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I'm trying to compute the probability of achieving a certain number of successes when rolling a die pool for both open-ended/exploding and closed tests. Success is defined as a roll above a certain (variable) threshold. If a roll is open-ended, any die which rolls the maximum value triggers an additional die to be rolled. This is for a role-playing game (Burning Wheel Gold).

Assume the use of a $n$-sided die. Additionally, we assume $d$ to be the number of dice, $r$ to be the required number of successes, and $c$ to be the minimum value accepted as a success. I'm breaking the question down into two blocks: one for open-ended rolls, and one for closed rolls.

Closed roll: For closed rolls, we can define a probability recurrence as follows:

$$Pr\left[r,d\right] = \begin{cases} \\1-\frac{c}{n}:Pr\left[d-1,r\right]\\\frac{c}{n}:Pr\left[d-1,r-1\right] \end{cases}$$

With probability $\frac{c}{n}$, a success will be met. However, I'm not sure how to solve this recurrence relation for any number of dice/required successes. There's probably an easier way to do this, but I haven't done much discrete probability.

Edit: It looks like a binomial probability distribution works for this part of the problem. This comes in the form of:

$${{d}\choose{r}}\left(\frac{c}{n}\right)^{r}\left(1-\frac{c}{n}\right)^{d-r}$$

Open ended roll: For open-ended rolls, we can find the equivalent number of rolled dice. This is going to be equal to:

$$d+\sum_{j=1}^{\infty}\left(\frac{d}{6^{j}} \right)=\frac{6d}{5}$$

This implies that, given a number of dice $n$, rolling open-ended results in approximately $\frac{6}{5}n$ dice rolled. If we had an easy way to solve the above for any number of dice $n$, it would be easy: we could just input the non-integer value $n$ and get our result. However, since the above is defined with a recurrence relation is a binomial distribution, and thus uses combinations, it won't work for a non-integer number. Thus, we can define a probability recurrence relation as follows:

$$Pr\left[ r,d\right] = \begin{cases} \\1-\frac{c}{n}:Pr\left[d-1,r\right]\\\frac{c-1}{n}:Pr\left[d-1,r-1\right]\\\frac{1}{n}:Pr\left[d,r-1\right] \end{cases}$$

The probability of no significant result is $1-\frac{c}{n}$, the probability of a regular success is $\frac{c-1}{n}$, and the probability of an 'exploding' or open-ended result is $\frac{1}{n}$. However, I don't know how to solve this recurrence relation, either.

So these are my questions:

  • For closed rolls, how do I find the probability of $n$ successes? I've looked at existing probability tables (PDF), but they don't actually contain the math for their generation, and it's non-intuitive to discern from the table. Edit: See above; the binomial formula applies.
  • For open rolls, can the closed roll formula be applied to a non-integer number $\frac{6n}{5}$? Edit: No. See above; the binomial formula uses combinations, thus requires integer numbers.
  • So, how do I solve the closed roll recurrence relation and/or find the open probability?

1 Answers1

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These can be solved at the same time, the closed roll is just a special case of the open ended.

Let

$$d = \text{Number of sides on each die}$$ $$n = \text{Number of dice rolled}$$ $$s = \text{Score required for success}$$ $$m = \text{number of successes required}$$

Also, let

$$s = n^k+r$$

where $$k\in 0,1,2,\dots$$ $$r\in 1,2,\dots,d$$

And, let

$$\begin{align} p_o &= \text{chance of getting open ended}\\ &= \begin{cases} \frac{1}{d}&,\text{if roll is open}\\ 0&,\text{otherwise}\\ \end{cases} \end{align}$$

So, for 1 die, we have a geometric distribution to get $k$ open ended results followed by a single roll to get $r$ (note that this is not exactly a geometric distribution since we need $k$ failures with $q=1-p_0$ followed by 1 success with a different probability). Also we will require $x^0=1$.

$$p_1=\left(p_0\right)^{k}\left(\frac{d-r+1}{d}\right)$$

To get $m$ success from $n$ dice, we have a binomial distribution as you noticed so

$$\begin{align} p_{m,n}&=\sum_{i=m}^n\binom{n}{i}p_1^i\left(1-p_1\right)^{n-i}\\ \end{align}$$

Some examples:

