Quick question on d6 dice probability

[Douche]

Lets say you've got a d6 system. You level your stats and this corresponds to the amount of d6's you get to roll.

Therefore, you have 5 strength, you get 5d6 on your attack roll.

Now, the system is based mostly on contested checks and successes are determined as such

1-3 = 0 success
4-5 = 1 success
6 = 2 success

The contested check is determined by who has more successes, rather than the raw number of the results. Therefore, even though rolling three 3's is a higher number (9), it is not as good as someone who just rolled a single 6.

So to determine your average chance to get a success - you have 4/6 chance at getting a success, right? Or is it 4/7?

Sorry but math has always been my worst subject. I just need a little help grasping probability for a mental exercise.

[Geddy2112]

Qualitatively, your chances of getting a success(a roll with any success) is 50%. But otherwise yes, the 6 is weighted as 2 successes meaning quantitatively your chances of getting a unit of success are 4 in 6. or 2 in 3.

When you roll multiple dice it gets crazier. on 2d6 you have a 1 in 3 chance of 2 successes, a 1 in 9 chance of 3, and a 1 in 36 chance of 4 successes.

[Douche]

Cool, thanks friendo

[Martin Greywolf]

In cases like these, you need to re-state the problem in your head. What you have here aren't typical d6 dice, it's six sided dice with 0 on three sides, 1 on two sides and a 2 on one side. Once you do this, you can pretty much start doing "normal" math on these - the average roll of one die is (0+0+0+1+1+2)/6 = 2/3, so you can conclude that someone with 5 strength will most likely roll 5*2/3 = 10/3, or 3 + 1/3, with 0 being the minimum and 10 being the maximum.

As you can see, this system heavily favours low rolls, with rare instances of overwhelming successes if thu hath not sinned and RNGsus blesses thy dice.

[Xervous]

If you want to visualize how the probability distribution plays out with various dice pool sizes you can look here at a small script in anydice. Simply adjust the Xd6 at the end to suit your taste.

[Guizonde]

sounds a lot like the kind of dice system used in star wars: edge of the empire. you've got those weird dice with either empire or rebel alliance symbols, you've got one side each with two symbols (two successes or fails), and four with either (single success or fail). so a single die goes from -2 to +2 and no zeroes that i remember. you throw the pool of dice according to your ability score and skill. you need to hit someone with a bat? strength of 4 (4 dice) plus your close quarters skill of let's say 3 (3 dice), meaning you'll throw 7 dice. in case of an opposed check, you need to have more of your symbol than your opponent has of theirs.

to add to the confusion, there are "good" dice and bad ones: the dm can give a player one with more than 3 of their symbols if the check is easy or boosted by rp and as the opposite when you're trying to bench-press a rancor, giving an edge to the likelier outcome.

i gave up thinking about it in terms of numbers and just visualized it. worked fairly accurately to estimate my chances, but holy crud did that game use buckets of proprietary dice that cost quite a bit (i think 5 sets cost my dm around the 10€ mark each, without shipping).

[Oerlaf]

In cases like these, you need to re-state the problem in your head. What you have here aren't typical d6 dice, it's six sided dice with 0 on three sides, 1 on two sides and a 2 on one side. Once you do this, you can pretty much start doing "normal" math on these - the average roll of one die is (0+0+0+1+1+2)/6 = 2/3, so you can conclude that someone with 5 strength will most likely roll 5*2/3 = 10/3, or 3 + 1/3, with 0 being the minimum and 10 being the maximum.

As you can see, this system heavily favours low rolls, with rare instances of overwhelming successes if thu hath not sinned and RNGsus blesses thy dice.

Yeah, that's right. More mathematically, we have a random variable X that attains values 0, 1 and 2 with probabilities 1/2, 1/3 and 1/6 respectively.

The average is 0*1/2+1*1/3+2*1/6=2/3, and the variance is 1/2*(0-2/3)^2+1/3*(1-2/3)^2+1/6*(2-2/3)^2=11/18.

If one character has x strength, and the other has y, the real random variable we are to research is x*X1-y*X2, where X1 and X2 are the rolls.

We need probability that x*X1-y*X2>0 (who has more successes). It is easy to calculate, because we only have 9 terms to sum. The worst case is X1=0 and X2=2 that leads us to -y*X2 and the best case is 2x that leads us to 2*X1. What are the probabilities?

X1/X2
0/0 - 1/2*1/2=1/4 -> 0
0/1 - 1/2*1/3=1/6 -> -y
0/2 - 1/2*1/6=1/12 -> -2y
1/0 - 1/3*1/2 = 1/6 -> y
1/1 - 1/3*1/3=1/9 -> x-y
1/2 - 1/3*1/6=1/18 -> x-2y
2/0 - 1/6*1/2=1/12 -> 2x
2/1 - 1/6*1/3=1/18 -> 2x-y
2/2 - 1/6*1/6=1/36 -> 2x-2y

The average of this distribution is equal to (x-y)/2.

Three results 0, -y and -2y are certainly not greater than 0 (note that x>0 and y>0 because we can't roll 0d6), so we can count them off, and we have estimate P{attack > 0} <= 6/9=2/3.
y and 2x are certainly positive, so P{attack > 0} >= 2/9.

Roughly, 2/9 <= P{attack > 0} <= 2/3 regardless of the corresponding stats.

Now, let us see the other parts.

x-y>0, x-2y>0, 2x-y>0, 2x-2y>0 holds iff x>2y -> then we have P=6/9.
x-y>0, x-2y>0, 2x-y>0, 2x-2y<0 never holds
x-y>0, x-2y>0, 2x-y<0, 2x-2y>0 never holds
x-y>0, x-2y<0, 2x-y>0, 2x-2y<0 holds iff y<x<2y - then we have P=5/9
x-y<0, x-2y>0, 2x-y>0, 2x-2y<0 never holds
It is impossible to have exactly two positive inequalities, so we never have P=4/9
P=3/9 holds iff y/2<x<y
And P=2/9 holds iff 0<x<y/2.

So, in conclusion we have P[Attack > 0] =
2/9 if 0<x<y/2
3/9 if y/2<x<y
5/9 if y<x<2y
2/3 if x>2y.

[Douche]

Yeah, that's right. More mathematically, we have a random variable X that attains values 0, 1 and 2 with probabilities 1/2, 1/3 and 1/6 respectively.

You saw the part where I said math is my worst subject, right? Hehe

Anyway sounds like you know what you're talking about... Would you mind checking out this thread I made in the homebrew board and tell me (in laymans terms) if my impression is right?
http://www.giantitp.com/forums/showt...9#post22682399

I probably should've just posted it here but I didn't want to break any rules and figured I wouldn't be upsetting anyone with a generalized question.