == "__main__":
all_comb, num_combinations_stats = probability_of_outcomes(range(1, 7) 4, subsum=slice(1, None, None))
in the same way I can compute:
abilities_comb, num_combinations_ability = probability_of_outcomes(range(3, 19) 6)
Now with the first computation I can already see that not every number is equally likely, but I am unsure where to add that probability in. Also the approach is not very elegant, as I simply try out every possible combination.
Any ideas?
probability combinatorics dice
edited 3 hours ago
Especially Lime
19.7k22353
asked 4 hours ago
meetaig
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Not everyone plays DnD, please explain in plain words what probability you're trying to compute.
– orlp
4 hours ago
1
You seem to be using the "roll four dice and ignore the smallest" method. Is that right?
– MJD
4 hours ago
@MJD yes that is the method I use in the first example with thesubsum. But when there is None specified, the code just takes all results (which is for the second call).
– meetaig
4 hours ago
@orlp I'll edit the question
– meetaig
4 hours ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I am trying to figure out how likely it is to roll a certain combination of 6 ability scores in Dungeons and Dragons.
edit:
Dungeons and Dragons is a tabletop role-playing game. You create a character in a fantasy world which has 6 main statistics, Strength, Dexterity, Constitution, Intelligence, Wisdom, and Charisma. Those ability scores determine the powers that a certain character can use and how strong they are.
The ability scores are rolled using 4 6-sided dice (4d6) and then using the 3 highest scores, giving you values $in [3, 18]$ for the possible values.
My goal is to compute the likelihood of getting a combination like (8, 8, 8, 8, 8, 8) or (15, 14, 13, 12, 10, 8)./edit
I already worked out the probabilities for rolling a single stat and a question on here confirmed that I did the math right:
3: $frac11296$, 4: $frac41296$, 5: $frac101296$, 6: $frac211296$, 7: $frac381296$, 8: $frac621296$, 9: $frac911296$, 10: $frac1221296$, 11: $frac1481296$, 12: $frac1671296$, 13: $frac1721296$, 14: $frac1601296$, 15: $frac1311296$, 16: $frac941296$, 17: $frac541296$, 18: $frac211296$
Now using the same reasoning I managed to work out the probabilities of combinations based on the sum of ability scores, but only if the outcomes are equally likely. I used Python to compute the probabilities like this:
def probability_of_outcomes(outcomes, repetitions, subsum=slice(None, None, None)):
probability_dict =
sum_combinations = 0
for c in combinations_with_replacement(outcomes, repetitions):
for p in set(permutations(c, repetitions)):
s = sum(sorted(p)[subsum])
if s in probability_dict:
probability_dict[s] += 1
else:
probability_dict[s] = 1
sum_combinations += 1
return probability_dict, sum_combinations
if __name__ == "__main__":
all_comb, num_combinations_stats = probability_of_outcomes(range(1, 7) 4, subsum=slice(1, None, None))
in the same way I can compute:
abilities_comb, num_combinations_ability = probability_of_outcomes(range(3, 19) 6)
Now with the first computation I can already see that not every number is equally likely, but I am unsure where to add that probability in. Also the approach is not very elegant, as I simply try out every possible combination.
Any ideas?
probability combinatorics dice
edited 3 hours ago
Especially Lime
19.7k22353
asked 4 hours ago
meetaig
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am trying to figure out how likely it is to roll a certain combination of 6 ability scores in Dungeons and Dragons.
edit:
Dungeons and Dragons is a tabletop role-playing game. You create a character in a fantasy world which has 6 main statistics, Strength, Dexterity, Constitution, Intelligence, Wisdom, and Charisma. Those ability scores determine the powers that a certain character can use and how strong they are.
The ability scores are rolled using 4 6-sided dice (4d6) and then using the 3 highest scores, giving you values $in [3, 18]$ for the possible values.
My goal is to compute the likelihood of getting a combination like (8, 8, 8, 8, 8, 8) or (15, 14, 13, 12, 10, 8)./edit
I already worked out the probabilities for rolling a single stat and a question on here confirmed that I did the math right:
3: $frac11296$, 4: $frac41296$, 5: $frac101296$, 6: $frac211296$, 7: $frac381296$, 8: $frac621296$, 9: $frac911296$, 10: $frac1221296$, 11: $frac1481296$, 12: $frac1671296$, 13: $frac1721296$, 14: $frac1601296$, 15: $frac1311296$, 16: $frac941296$, 17: $frac541296$, 18: $frac211296$
Now using the same reasoning I managed to work out the probabilities of combinations based on the sum of ability scores, but only if the outcomes are equally likely. I used Python to compute the probabilities like this:
def probability_of_outcomes(outcomes, repetitions, subsum=slice(None, None, None)):
probability_dict =
sum_combinations = 0
for c in combinations_with_replacement(outcomes, repetitions):
for p in set(permutations(c, repetitions)):
s = sum(sorted(p)[subsum])
if s in probability_dict:
probability_dict[s] += 1
else:
probability_dict[s] = 1
sum_combinations += 1
return probability_dict, sum_combinations
if __name__ == "__main__":
all_comb, num_combinations_stats = probability_of_outcomes(range(1, 7) 4, subsum=slice(1, None, None))
in the same way I can compute:
abilities_comb, num_combinations_ability = probability_of_outcomes(range(3, 19) 6)
Now with the first computation I can already see that not every number is equally likely, but I am unsure where to add that probability in. Also the approach is not very elegant, as I simply try out every possible combination.
