# Dice pool success probability with one exploding die

I'm looking for the math behind the following:

Players have a pool of 3–11 d6s, and have to get from 1–4 successes, "success" being a roll of 5+. One of these dice is special (represented with another color, probably) and may explode on a roll of 6, but only once.

Everything I've found on exploding dice and math involves all dice exploding infinitely or die face summation, when I'm looking for one and successes, resp. How can I model these success percentages to be put in a nifty little table for 1) playtesting and 2) other GMs? (Even though a table would be nice, there are other mechanics at work beyond these so I'd appreciate some actual math/formulae I can play with later.)

• Clarification: "Explode" here means roll a second time, so it can get 0, 1 or 2 successes? Oct 19, 2014 at 7:43
• @NeilSlater - I asked almost the same thing yesterday (comments were probably deleted after the re-edit) - Metalgearmaycry answered positively - if the re-roll comes up as a 5 or 6, it's an additional success. This confused me too as afik, normally exploding dice are summed for a bigger number (so you can roll an exploding 1d6 and get an 11), but since the target number in this question is 5, there won't be any meaningful difference between a 6 and an 11. Oct 19, 2014 at 16:58

Let's discuss all of the "ingredients" separately, and then add them:
The base odds for rolling any given number on a d6 are $\frac{1}{6}$.
The base odds for rolling a 5 or 6 on a d6 are $\frac{1}{3} = 0.33$ (2 out of 6 making $\frac{2}{6} = \frac{1}{3}$).
Similarly, the odds for rolling 1-4 on a d6 are $\frac{2}{3} = 0.67$.
The odds for a successful "single-shot" explosion are $\frac{1}{6} \times \frac{1}{3} = \frac{1}{18} = 0.06$.
Similarly, the odds for a failed explosion are $\frac{1}{6} \times \frac{2}{3} = \frac{4}{18} = 0.22$.

Combining these for a single die yields the following:

\begin{array}{l|lll} \text{Number of successes} & 0 & 1 & 2 \\ \hline \text{Odds using exploding die} & 0.67 & 0.28 & 0.06 \\ \text{Odds using normal die} & 0.67 & 0.33 & 0 \end{array}

There are a number of ways to combine the various rolls, but I think the most straight forward one is to see how the "normal" dice behave, and then account for what the "special" (exploding) die does and how it effects the results.
For example, if we know that the odds of $N$ normal dice to produce 2 successes is $P$, then we'll get:

• 2 successes at probability $P \times 0.67$ (no success from special die)
• 3 successes at probability $P \times 0.28$ (normal success from special die)
• 4 successes at probability $P \times 0.06$ (double success from special die)

So, the normal dice odds for various successes are:

\begin{array}{l|lllll} \text{# successes} & 0 & 1 & 2 & 3 & 4+ \\ \hline \text{2d6} & 0.44 & 0.44 & 0.11 & 0 & 0 \\ \text{3d6} & 0.30 & 0.44 & 0.22 & 0.04 & 0 \\ \text{4d6} & 0.20 & 0.40 & 0.30 & 0.10 & 0.01 \\ \text{5d6} & 0.13 & 0.33 & 0.33 & 0.16 & 0.05 \\ \text{6d6} & 0.09 & 0.26 & 0.33 & 0.22 & 0.10 \\ \text{7d6} & 0.06 & 0.20 & 0.31 & 0.26 & 0.17 \\ \text{8d6} & 0.04 & 0.16 & 0.27 & 0.27 & 0.26 \\ \text{9d6} & 0.03 & 0.12 & 0.23 & 0.27 & 0.35 \\ \text{10d6} & 0.02 & 0.09 & 0.20 & 0.26 & 0.44 \end{array}

And, the odds for results with one special die are:

\begin{array}{l|lllll} \text{# successes} & 0 & 1 & 2 & 3 & 4+ \\ \hline \text{special + 2d6} & 0.30 & 0.42 & 0.22 & 0.06 & 0.01 \\ \text{special + 3d6} & 0.20 & 0.38 & 0.29 & 0.11 & 0.02 \\ \text{special + 4d6} & 0.13 & 0.32 & 0.32 & 0.17 & 0.06 \\ \text{special + 5d6} & 0.09 & 0.26 & 0.32 & 0.22 & 0.12 \\ \text{special + 6d6} & 0.06 & 0.20 & 0.30 & 0.25 & 0.19 \\ \text{special + 7d6} & 0.04 & 0.15 & 0.27 & 0.27 & 0.28 \\ \text{special + 8d6} & 0.03 & 0.11 & 0.23 & 0.27 & 0.36 \\ \text{special + 9d6} & 0.02 & 0.09 & 0.19 & 0.25 & 0.45 \\ \text{special + 10d6} & 0.01 & 0.06 & 0.16 & 0.23 & 0.54 \end{array}

Hope this helps!

P.S. - If anybody spots a miscalculation, or an error in the general approach, please correct or comment...

## Addendum - more on probability calculations:

So would you be adding or multiplying those probabilities together?
– Metalgearmaycry

Actually, we do some of both: When calculating the odds for complex scenarios, there are two ways to combine the probabilities of the basic elements of these scenarios together:

• For the odds of both A and B occurring together, Multiply their probabilities
$P(A\;and\;B) = P(A) \times P(B)$

• For the odds of either A or B occurring, Add their probabilities
$P(A\;or\;B) = P(A) + P(B)$

As an example, let's consider the calculation for the odds of getting a single success with your "one-shot" exploding die. There are two ways to get (only) a single success:

1. Roll a 5 (mark this as event A).
OR
2. Roll a 6 (mark this as event B) and then roll 1-4 on the "exploding roll" (mark this as event C).