  • $d=6$, $n=1$, $s=4$, $m=1$ & open ended; so $k=0$, $r=4$

$$\begin{align} p_0=\frac{1}{6}\\ \end{align}$$

$$\begin{align} p_1&=\left(\frac{1}{6}\right)^{0}\left(\frac{6-4+1}{6}\right)\\ &=1\left(\frac{3}{6}\right)\\ &=1\times\frac{1}{2}\\ &=\frac{1}{2}\\ \end{align}$$

$$\begin{align} p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\ &=\binom{1}{1}\left(\frac{1}{2}\right)^{1}\left(1-\frac{1}{2}\right)^{1-1}\\ &=1\times\frac{1}{2}\times{1}\\ &=\frac{1}{2}\\ \end{align}$$

  • $d=6$, $n=1$, $s=10$, $m=1$ & open ended; so $k=1$, $r=4$

$$\begin{align} p_0=\frac{1}{6}\\ \end{align}$$

$$\begin{align} p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\ &=\frac{1}{6}\times\frac{3}{6}\\ &=\frac{1}{12}\\ \end{align}$$

$$\begin{align} p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\ &=\binom{1}{1}\left(\frac{1}{12}\right)^{1}\left(1-\frac{1}{12}\right)^{1-1}\\ &=1\times\frac{1}{12}\times{1}\\ &=\frac{1}{12}\\ \end{align}$$

  • $d=6$, $n=3$, $s=4$, $m=2$ & closed; so $k=0$, $r=4$

$$\begin{align} p_0=0\\ \end{align}$$

$$\begin{align} p_1&=\left(0\right)^{0}\left(\frac{6-4+1}{6}\right)\\ &=1\left(\frac{3}{6}\right)\\ &=\frac{1}{2}\\ \end{align}$$

$$\begin{align} p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\ &=\binom{3}{2}\left(\frac{1}{2}\right)^{2}\left(1-\frac{1}{2}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{2}\right)^{3}\left(1-\frac{1}{2}\right)^{3-3}\\ &=3\times\frac{1}{4}\times\frac{1}{2}+1\times\frac{1}{8}\times1\\ &=\frac{4}{8}\\ &=\frac{1}{2}\\ \end{align}$$

  • $d=6$, $n=3$, $s=10$, $m=2$ & open ended; so $k=1$, $r=4$

$$\begin{align} p_0=\frac{1}{6}\\ \end{align}$$

$$\begin{align} p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\ &=\left(\frac{1}{6}\right)\left(\frac{3}{6}\right)\\ &=\frac{1}{12}\\ \end{align}$$

$$\begin{align} p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\ &=\binom{3}{2}\left(\frac{1}{12}\right)^{2}\left(1-\frac{1}{12}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{12}\right)^{3}\left(1-\frac{1}{12}\right)^{3-3}\\ &=3\times\frac{1}{144}\times\frac{11}{12}+1\times\frac{1}{1,728}\times1\\ &=\frac{34}{1,728}\\ &=\frac{17}{864}\\ &\approx0.02\\ \end{align}$$

Dale M
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  • I'm not sure what you're setting $r$ and $k$ to; could you please clarify? (I'm generally confused by this; it really looks like a bunch of formulas) –  May 15 '13 at 03:56
  • The formula is given above - essentially k is how many open ended rolls you need and r is the result you need once you get those open ended rolls. – Dale M May 15 '13 at 03:58
  • But what if no open-ended rolls are actually needed? This is probabalistic - if I roll 6 dice and need 5 successes, I don't actually need any open ended rolls. –  May 15 '13 at 03:59
  • Then $k=0$, $r=\text{whatever result} \le d \text{ you need}$. – Dale M May 15 '13 at 04:04
  • Remember, $k$ & $r$ are only used to work out the chance of succeeding with 1 die. The next part of the calculation deals with how many successes you require. – Dale M May 15 '13 at 04:05
  • I think there's an easier way to do it without introducing two extra variables, but this now that I understand it, this is a pretty good explanation too. –  May 16 '13 at 04:07
  • I'm a bit confused by your answer. You state $s=n^k + r$, but that equation doesn't hold for your example. Also what is $s$ supposed to be? The question was, what is the probability to get exactly $m$ successes in a role of $n$ identical, fair, $d$-sided dice, where every roll equal or above the threshold $r$ counts as a success and every die is re-rolled, when it shows its maximum until it doesn't, its successes counted to the total successes. How would I calculate the other values ($s$, $k$) given I have $d$, $n$, $m$ and $r$ with your formula? – con-f-use Feb 07 '16 at 20:10