Ability Score Probabilities [Archive] - Wizards Community

Ability Score Probabilities [Archive] - Wizards Community

Post/Author/DateTime Post
StoneStokes
12-19-06, 06:03 PM There are 1296 ways of rolling 4d6. Using the 4d6-drop-lowest method for ability scores, we get the following probabilities. Column 1 is the score, Column 2 is the number of ways to roll that score with 4d6-drop-lowest, out of 1296.

18 21
17 54
16 94
15 131
14 160
13 172
12 167
11 148
10 122
9 91
8 62
7 38
6 21
5 10
4 4
3 1

Using these probabilities we can make a few comments:

The most likely score is 13
The expected score is 12.2
Rolling six scores, the expected total modifier is +5.4 (add the modifiers of all your abilities, and you should expect +5.4).

Cheers,
StoneStokes
TheChilliGod
12-19-06, 06:56 PM Well done.
May I ask how you made such calculations? For comparison, see; I came up with pretty much the same probabilities as you about half a year ago, but I manually worked out every one of the 1296 combinations, and I'm wondering if you did the same thing or smartly used some formula.

One thing I find interesting about the results, you have an equal probability (1.62%) of rolling an 18 as you do a 6. And that is the only pair of equal probabilities.
Gilean
12-20-06, 03:36 AM So, let us see...

Mode (the most likely result): 13
Median (the middle result): 12
Expected value (average): 15869/1296 (~12,24)
Variance: ~8,10
Standard deviation: ~2,85
Lower and upper quartile (half of results are between these, inclusive): 10, 14
Entropy (scaled): ~0,89

I'm not bored enough to calculate skewness or curtosis.
Nihzlet
12-20-06, 04:16 AM interesting. probabilities dont stop player from rolling ridiculously good(or crappy) sets of stats using that system, however :P
StoneStokes
12-21-06, 01:20 PM Well done.
May I ask how you made such calculations? For comparison, see; I came up with pretty much the same probabilities as you about half a year ago, but I manually worked out every one of the 1296 combinations, and I'm wondering if you did the same thing or smartly used some formula.

One thing I find interesting about the results, you have an equal probability (1.62%) of rolling an 18 as you do a 6. And that is the only pair of equal probabilities.

I counted them in units. A unit is a sequence: either abcc or abcx, where a,b,c are known with a >= b >= c > x. Then multiply by the number of ways you can permute abcc or abcx. It didn't take too long, but would have been easier if I had a calculator when I computed it (the first run through, and I had a total of 1295 possibilities!).
Solaris
12-21-06, 07:38 PM I counted them in units. ... Then multiply by the number of ways you can permute

Not bad; I hadn't thought of doing it that way. This is a little harder to implement than a simple brute-force search, but since this generates only C(dice + sides - 1, dice) results, while brute-force has to generate all sides^dice possibilities, there's a substantial savings of computation time as dice and sides increase, making larger numbers more accessible.
darkhunter2007
12-22-06, 04:36 AM how bored you are to do such amezing thing?
Ryusage
12-22-06, 09:59 AM how bored you are to do such amezing thing?

How bored you are to play D&D?
Azezel
12-22-06, 12:07 PM StoneStokes, excellent resource - I have wondered about that myself.

I have put your information into visual form for anyone who wants it.

I grant permission to download, reproduce, distribute or modify the below image without limit.

http://xs510.xs.to/xs510/06515/Ability_Generation.JPG
MindLich
12-22-06, 12:23 PM I'm starting a campaign with a 3d6 roll for ability scores. Is anyone sufficiently bored to calculate the probabilities? Please, and thanks.
Nihzlet
12-22-06, 12:55 PM Posted by Ryusage:
How bored you are to play D&D?since when was being bored a prerequisite to playing D&D?
David5514
12-22-06, 01:10 PM There are 1296 ways of rolling 4d6. Using the 4d6-drop-lowest method for ability scores, we get the following probabilities. Column 1 is the score, Column 2 is the number of ways to roll that score with 4d6-drop-lowest, out of 1296.

18 21

Cheers,
StoneStokes

Sorry I may be missing some thing here but surrly the only way to get 18 on 4D6 drop lowest is 3 six's? I may be bad at maths but that is the only way I can see you geting an 18? I would love to see how you got 21. Cheers
Sarella Starshine
12-22-06, 01:26 PM there are 21 different combinations of dice that lead to a score of 18.
mathogre
12-22-06, 01:43 PM Sorry I may be missing some thing here but surrly the only way to get 18 on 4D6 drop lowest is 3 six's? I may be bad at maths but that is the only way I can see you geting an 18? I would love to see how you got 21. Cheers

You have four dice. Here are all of the unique rolls. Drop the lowest die.

