Probability Distribution of Dice

[LazarusLong42]

With a huge thanks to Nathan for pointing out what I'd missed... here are final equations.

In all of these:

R is the desired result.
k is the total number of die types.
n[z] is the number of dice of type z, 1<=z<=k.
d[z] is the number of sides on a die of type z.
N is the total number of dice.
Certain sums used in the k > 2 equation are defined below the equation.

(z is never actually used in the equations, it just makes a good neutral index.)

For two die types (k=2, e.g. 5d4 + 2d8):

For n die types (k<2, e.g. 2d4 + 3d6 + 5d10 + 1d12...):

Remember that this gives only the numerator; the denominator is the iterated product of d[z]^n[z].

My apologies for the large pics, but there's a bunch of information there. If anyone finds discrepancies, let me know... otherwise, I think my work here is done

I think I need to put the k=2 equation in my sig. Iconic Probability equation or some such.

LL

 

[Graf]

This is a great thread. The kind of thread that makes you wish you knew more math.

I'm particularly interested in the stuff that Xeriar brought up at the begining. (I.e. social impications you can draw from the numbers)

Can any tell me what percentile the standard array of attributes is (using 3d6)?

how likely is someone to have a 14, a 16 or a 18?

(sorry to interupt a serious math conversation with fluff, btw)

looking forward to seeing what you guys make

Over 2/3rds of the population is within one Standard deviation of the mean (7.55-13.45), another 2/7ths are between one and two deviations (4.6-7.55 and 13.45-16.4), 2/46ths (1.65-4.6, 16.4-19.35)...

As you can see, this is an approximation, and it also extrapolates beyond what the dice can physically represent. This can be useful for determining what a given ability score should mean (if you're paying careful attention to detail )

As more deviations get factored in, the probabilities get drastically smaller, of course - one in a billion for being six standard deviations above the mean, even (which would equate to a half dozen people on earth with an intelligence score of 28 - maybe one with a 29).

 

[LazarusLong42]

Can any tell me what percentile the standard array of attributes is (using 3d6)?

how likely is someone to have a 14, a 16 or a 18?

Heh, that one's *easy* As follows:

[COLOR=silver]
3	1	0.4630	0.4630	100.000
4	3	1.3889	1.8519	99.5370
5	6	2.7778	4.6296	98.1481
6	10	4.6296	9.2593	95.3704
7	15	6.9444	16.2037	90.7407
8	21	9.7222	25.9259	83.7963
9	25	11.5741	37.5000	74.0741
10	27	12.5000	50.0000	62.5000
11	27	12.5000	62.5000	50.0000
12	25	11.5741	74.0741	37.5000
13	21	9.7222	83.7963	25.9259
14	15	6.9444	90.7407	16.2037
15	10	4.6296	95.3704	9.2593
16	6	2.7778	98.1481	4.6296
17	3	1.3889	99.5370	1.8519
18	1	0.4630	100.000	0.4630
[/COLOR]

Column one is the roll of 3d6; column 2 is the number of rolls (out of 216 possible) that will produce that result. Column three is the percent of rolls that produce that result. Column 4 is the percent of rolls with a result equal to that number or less; column 5 is the percent equal to that number of higher.

So, 74.0741% of 3d6 rolls will be 12 or lower, and the same percentage will be 9 or higher.

Someone floated results for 4d6 drop lowest a while ago, and I have them somewhere, but they can't be calculated using the formulae in this thread.

 

[ichabod]

Can any tell me what percentile the standard array of attributes is (using 3d6)?

how likely is someone to have a 14, a 16 or a 18?

I've actually been obsessing about that sort of question for the past week or so. This thread reignited my interest in finding the ordinal probabilities for various dice rolling methods. That is, what are the odds of various numbers being your second highest roll out of six, when using 4d6 drop low or some other method. I finally have a spreadsheet that can calculate those probabilities given any base distribution from 3-18 (without reroll conditions). However, I don't have access to it at the moment, and I'm about to be swamped with work. Keep your eye out, though, I should have something posted either here or in house rules by early next week.

 

[frisbeet]

Good work LL42!

