What is the statistically superior character creation method, twelve 3d6 or six 4d6?

Question

The D&D 3.5 Player's Handbook gives (among others) two preferred methods to roll ability scores for a new character: a) roll 3d6 twelve times and keep the preferred six results, or b) roll 4d6 and drop the lowest die, six times.

What is the statistically better method in terms of total modifiers?

In addition, IIRC the core rulebook excludes characters whose total bonus is lower than +3, because adventurers are assumed to be exceptional people.

Answer 1

If you actually get a standard distribution from the dice in the 3d6 x12 method, it will be slightly better than a standard distribution of results from the 4d6 method. The more samples you take, the more likely it is that you will get something approaching average or a standard distribution. The fewer samples you take, the more likely the results will just be random.

Answer 2

D&D players are quick on the math. Agreed.

Want a lot of 14-16, go with 4d6. If you want more 17 & 18 go with the 3d6 method. So are you building a Monk or a Wizard?

Answer 3

Writing a program to brute force it looks like that the difference is slight

I added up all six attributes and counted the number of times that total appears.

The 3d6 six times method clusters around a total of 72, The 4d6 drop low clusters around a total of 74

A straight 3d6 roll clusters around a total of 63.

3d6 six time is more tightly clustered and ranges from 56 to 95 while 4d6 drop low ranges from 40 to 100.

Here is the source code for Visual Basic

Option Explicit
Dim Result1(1 To 18 * 6) As Long
Dim Result2(1 To 18 * 6) As Long
Dim Result3(1 To 18 * 6) As Long

Private Sub Command1_Click()
    Dim I As Long
    Dim R1 As Long
    Dim R2 As Long
    Dim R3 As Long
    Cls
    For I = 1 To 100000
        R1 = RollStat6TimesTakeBest
        R2 = RollStat4
        R3 = RollStat
        Result1(R1) = Result1(R1) + 1
        Result2(R2) = Result2(R2) + 1
        Result3(R3) = Result3(R3) + 1
    Next I

    Dim F As FileSystemObject
    Set F = New FileSystemObject
    Dim T As TextStream
    Set T = F.CreateTextFile("C:\test.csv", True)
    T.WriteLine "Total,3d6 6 times , 4d6 drop one , straight 3d6"
    For I = 1 To 18 * 6
        T.WriteLine CStr(I) & "," & CStr(Result1(I)) & "," & CStr(Result2(I)) & "," & CStr(Result3(I))
    Next I
    T.Close
    MsgBox "Done"
End Sub

Private Function D(Roll As Integer) As Integer
    Dim Result As Long
    Dim Test As Double
    Result = Rnd * 1000000000
    D = Result Mod Roll + 1
End Function

Private Function Roll3D6() As Integer
    Roll3D6 = D(6) + D(6) + D(6)
End Function

Private Function RollStat() As Integer
    Dim Total As Integer
    Dim I As Long
    For I = 1 To 6
        Total = Total + Roll3D6
    Next I
    RollStat = Total
End Function

Private Function RollStat6TimesTakeBest() As Integer
    Dim Best As Integer
    Dim I As Long
    Dim Roll(1 To 6) As Integer
    For I = 1 To 6
        Roll(I) = RollStat
    Next I
    Best = Roll(1)
    For I = 2 To 6
        If Best < Roll(I) Then Best = Roll(I)
    Next I
    RollStat6TimesTakeBest = Best
End Function



Private Function Roll4D6DropLow() As Integer
    Dim Roll(1 To 4) As Integer
    Dim Low As Integer
    Dim I As Integer
    Dim Total As Integer
    Roll(1) = D(6)
    Roll(2) = D(6)
    Roll(3) = D(6)
    Roll(4) = D(6)
    Low = 1
    For I = 2 To 4
        If Roll(I) < Roll(Low) Then Low = I
    Next I
    For I = 1 To 4
        If I <> Low Then Total = Total + Roll(I)
    Next I
    Roll4D6DropLow = Total
End Function

Private Function RollStat4() As Integer
    Dim Total As Integer
    Dim I As Long
    For I = 1 To 6
        Total = Total + Roll4D6DropLow
    Next I
    RollStat4 = Total
End Function

Answer 4

I appears that bySwarm is right. Here are the results:

along the X axis is the total bonus over the six ability scores. Along the Y axis, the probability, obtained from 1 million runs. Results below a total bonus of +3 have been purged from the count, so the grand total of runs is less than the original 1 million.

It appears that the twelve 3d6 statistically produces a better total bonus than the 4d6 method.