Any ideas?
probability combinatorics dice
probability combinatorics dice
edited 3 hours ago
Especially Lime
19.7k22353
asked 4 hours ago
meetaig
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Especially Lime
19.7k22353
asked 4 hours ago
meetaig
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Especially Lime
19.7k22353
edited 3 hours ago
Especially Lime
19.7k22353
edited 3 hours ago
Especially Lime
19.7k22353
19.7k22353
asked 4 hours ago
meetaig
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
meetaig
1335
asked 4 hours ago
meetaig
1335
1335
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
meetaig is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Not everyone plays DnD, please explain in plain words what probability you're trying to compute.
– orlp
4 hours ago
1
You seem to be using the "roll four dice and ignore the smallest" method. Is that right?
– MJD
4 hours ago
@MJD yes that is the method I use in the first example with thesubsum. But when there is None specified, the code just takes all results (which is for the second call).
– meetaig
4 hours ago
@orlp I'll edit the question
– meetaig
4 hours ago
add a comment |Â
Not everyone plays DnD, please explain in plain words what probability you're trying to compute.
– orlp
4 hours ago
1
You seem to be using the "roll four dice and ignore the smallest" method. Is that right?
– MJD
4 hours ago
@MJD yes that is the method I use in the first example with thesubsum. But when there is None specified, the code just takes all results (which is for the second call).
– meetaig
4 hours ago
@orlp I'll edit the question
– meetaig
4 hours ago
Not everyone plays DnD, please explain in plain words what probability you're trying to compute.
– orlp
4 hours ago
Not everyone plays DnD, please explain in plain words what probability you're trying to compute.
– orlp
4 hours ago
1
1
You seem to be using the "roll four dice and ignore the smallest" method. Is that right?
– MJD
4 hours ago
You seem to be using the "roll four dice and ignore the smallest" method. Is that right?
– MJD
4 hours ago
@MJD yes that is the method I use in the first example with the
subsum. But when there is None specified, the code just takes all results (which is for the second call).– meetaig
4 hours ago
@MJD yes that is the method I use in the first example with the
subsum. But when there is None specified, the code just takes all results (which is for the second call).– meetaig
4 hours ago
@orlp I'll edit the question
– meetaig
4 hours ago
@orlp I'll edit the question
– meetaig
4 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
First order the combination, e.g. $(8, 10, 12, 13, 14, 15)$. Then calculate the exact probability of rolling this combination in this order. This calculation is simple, just multiply the probabilities of the individual rolls you calculated earlier:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15)$$
But we're undercounting. After all, it's possible that you rolled $15$ first instead of last. So we multiply by $6!$ to account for all permutations:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15) times 6!$$
In this case we're done, because all attributes are unique. When this isn't the case, we're now overcounting. Because in a two-dice example, if you swap the order of rolling two sixes, nothing changes.
So to finish the calculation you need to calculate the number of orderings considering duplicates.
If $a$ is the multiset of attributes, $a_1$ is the first attribute, $a_2$ the second, etc, $|a_1|$ is the number of times the first attribute occurs, and $k$ is the number of distinct attributes, the full formula is:
$$p(a) = p(a_1)^a_1p(a_2)^cdots p(a_k)^ cdot frack!!cdots$$
Note that in our above example where all attributes were distinct the denominator went away because it was simply $1! cdot 1! cdots 1! = 1$.
And in the other extreme, where the multiset only has a single element repeated, like in $(8, 8, 8, 8, 8, 8)$, we get:
$$p(8)^6cdot frac6!6! = p(8)^6$$
As expected.
answered 4 hours ago
orlp
6,9981229
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
First order the combination, e.g. $(8, 10, 12, 13, 14, 15)$. Then calculate the exact probability of rolling this combination in this order. This calculation is simple, just multiply the probabilities of the individual rolls you calculated earlier:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15)$$
But we're undercounting. After all, it's possible that you rolled $15$ first instead of last. So we multiply by $6!$ to account for all permutations:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15) times 6!$$
In this case we're done, because all attributes are unique. When this isn't the case, we're now overcounting. Because in a two-dice example, if you swap the order of rolling two sixes, nothing changes.
So to finish the calculation you need to calculate the number of orderings considering duplicates.
If $a$ is the multiset of attributes, $a_1$ is the first attribute, $a_2$ the second, etc, $|a_1|$ is the number of times the first attribute occurs, and $k$ is the number of distinct attributes, the full formula is:
$$p(a) = p(a_1)^a_1p(a_2)^cdots p(a_k)^ cdot frack!!cdots$$
Note that in our above example where all attributes were distinct the denominator went away because it was simply $1! cdot 1! cdots 1! = 1$.
