# Talk:Dice pool

## Shadowrun Glitches[edit]

Excuse me, omae, but your math seems to be wrong. Shadowrun dice pool rolls are considered glitches when half the dice come up with ones - that makes the probability of glitch for one die 16,6%, and for two dice - as much as 33,3%. Fatum 05:44, 29 October 2009 (UTC)

- You are correct, sir. I totally expect someone to fix my mistakes when I'm writing on a wiki. I also expect someone to eventually replace my math with "SHITCOCKSSHITCOCKSSHITCOCKS" someday. - NotBrandX
- Hm, come to think of it, I am not. The math should go as follows: one die, one side out of six; 16,6%, yes.
- Now, two dice are not independent events, so the probability is not 33%. The probability for one die to come with NOT a 1 is 5/6, right? Now, for two dice to come with NO ones it's (5/6)^2, or 25/36; thus, the probability of at least one die rolling a one is 1-25/36, making it 30%.
- Oh, and hey, there's always the undo function! Fatum 05:03, 30 October 2009 (UTC)
- The Shadowrun 4e rule is "if at least half of the dice you rolled are showing ones, then something goes wrong; a "glitch." If you have a glitch and no successes, it's a critical failure.

This is how notBrandX was doing it wrong 1d size(1)/6^1 1/6 .1666~ 2d size(1?, ?1, 11)/6^2 3/36 0.08333~ 3d size(11?, 1?1, ?11, 111)/6^3 4/216 0.0185185~ 4d size(11??, 1?1?, ?11?, 1??1, ?1?1, ??11, 111?, 11?1, 1?11, ?111, 1111)/6^4 11/1296 0.00848765...

- (sorry about the brute-force, my combinatorics professor was a bad teacher.) --NotBrandX
- Seriously, use normal math notation, not the functions, would you?
- Also, which part of my reasoning don't you understand or don't agree with, when I say that the probability of two dice coming up with at least one 1 is 11/36?
- Nm, got what you had to say with those size(x). Except don't you notice that the 1? case covers not one but six possible outcomes, 11, 12, 13, 14, 15 and 16? Same with ?1: 61, 51, 41, 31, 21, 11. That 11 is a double, so of 6^2 (36) total possible cases 11 work for us.
- Same with the other cases, for 3d it'd be, what, 16 out of 6^3, not 4. Fatum 21:59, 30 October 2009 (UTC)

*"which part of my reasoning don't you understand..."*the part where you were calculating probability of a roll without glitches, and then subtracted that from 1.000.*"use normal math notation, not the functions"*Can't do n-choose-m in ascii.- Unlike most /b/tards, I'm totally okay with being proven wrong and corrected. My ego can go fuck itself; just make sure the article is correct, please. --NotBrandX 00:36, 31 October 2009 (UTC)
- Oh, so, you are also a /b/tard. Sweet. Just what we lacked around here.
- As a matter of fact, the sum of all probabilities is always 100%. It seems rather obvious to me. This way, if you know the probability of something NOT happening (for example, a couple of dice NOT rolling a glitch), the probability of the same thing
*happening*is one hundred percent minus the first probability (well, if you don't have some half-assed variants, of course). Fatum 02:55, 31 October 2009 (UTC)

- Put your cock away; I already know yours is bigger. I guess I'm not much of a /b/tard if someone else is more arrogant than I am.
- Well, maybe you didn't know it - calling yourself a /b/tard is basically admitting to be a stupid teenage troll nowadays. If you want people on /tg/ to treat you like a human being, don't call yourself that (and I have no idea why would anyone willingly go to /b/ anyway).

- (sorry about the brute-force, my combinatorics professor was a bad teacher.) --NotBrandX

- So, trying again.
- 1d6 glitch = {1} in {1,2,3,4,5,6} = 1/(6^1)
- 2d6 glitch = {(1,(2,3,4,5,6)),((2,3,4,5,6),1),(1,1)} = ((5+5)+1)/(6^2)
- 3d6 glitch = {(1,1,?),(1,?,1),(1,1,?),(1,1,1)} = ((5+5+5)+1)/(6^3)
- 4d6 glitch = {(1,1,?,?),(?,1,1,?),(?,?,1,1),(1,?,?,1),(1,?,1,?),(?,1,?,1),(1,1,1,?),(1,1,?,1),(1,?,1,1),(?,1,1,1),(1,1,1,1)}
- = ((5^2+5^2+5^2+5^2+5^2+5^2)+(5+5+5+5)+1)/(6^4)
- 5d6 glitch = ((choose(5,3)*5^2)+(choose(5,4)*5^1)+(choose(5,5)*5^0))/(6^5)
- (where the choose(n,k) function is n-choose-k, or [n!/(k!(n-k)!)]) because I can't do binomial coefficient notation in this wiki.)
- 6d6 glitch = ((choose(6,3)*5^3)+(choose(6,4)*5^2)+(choose(6,5)*5^1)+(choose(6,6)*5^0))/(6^6)
- ... am I on the right track this time? --NotBrandX 01:43, 1 November 2009 (UTC)
- Yeah, seems to be right this time to me. Fatum 07:42, 1 November 2009 (UTC)

1d6 glitches: 0.16667 or 1/6

2d6 glitches: 0.30556 or 11/36

3d6 glitches: 0.07407 or 16/216

4d6 glitches: 0.13194 or 171/1296

5d6 glitches: 0.03549 or 276/7776

6d6 glitches: 0.06229 or 2906/46656

7d6 glitches: 0.01763 or 4936/279936

8d6 glitches: 0.03066 or 51491/1679616

9d6 glitches: 0.00895 or 90196/10077696

--NotBrandX 14:55, 1 November 2009 (UTC)#!/usr/bin/env python def fact(x): if x == 0 or x == 1: return 1 return reduce(lambda x, y: x*y, xrange(2, x+1)) def choose(n,k): return (fact(n)/(fact(k)*fact(n-k))) sides = 6 for dice in range(1,10): numer = 0 half = int(round(dice/2)) for x in range(0,half+1): numer += choose(dice,(dice-x))*((sides-1)**x) denom = sides**dice prob = round((float(numer)/denom),5) print "%dd%d glitches: %.5f or %d/%d" % (dice,sides,prob,numer,denom) print "done."

## Legend of the 5 Rings[edit]

Note to self: another mathfag did the "roll Xd10, keep Y, exploding 10s" odds used by L5R. I'll come back later to chartify this and maybe turn the C code into easier-to-read python or perl. L5R uses 5/10/15/20 for target numbers, and max rolled allowed is "10d10 keep 10" --NotBrandX 20:53, 2 November 2009 (UTC)