So, we are looking at the scenario "(roll a 5) or (roll a 6 and then 1-4)",
which may be written as $P(one\;success) = P(A) + (P(B) \times P(C))$.
The odds for a die to come up on any given number are $\frac{1}{6}$, plugging that into the equation yields:

$$P(one\;success) = P(A) + (P(B) \times P(C)) = \dfrac{1}{6} + \left(\dfrac{1}{6} \times \dfrac{4}{6}\right) = \dfrac{6}{36} + \dfrac{4}{36} = \dfrac{10}{36} = 0.28$$

-- That's how the small table above was calculated.

### Generating the "normal" dice table

When working with large groups of scenarios, there are shortcuts which work better than using only addition and multiplication. For example, consider the odds for rolling 5d6 normal dice and getting exactly 2 successes. In order to calculate that, we need to combine the following:

1. Dice 1 & 2 rolled a success and dice 3, 4, 5 rolled a failure.
2. Dice 1 & 3 rolled a success and dice 2, 4, 5 rolled a failure.
3. Dice 1 & 4 rolled a success and dice 2, 3, 5 rolled a failure.
4. Dice 1 & 5 rolled a success and dice 2, 3, 4 rolled a failure.
5. Dice 2 & 3 rolled a success and dice 1, 4, 5 rolled a failure.
6. And so on and so forth...

Note that for each of these scenarios, the odds are the same (2 dice succeed and 3 dice fail). So, we can write them down and sum them all, but it would be easier to just count them and multiply the result by the odds for one such scenario.
For this we can use another tool, which comes from basic combinatorics (you can read more about it here) - it is used to find how many different combinations are there for $K$ elements chosen from an $N$ sized collection.
And its formula is:

$$\dfrac{N!}{(K!) \times (N-K)!}$$

For example, to calculate all the combinations of choosing the 2 successful dice out of the total of 5, the calculation is:

$$\dfrac{5!}{2! \times 3!} = \dfrac{120}{2 \times 6} = 10$$

Good, so we know there are 10 scenarios, and each with the odds of $P(success)^2 \times P(failure)^3 = \left(\frac{2}{6}\right)^2 \times \left(\frac{4}{6}\right)^3 = 0.033$ So the odds of rolling 5d6 and getting exactly 2 successes are $10 \times 0.033 = 0.33$. That's how the "normal" dice table was generated.

The general formula for $K$ successes rolling $N$ "normal" dice is:

$$\dfrac{N!}{K! \times (N-K)!} \times \left(\dfrac{2}{6}\right)^K \times \left(\dfrac{4}{6}\right)^{N-K}$$

### Generating the final table

Still here? well done!
We already know how a bunch of normal dice behave, in order to generate the final table, we need to account for what the "special" one-shot-explosion die can add to the result.
Reminder:

\begin{array}{l|lll} \text{Number of successes} & 0 & 1 & 2 \\ \text{Odds using exploding die} & 0.67 & 0.28 & 0.06 \end{array}

Let's continue with the 5d6 and 2 successes example from before, we now add a sixth exploding die to the mix. So, there are 3 different scenarios to take into account:

1. The 5d6 yield no successes, and the special die explodes for 2 successes.
2. The 5d6 yield 1 success, and the special die yield another 1 success.
3. The 5d6 yield 2 successes, and the special die yields no successes.

In order to calculate their odds, we just multiply the value of the relevant cell in the "normal" dice table (for the 5d6 part) with the relevant cell in the mini-table above (for the exploding die part). So we'll get:

\begin{align} &\text{1. } P(\text{5d6 with 0 successes}) &\times &P(\text{exploding 2 successes}) &= 0.13 \times 0.06 &= 0.0078 \\ &\text{2. } P(\text{5d6 with 1 success}) &\times &P(\text{exploding 1 success}) &= 0.33 \times 0.28 &= 0.092 \\ &\text{3. } P(\text{5d6 with 2 successes}) &\times &P(\text{exploding 0 successes}) &= 0.33 \times 0.67 &= 0.22 \end{align}

Summing all three results give the odds for getting exactly 2 successes for 5d6 + special, which are $0.0078 + 0.092 + 0.22 = 0.32$

And that's how the final table is generated!

• So would you be adding or multiplying those probabilities together? Oct 19, 2014 at 1:39
• @Metalgearmaycry with probabilities, you multiply the odds of a combined result so P(A and also B) = P(A) x P(B). And you sum the odds for alternative scenarios, so P(A or B) = P(A) + P(B). For example, a single success with the exploding die is the result of rolling a 5 or rolling a 6 and then rolling 1-4. You'll calculate the odds as P(rolling a 5) + P(rolling a 6) x P(rolling 1-4), which is 1/6 + 1/6 x 4/6 = 10/36 = 0.28. Oct 19, 2014 at 2:16
• @GoBLiN, I was able to basically understand what you've got, and replicate the "normal dice table" in Excel, but I'm at a loss as to what to do to achieve the numbers in the "special die table." Also, would the odds in your first comment not be (p(a) + p(b))*p(c) instead of just p(a) + p(b) * p(c)? Oct 19, 2014 at 20:14
• @Metalgearmaycry - regarding the first comment, marking (a) as rolling a 5, (b) as rolling a 6 and (c) as rolling 1-4. The two scenarios we need to consider are either [a] or [b and then c] (because one success means the first roll is a 5 or a 6, and if it's a 6, the second role must be 1-4). So the odds are P(a) + (P(b) x P(c)). The variant you suggested (P(a) + P(b)) x P(c) is equivalent to (P(a) x P(c)) + (P(b) x P(c)). Which may be described in words as "the odds for rolling 5 or 6 and then rolling 1-4", this is not consistent with an "explosion" occurring only after rolling a 6. Oct 19, 2014 at 21:03
• You only multiply probabilities if the results are independent as they are here Jun 5, 2016 at 9:02