Dice Pool Probabilities (Help!)

[PPoS]

Hey ! Having trouble wrapping my head around the probabilities of the following dice pool system:

Roll a number of D6 (3-4 is normal for a standard character), each 4+ counts as one success (with the target being to roll as many as possible, the difficulty of any given task is measured in the number of successes needed, 2 successes being 'normal/average' difficulty).

Each roll of a natural '6' counts as a success, but can also be re-rolled for additional successes (i.e. the "exploding" dice effect).

Now figuring out the probabilities for rolling a set number of successes on a set number of dice is simple, but when you add in the "exploding" dice effect things get more complicated. In fact, so complicated that I cannot wrap my brain around it.

So, I though I would just throw this out here and see if anyone could be so kind as to just make a table with the percentage chances of rolling 0, 1, 2, 3, 4, 5, and 6 successes on the same number of dice (i.e. 1 success on up to 6 dice, etc.) I would be eternally grateful for that. No need to really explain how the probabilities work, I've skimmed through what feels like 100 forums dealing with just this. Some day I will perhaps learn, right now I'm just interested in the percentages.

Thanks !

[RogerBW]

Methodology: probability table for one die is (3/6, 2/6, 1/6×3/6, 1/6×2/6, 1/6×1/6×3/6, 1/6×1/6×2/6) then 1 - the sum of the others for the 6+ probabilities.

To combine, multiple the two probability tables together (sum of p(n+m) = p(n)*p(m) for all valid n, m).

Rounded to nearest whole-number percentage:

1 die
0 successes: 50%
1: 33%
2: 8%
3: 6%
4: 1%
5: 1%
6+: 0%

2 dice
0: 25%
1: 33%
2: 19%
3: 11%
4: 6%
5: 3%
6+: 3%

3
0: 13%
1: 25%
2: 23%
3: 16%
4: 10%
5: 6%
6+: 7%

4
0: 6%
1: 17%
2: 21%
3: 19%
4: 14%
5: 10%
6+: 14%

5
0: 3%
1: 10%
2: 16%
3: 18%
4: 16%
5: 12%
6+: 24%

[PPoS]

Originally Posted by RogerBW

Text

Thank you very much ! So, just to see if I get this straight. If I wanted to know what the percentage is of getting either 2 successes, or say, at least 1 success, it would look something like this:

1 Dice:
Chance of at least 1 success = 50%
Chance of 2 successes = 41%

2 Dice
Chance of at least 1 success = 78%
Chance of 2 successes = 52%

Correct ?

[Anthony]

This should be doable with AnyDice, but might require some sophisticated coding. Doing a rather stupid monte carlo, I get:

Result  0       1       2       3       4       5       6
1 Die   49%     41%      6%      1%      0%      0%      0%
2 Dice  25%     40%     24%      7%      1%      0%      0%
3 Dice  12%     31%     31%     16%      6%      1%      0%
4 Dice   6%     21%     28%     23%     12%      5%      2%
5 Dice   2%     13%     24%     25%     18%      9%      6%
6 Dice   1%      8%     18%     23%     21%     14%     12%

[PPoS]

Originally Posted by Anthony

This should be doable with AnyDice, but might require some sophisticated coding.

Yeah, most likely. I've tried to do something similar with it, but alas ... Maybe if I had time to learn all the functions I could do it.

[Andrew Hackard]

The exploding die makes things complicated if you want to sum the probabilities directly. MUCH easier to find the total probability of failure, which will be a relatively easy calculation, and subtract from 1 (100%).

For instance, if you want at least 2 successes, then

P(2+) = 1 - [P(0) + P(1)]

Let's say you're rolling 4 dice and a 4, 5, or 6 counts as a success, with 6 exploding.

P(0) means you roll 1-3 on all four dice: (3/6)^4 = 1/16 = 6.25%.

P(1) has two possibilities:

a) 4 or 5 on one die, 1-3 on the other three: P(1_4,5) = 2/6 * (3/6)^3 * 4 = 1/6 = about 16.7%
b) 6 on one die followed by 1-3 on that die, 1-3 on the other three: P(1_6) = 1/6 * 3/6 * (3/6)^3 * 4 = 1/24 = about 4.2%

(You multiply by 4 because the success can be any one of the four dice.)

So the total probability of *failure* is P(0) + P(1_4,5) + P(1_6) = 1/16 + 1/6 + 1/24 = 13/48 = about 27.1%. P(success) = 1 - P(failure) = about 62.9%.

[PPoS]

Originally Posted by Andrew Hackard

The exploding die makes things complicated if you want to sum the probabilities directly.

Which is what I'm after. The probability of rolling say 1 or 2 successes on a # of dice.

For instance, if you want at least 2 successes, then

P(2+) = 1 - [P(0) + P(1)]

Let's say you're rolling 4 dice and a 4, 5, or 6 counts as a success, with 6 exploding.

P(0) means you roll 1-3 on all four dice: (3/6)^4 = 1/16 = 6.25%.

This is how far I understand. This makes sense to me.

P(1) has two possibilities:

a) 4 or 5 on one die, 1-3 on the other three: P(1_4,5) = 2/6 * (3/6)^3 * 4 = 1/6 = about 16.7%
b) 6 on one die followed by 1-3 on that die, 1-3 on the other three: P(1_6) = 1/6 * 3/6 * (3/6)^3 * 4 = 1/24 = about 4.2%

(You multiply by 4 because the success can be any one of the four dice.)

So the total probability of *failure* is P(0) + P(1_4,5) + P(1_6) = 1/16 + 1/6 + 1/24 = 13/48 = about 27.1%. P(success) = 1 - P(failure) = about 62.9%.

Stuff like this makes my eyes bleed though. I understand that this is the way to calculate it, I'm just having trouble wrapping my brain around equations. Math has never been my strong side. But thanks for explaining anyway !

[Anthony]

You might have missed it since I edited after the fact, but I did a monte carlo with just rolling 10,000 times in my post above. It's not perfect, but it's good enough for gaming.

[whswhs]

If you didn't have the rule about rerolls, then rolling one die would give you 0.5 successes. With the rerolls, it goes up to 0.6 successes.

The dice you roll are effectively independent, so for N dice, you get 0.6N successes.

That's what I would look at, if I were using this system; I don't find the distribution of numbers of successes as helpful, though there are cases where it would have its uses.

[Anthony]

Doing the monte carlo slightly differently, what you probably want is the odds of N or more successes, not exactly N successes. so we get:

Result   1+      2+      3+      4+      5+      6+      7+      8+
1 Die   50%      8%      1%      0%      0%      0%      0%      0%
2 Dice  74%     34%      9%      2%      0%      0%      0%      0%
3 Dice  87%     56%     25%      8%      2%      0%      0%      0%
4 Dice  93%     72%     43%     19%      7%      2%      0%      0%
5 Dice  97%     83%     59%     33%     15%      6%      1%      0%
6 Dice  98%     90%     72%     48%     27%     12%      4%      1%