This is the code to run the 4d6 case (in Python)

import sys
import random

count = {}
for i in xrange(1,1000000):
    collection = []
    for j in xrange(0,6):
        extraction = [random.randint(1,6), random.randint(1,6), random.randint(1,6), random.randint(1,6) ]
        #print extraction
        collection.append( sum( sorted( extraction )[1:] ) )
    #print collection 
    bonuses = map(lambda x: (x-10)/2, collection)
    #print bonuses
    total_bonus = sum(bonuses)
    #print total_bonus

    if total_bonus < 3:
        #print "too low, excluded"
        continue

    if not count.has_key(total_bonus):
        count[total_bonus]=0
    count[total_bonus] += 1

total_extractions = sum(count.values())
for bonus,occurrences in sorted(count.items()):
    print bonus,occurrences/float(total_extractions)

This is for the twelve 3d6 case:

import random

count = {}
for i in xrange(1,1000000):
    collection = []
    for j in xrange(0,12):
        extraction = [random.randint(1,6), random.randint(1,6), random.randint(1,6) ] 
        collection.append( sum( extraction ) )
    ##print collection
    collection = sorted(collection)[6:]
    #print collection

    bonuses = map(lambda x: (x-10)/2, collection)
    #print bonuses
    total_bonus = sum(bonuses)
    #print total_bonus

    if total_bonus < 3:
        #print "too low, excluded"
        continue

    if not count.has_key(total_bonus):
        count[total_bonus]=0
    count[total_bonus] += 1

total_extractions = sum(count.values())
for bonus,occurrences in sorted(count.items()):
    print bonus,occurrences/float(total_extractions)

Answer 5

As everyone else has stated, 12 rolls of 3d6 is better.

You guys writing hundreds of lines of dice code... I love the web-based tool for Troll for dice calculations.

Here's Troll code for best six of 12 rolls of 3d6:

sum (largest 6 12#(sum 3d6))

This averages in a total score around 76.4.

And the Troll code for six rolls of 4d6 keeping the best 3d6:

sum 6#(sum largest 3 4#d6)

This averages in a total score around 73.5.

Answer 6

It's no use me repeating the brilliance of the master statitician's above as the answer has been given already.

But to be fair to the PC's in the group, there is nothing worse then when someone rolls 3 x 18's a 17 and 2 x 16's for their stats and another player rolls 1 x 17, and the rest 16's or worse. It is not very inspiring.

I think to give all your players a feeling of "joy" is to perhaps start with a point buy, and then allow then an additional dice roll (perhaps 2d4 or 1d6) to add to the base point buy.

This way they get to have characters of "similar" abilities, they get to balance out the stats where they want them. If they want a 20 Strength, they can sacrifice int or wis or cha.

I know rolling dice is fun, but as stats are quite important, you could try this method -- remember, playing the game is about fun, not always about rules :)

Answer 7

My answer contains some LaTeX markups, so I've posted a nicely formatted PDF of it that you can read/print/download over at http://www.scribd.com/doc/37700790

When you roll 4d6k3, each of your 6 ability scores follows the exact same probability distribution. In statistics lingo, your 6 ability scores are i.i.d.---independent and identically distributed---random variables. Call one of these i.i.d. random variables Y. The mean of Y is E[Y]=12.2445987654321, and its standard deviation is $\sigma_Y=2.8468444453115$. Additionally, this distribution is skewed to the left; its skewness is -0.283507652977282.

By comparison, when you roll 3d6 once, you get a random variable X, with E[X]=10.5, and standard deviation $\sigma_X=2.95803989155$. Additionally, it is symmetric, so its skewness is 0.

However, when you roll 3d6 12 times and keep the highest 6, you get 6 different random variables (not i.i.d.), called the 7th through 12th order statistics, denoted $X_{(7)},\ldots,X_{(12)}$. E.g., $X_{(12)}$ is the maximum of the 12 rolls. Each order statistic has its own mean, standard deviation, and skewness:

\begin{tabular}{ l | r r r } Order statistic & mean & standard deviation & skewness \ \hline $X_{(7)}$ & 10.8184 & 1.1411 & -0.00564534 \ $X_{(8)}$ & 11.4663 & 1.14865 & -0.00978342 \ $X_{(9)}$ & 12.1517 & 1.1693 & 0.00712255 \ $X_{(10)}$ & 12.919 & 1.2154 & 0.0435863 \ $X_{(11)}$ & 13.8598 & 1.30455 & 0.0503448 \ $X_{(12)}$ & 15.2263 & 1.44603 & -0.125062 \ \end{tabular}

Of course, you can easily find the average of the means of the 7th through 12th order statistics: $\mu=\frac{\sum_{i=7}^{12} E[X_{(i)}]}{6}=12.7403$, so $\mu > E[Y]$ by about a $\frac{1}{2}$ point. But note that $E[X_{(7)}] < E[X_{(8)}] < E[X_{(9)}] < E[Y]$, meaning the expected value of each of the 6 ability scores generated with 4d6k3 is greater than what you can expect from half the ability scores generated by the largest 6 of 12x(3d6). So the answer isn't so simple.

Answer 8

This may sound crazy, but I took to using 5d4. Why? I noticed that with either 3d6 and 4d6, characters tended to gain extreme rolls - lots of 17's and 18's, but also 3's, 4's, and 5's. This made crazy 'Rainman' characters - great at some things, retarded at others. Such characters are unplayable which led to a lot of re-rolling, or worst, encouraged DM's to cave and let the players drop the 4's, but keep the 17's.

In life, most people are more generally average. Roll 5 4-sided dice and you'll get a lot of 9 - 13's with perhaps one better and one worst roll. Easier to play and more realistic.