Probability Distribution of Dice

[morpheous1777]

Bump........this is a really good thread, i didnt want it to fade away into obscurity

 

[frisbeet]

anyone want to tackle 4d6 drop lowest

Bump, and

The generalized drop formula.

n = # of dice
s = sides of die
d = # of dropped dice

What is the # of ways of rolling an outcome?

Or, what is the probability of rolling an outcome?

Anyone wanna crack this with elegance?

 

[ichabod]

The generalized drop formula.


Anyone wanna crack this with elegance?

I have been doing a lot of work on this without success. I have a method which should work but isn't working, and I can't figure out what the error is. However, I'm a bit crunched for time right now with the end of the semester (and the end of my degree) coming up. As soon as I get some time, I'll post what I've been doing and see if anyone has ideas as to where the problem is.

 

[Zad]

What you want is a chi squared test which will measure how random your dice are. Dragon magazine published one many many years ago and if you search the archives you can probably dig up something on it.

 

[Xeriar]

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

No, more like top 10% of the population, which is also what 4d6 will produce on average (rather neat, that).

I wrote a program which makes millions of people, but unfortunately it sorts them according to raw stats rather than point buy, so I can only confirm that 25 point buy usually equals the 10th percentile (give or take a couple).

Above the top .1% is no less than 36 point buy - usually more.

 

[frisbeet]

double post

 

[frisbeet]

So here are some numbers to start with.

In this example I consider n=3,4,&5; s=6, and d = 0,1,or 2. I've been thinking along the lines of:

1). Count the ways of rolling the outcome with 3 dice. We've done this already in this thread.

2). For each way of rolling an outcome, count the number of ways of replacing the highest dice with the discarded dice.

Add, not multiply, the results of 1 & 2.

So # ways of getting a 6 by 4d6 drop 1 = #ways of 3d6 getting a 6 + comb(3,0) + comb(3,0) + comb (3,1) + comb (3,1) + comb(3,2) = 21

By a more brutal but accurate method I've determined the distributions for 4d6 drop1 and 5d6 drop 2.

oc      3d6     4d6     5d6     3d6 P           4d6 P           5d6 P
18	1	21	276	0.004630	0.016204	0.035494
17	3	54	610	0.013889	0.041667	0.078447
16	6	94	935	0.027778	0.072531	0.120242
15	10	131	1111	0.046296	0.101080	0.142876
14	15	160	1155	0.069444	0.123457	0.148534
13	21	172	1055	0.097222	0.132716	0.135674
12	25	167	881	0.115741	0.128858	0.113297
11	27	148	665	0.125000	0.114198	0.085520
10	27	122	470	0.125000	0.094136	0.060442
9	25	91	296	0.115741	0.070216	0.038066
8	21	62	170	0.097222	0.047840	0.021862
7	15	38	90	0.069444	0.029321	0.011574
6	10	21	41	0.046296	0.016204	0.005273
5	6	10	15	0.027778	0.007716	0.001929
4	3	4	5	0.013889	0.003086	0.000643
3	1	1	1	0.004630	0.000772	0.000129

 

[frisbeet]

bumping for needed reference.