1. 1,6,6,6
2. 2,6,6,6
3. 3,6,6,6
4. 4,6,6,6
5. 5,6,6,6
6. 6,6,6,6
7. 6,1,6,6
8. 6,2,6,6
9. 6,3,6,6
10. 6,4,6,6
11. 6,5,6,6
12. 6,6,1,6
13. 6,6,2,6
14. 6,6,3,6
15. 6,6,4,6
16. 6,6,5,6
17. 6,6,6,1
18. 6,6,6,2
19. 6,6,6,3
20. 6,6,6,4
21. 6,6,6,5

Hope this helps!
mathogre
12-22-06, 01:57 PM Btw, for anyone so inclined in the ways of Python, here's some console mode code that you can run.

Run this first, and hit return afterwards. There's no output.

rolls = []
for i in range (1,7) :
for j in range (1,7):
for k in range (1,7):
for l in range (1,7):
tempRoll = [i,j,k,l]
tempRoll.sort()
del tempRoll[0]
rolls += [sum(tempRoll)]

Then run this and hit return.

for i in range (3, 19):
print i, rolls.count(i)

Here's how it looks when you run it.

[mathogre@comp64-7 ~]$ python
Python 2.5 (r25:51908, Oct 9 2006, 16:09:22)
[GCC 3.4.6 20060404 (Red Hat 3.4.6-3)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> rolls = []
>>> for i in range (1,7) :
... for j in range (1,7):
... for k in range (1,7):
... for l in range (1,7):
... tempRoll = [i,j,k,l]
... tempRoll.sort()
... del tempRoll[0]
... rolls += [sum(tempRoll)]
...
>>> for i in range (3, 19):
... print i, rolls.count(i)
...
3 1
4 4
5 10
6 21
7 38
8 62
9 91
10 122
11 148
12 167
13 172
14 160
15 131
16 94
17 54
18 21
>>>

:mage:
Nom
12-22-06, 03:50 PM If the archives were working, you could also search to find http://users.bigpond.net.au/nom/d20/tools/, which contains a perl script to calculate the probabilities for any arbitrary set of dice.
StoneStokes
12-28-06, 08:39 PM Here is some more interesting statistics.

The probability that all of your scores are at least "average" of 10 is 31.5%.

The probability that all of your scores are at least "median" of 12 is 5.5%.

The probability that all of your scores are at least "mode" of 13 is only 1.345%.

The probability that three of your scores are at least 11, and three of your scores are at least 10 (the standard array) is about 22%

The probability that one of your scores is at least 15, one is at least 14, one is at least 13, one is at least 12, one is at least 10, and the last is at least 8 is about 1.9%.

A score-set that resembles the elite array (eg., 15, 14, 12, 10, 8) I will call an elite set. Likewise, sets such as 13, 13, 13, 13, 13, 12 will be called mode score-sets. Looking at the above data, it seems that elite score-sets should occur about as frequently as mode score-sets.