And in the other extreme, where the multiset only has a single element repeated, like in $(8, 8, 8, 8, 8, 8)$, we get:
$$p(8)^6cdot frac6!6! = p(8)^6$$
As expected.
answered 4 hours ago
orlp
6,9981229
add a comment |Â
up vote
4
down vote
accepted
First order the combination, e.g. $(8, 10, 12, 13, 14, 15)$. Then calculate the exact probability of rolling this combination in this order. This calculation is simple, just multiply the probabilities of the individual rolls you calculated earlier:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15)$$
But we're undercounting. After all, it's possible that you rolled $15$ first instead of last. So we multiply by $6!$ to account for all permutations:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15) times 6!$$
In this case we're done, because all attributes are unique. When this isn't the case, we're now overcounting. Because in a two-dice example, if you swap the order of rolling two sixes, nothing changes.
So to finish the calculation you need to calculate the number of orderings considering duplicates.
If $a$ is the multiset of attributes, $a_1$ is the first attribute, $a_2$ the second, etc, $|a_1|$ is the number of times the first attribute occurs, and $k$ is the number of distinct attributes, the full formula is:
$$p(a) = p(a_1)^a_1p(a_2)^cdots p(a_k)^ cdot frack!!cdots$$
Note that in our above example where all attributes were distinct the denominator went away because it was simply $1! cdot 1! cdots 1! = 1$.
And in the other extreme, where the multiset only has a single element repeated, like in $(8, 8, 8, 8, 8, 8)$, we get:
$$p(8)^6cdot frac6!6! = p(8)^6$$
As expected.
answered 4 hours ago
orlp
6,9981229
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
First order the combination, e.g. $(8, 10, 12, 13, 14, 15)$. Then calculate the exact probability of rolling this combination in this order. This calculation is simple, just multiply the probabilities of the individual rolls you calculated earlier:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15)$$
But we're undercounting. After all, it's possible that you rolled $15$ first instead of last. So we multiply by $6!$ to account for all permutations:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15) times 6!$$
In this case we're done, because all attributes are unique. When this isn't the case, we're now overcounting. Because in a two-dice example, if you swap the order of rolling two sixes, nothing changes.
So to finish the calculation you need to calculate the number of orderings considering duplicates.
If $a$ is the multiset of attributes, $a_1$ is the first attribute, $a_2$ the second, etc, $|a_1|$ is the number of times the first attribute occurs, and $k$ is the number of distinct attributes, the full formula is:
$$p(a) = p(a_1)^a_1p(a_2)^cdots p(a_k)^ cdot frack!!cdots$$
Note that in our above example where all attributes were distinct the denominator went away because it was simply $1! cdot 1! cdots 1! = 1$.
And in the other extreme, where the multiset only has a single element repeated, like in $(8, 8, 8, 8, 8, 8)$, we get:
$$p(8)^6cdot frac6!6! = p(8)^6$$
As expected.
answered 4 hours ago
orlp
6,9981229
First order the combination, e.g. $(8, 10, 12, 13, 14, 15)$. Then calculate the exact probability of rolling this combination in this order. This calculation is simple, just multiply the probabilities of the individual rolls you calculated earlier:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15)$$
But we're undercounting. After all, it's possible that you rolled $15$ first instead of last. So we multiply by $6!$ to account for all permutations:
$$p(8) times p(10) times p(12) times p(13) times p(14) times p(15) times 6!$$
In this case we're done, because all attributes are unique. When this isn't the case, we're now overcounting. Because in a two-dice example, if you swap the order of rolling two sixes, nothing changes.
So to finish the calculation you need to calculate the number of orderings considering duplicates.
If $a$ is the multiset of attributes, $a_1$ is the first attribute, $a_2$ the second, etc, $|a_1|$ is the number of times the first attribute occurs, and $k$ is the number of distinct attributes, the full formula is:
$$p(a) = p(a_1)^a_1p(a_2)^cdots p(a_k)^ cdot frack!!cdots$$
Note that in our above example where all attributes were distinct the denominator went away because it was simply $1! cdot 1! cdots 1! = 1$.
And in the other extreme, where the multiset only has a single element repeated, like in $(8, 8, 8, 8, 8, 8)$, we get:
$$p(8)^6cdot frac6!6! = p(8)^6$$
As expected.
answered 4 hours ago
orlp
6,9981229
edited 3 hours ago
answered 4 hours ago
orlp
6,9981229
answered 4 hours ago
orlp
6,9981229
answered 4 hours ago
orlp
6,9981229
6,9981229
add a comment |Â
add a comment |Â
meetaig is a new contributor. Be nice, and check out our Code of Conduct.
meetaig is a new contributor. Be nice, and check out our Code of Conduct.
meetaig is a new contributor. Be nice, and check out our Code of Conduct.
meetaig is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2939250%2fdungeons-and-dragons-ability-score-combinations%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Not everyone plays DnD, please explain in plain words what probability you're trying to compute.
– orlp
4 hours ago
1
You seem to be using the "roll four dice and ignore the smallest" method. Is that right?
– MJD
4 hours ago
@MJD yes that is the method I use in the first example with the
subsum. But when there is None specified, the code just takes all results (which is for the second call).– meetaig
4 hours ago
@orlp I'll edit the question
– meetaig
4 hours ago