At a glance this looks right, but to honor your formulae I'm gonna work out a couple examples when time permits.

I'm 44 reg.

 

[frisbeet]

I've actually been obsessing about that sort of question for the past week or so. This thread reignited my interest in finding the ordinal probabilities for various dice rolling methods. That is, what are the odds of various numbers being your second highest roll out of six, when using 4d6 drop low or some other method. I finally have a spreadsheet that can calculate those probabilities given any base distribution from 3-18 (without reroll conditions). However, I don't have access to it at the moment, and I'm about to be swamped with work. Keep your eye out, though, I should have something posted either here or in house rules by early next week.

This is not a general solution (equation), since I did a lot of counting for only the 4d6 drop lowest and 5d6 drop 2 lowest methods, but here at least are these results. To check with.

# ways:

oc___3d6____4d6____5d6

18____1_____21_____276
17____3_____54_____610
16____6_____94_____935
15____10____131____1111
14____15____160____1155
13____21____172____1055
12____25____167____881
11____27____148____665
10____27____122____470
9_____25____91_____296
8_____21____62_____170
7_____15____38_____90
6_____10____21_____41
5_____6_____10_____15
4_____3_____4______5
3_____1_____1______1


probability:

oc___3d6________4d6_________5d6

18___0.004630___0.016204___0.035494
17___0.013889___0.041667___0.078447
16___0.027778___0.072531___0.120242
15___0.046296___0.101080___0.142876
14___0.069444___0.123457___0.148534
13___0.097222___0.132716___0.135674
12___0.115741___0.128858___0.113297
11___0.125000___0.114198___0.085520
10___0.125000___0.094136___0.060442
9____0.115741___0.070216___0.038066
8____0.097222___0.047840___0.021862
7____0.069444___0.029321___0.011574
6____0.046296___0.016204___0.005273
5____0.027778___0.007716___0.001929
4____0.013889___0.003086___0.000643
3____0.004630___0.000772___0.000129

average

3d6: 10.50

4d6: 12.24

5d6: 13.43

98% sure of these results.

 

[ichabod]

frisbeet: Those are exactly what I'm getting for 5d6, which I hadn't been able to check, so it looks like at least the nd6 drop low section of my spreadsheet works right. The main part with the ordinal probabilities are harder to test. But that does bode well for the formulas I'm working on for nd6 drop high and low.

 

[Syunsuke]

Excuse me about asking this, but can anyone make some program with which we can calculate LazarusLong42's formula? We input dice type and number of them, and we get the probability distribution.
Or, is there such thing already?
I'm not very good at math or computers (or English )

 

[Graf]

Heh, that one's *easy* As follows:

cheers, I think I can do it from here.

Assuming a world where everyone is generated using 3d6 and someone with the default array (15 14 13 12 10 8) is pretty exceptional.

The number of people out of 100 (if they're a perfect sample) who have a 15 or better is 9.25. The number of those who have a 14 or better is 1.49 (.162 times 9.25) and so on.
So PCs with the standard array are .0765% of the population or on in 1300 or so?
(assuming that most of the pop is below 4th level and therefor doesn't have any attribute bonuses).

pretty sick really

 

[Lord Zardoz]

Math vs Brute Force

I am one of those happy individuals with very strong math skills (compared to average people) and very strong programming skills.

Unfortunately, my math skills are in the realm of geometry type questions (Vector and Matrix multiplication) rather then statistics.

This is one of those probems that can either be worked out through math equasions, or by brute force. For calculating the odds of most dice based roll results, I tend to rely on brute force algorthims written in C / C++.

The one thing about simply working out the equasions mathematically is that when you add a "Drop the lowest" type of rolling rule, as in the standard 4d6 drop character generation method, you end up with, simple high school algerbra and Calculus just dont cover it.

The nice thing about brute force checking is that you dont need to worry about errors in your math. And for most dice combinations, you can brute force them on a "typical" PC in a very reasonable amount of time. Sure, its a boring and repetitive task to manually calculate this, but computers were invented to handle boring and repetitive tasks.

END COMMUNICATION