Cheers,
StoneStokes

Post/Author/DateTime	Post
StoneStokes 12-19-06, 06:03 PM	There are 1296 ways of rolling 4d6. Using the 4d6-drop-lowest method for ability scores, we get the following probabilities. Column 1 is the score, Column 2 is the number of ways to roll that score with 4d6-drop-lowest, out of 1296. 18 21 17 54 16 94 15 131 14 160 13 172 12 167 11 148 10 122 9 91 8 62 7 38 6 21 5 10 4 4 3 1 Using these probabilities we can make a few comments: The most likely score is 13 The expected score is 12.2 Rolling six scores, the expected total modifier is +5.4 (add the modifiers of all your abilities, and you should expect +5.4). Cheers, StoneStokes
TheChilliGod 12-19-06, 06:56 PM	Well done. May I ask how you made such calculations? For comparison, see; I came up with pretty much the same probabilities as you about half a year ago, but I manually worked out every one of the 1296 combinations, and I'm wondering if you did the same thing or smartly used some formula. One thing I find interesting about the results, you have an equal probability (1.62%) of rolling an 18 as you do a 6. And that is the only pair of equal probabilities.
Gilean 12-20-06, 03:36 AM	So, let us see... Mode (the most likely result): 13 Median (the middle result): 12 Expected value (average): 15869/1296 (~12,24) Variance: ~8,10 Standard deviation: ~2,85 Lower and upper quartile (half of results are between these, inclusive): 10, 14 Entropy (scaled): ~0,89 I'm not bored enough to calculate skewness or curtosis.
Nihzlet 12-20-06, 04:16 AM	interesting. probabilities dont stop player from rolling ridiculously good(or crappy) sets of stats using that system, however :P
StoneStokes 12-21-06, 01:20 PM	Well done. May I ask how you made such calculations? For comparison, see; I came up with pretty much the same probabilities as you about half a year ago, but I manually worked out every one of the 1296 combinations, and I'm wondering if you did the same thing or smartly used some formula. One thing I find interesting about the results, you have an equal probability (1.62%) of rolling an 18 as you do a 6. And that is the only pair of equal probabilities. I counted them in units. A unit is a sequence: either abcc or abcx, where a,b,c are known with a >= b >= c > x. Then multiply by the number of ways you can permute abcc or abcx. It didn't take too long, but would have been easier if I had a calculator when I computed it (the first run through, and I had a total of 1295 possibilities!).
Solaris 12-21-06, 07:38 PM	I counted them in units. ... Then multiply by the number of ways you can permute Not bad; I hadn't thought of doing it that way. This is a little harder to implement than a simple brute-force search, but since this generates only C(dice + sides - 1, dice) results, while brute-force has to generate all sides^dice possibilities, there's a substantial savings of computation time as dice and sides increase, making larger numbers more accessible.
darkhunter2007 12-22-06, 04:36 AM	how bored you are to do such amezing thing?
Ryusage 12-22-06, 09:59 AM	how bored you are to do such amezing thing? How bored you are to play D&D?
Azezel 12-22-06, 12:07 PM	StoneStokes, excellent resource - I have wondered about that myself. I have put your information into visual form for anyone who wants it. I grant permission to download, reproduce, distribute or modify the below image without limit. http://xs510.xs.to/xs510/06515/Ability_Generation.JPG
MindLich 12-22-06, 12:23 PM	I'm starting a campaign with a 3d6 roll for ability scores. Is anyone sufficiently bored to calculate the probabilities? Please, and thanks.
Nihzlet 12-22-06, 12:55 PM	Posted by Ryusage: How bored you are to play D&D?since when was being bored a prerequisite to playing D&D?
David5514 12-22-06, 01:10 PM	There are 1296 ways of rolling 4d6. Using the 4d6-drop-lowest method for ability scores, we get the following probabilities. Column 1 is the score, Column 2 is the number of ways to roll that score with 4d6-drop-lowest, out of 1296. 18 21 Cheers, StoneStokes Sorry I may be missing some thing here but surrly the only way to get 18 on 4D6 drop lowest is 3 six's? I may be bad at maths but that is the only way I can see you geting an 18? I would love to see how you got 21. Cheers
Sarella Starshine 12-22-06, 01:26 PM	there are 21 different combinations of dice that lead to a score of 18.
mathogre 12-22-06, 01:43 PM	Sorry I may be missing some thing here but surrly the only way to get 18 on 4D6 drop lowest is 3 six's? I may be bad at maths but that is the only way I can see you geting an 18? I would love to see how you got 21. Cheers You have four dice. Here are all of the unique rolls. Drop the lowest die. 1. 1,6,6,6 2. 2,6,6,6 3. 3,6,6,6 4. 4,6,6,6 5. 5,6,6,6 6. 6,6,6,6 7. 6,1,6,6 8. 6,2,6,6 9. 6,3,6,6 10. 6,4,6,6 11. 6,5,6,6 12. 6,6,1,6 13. 6,6,2,6 14. 6,6,3,6 15. 6,6,4,6 16. 6,6,5,6 17. 6,6,6,1 18. 6,6,6,2 19. 6,6,6,3 20. 6,6,6,4 21. 6,6,6,5 Hope this helps!
mathogre 12-22-06, 01:57 PM	Btw, for anyone so inclined in the ways of Python, here's some console mode code that you can run. Run this first, and hit return afterwards. There's no output. rolls = [] for i in range (1,7) : for j in range (1,7): for k in range (1,7): for l in range (1,7): tempRoll = [i,j,k,l] tempRoll.sort() del tempRoll[0] rolls += [sum(tempRoll)] Then run this and hit return. for i in range (3, 19): print i, rolls.count(i) Here's how it looks when you run it. [mathogre@comp64-7 ~]$ python Python 2.5 (r25:51908, Oct 9 2006, 16:09:22) [GCC 3.4.6 20060404 (Red Hat 3.4.6-3)] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> rolls = [] >>> for i in range (1,7) : ... for j in range (1,7): ... for k in range (1,7): ... for l in range (1,7): ... tempRoll = [i,j,k,l] ... tempRoll.sort() ... del tempRoll[0] ... rolls += [sum(tempRoll)] ... >>> for i in range (3, 19): ... print i, rolls.count(i) ... 3 1 4 4 5 10 6 21 7 38 8 62 9 91 10 122 11 148 12 167 13 172 14 160 15 131 16 94 17 54 18 21 >>> :mage:
Nom 12-22-06, 03:50 PM	If the archives were working, you could also search to find http://users.bigpond.net.au/nom/d20/tools/, which contains a perl script to calculate the probabilities for any arbitrary set of dice.
StoneStokes 12-28-06, 08:39 PM	Here is some more interesting statistics. The probability that all of your scores are at least "average" of 10 is 31.5%. The probability that all of your scores are at least "median" of 12 is 5.5%. The probability that all of your scores are at least "mode" of 13 is only 1.345%. The probability that three of your scores are at least 11, and three of your scores are at least 10 (the standard array) is about 22% The probability that one of your scores is at least 15, one is at least 14, one is at least 13, one is at least 12, one is at least 10, and the last is at least 8 is about 1.9%. A score-set that resembles the elite array (eg., 15, 14, 12, 10, 8) I will call an elite set. Likewise, sets such as 13, 13, 13, 13, 13, 12 will be called mode score-sets. Looking at the above data, it seems that elite score-sets should occur about as frequently as mode score-sets. Cheers, StoneStokes