# What is the average roll of a die with an expanded explosion threshold as compared to one with an expanded success threshold?

This is going to be a bit of a weird question, but I'm figuring dice pool probabilities for a homebrew system and I want to know how certain abilities should be priced regarding their use. The basic roll of this system is roll a pool of d6's, count the dice that read 4, 5, or 6 as a "plus" (similar to Burning Wheel or Shadowrun), and explode dice that show 6. Pretty standard so far, but where I'm running into issues is regarding two abilities.

The first ability expands the threshold of success to threes on the d6, but otherwise the explosion threshold remains the same.

The second ability is the more difficult one; it expands the explosion threshold to fives and sixes on the initial roll but not any rerolls from exploding dice; the success threshold remains the same.

My question is this: What do the statistics for all three of these look like? I've attempted to use Anydice to answer this question but it doesn't appear to have a way of looking at the side of the dice rather than the number printed on it, so I don't think that can answer my question. I'm just looking for the average of total "hits" or "successes" per roll here, not the actual total number. Any detailed answer would be much appreciated.

• As an aside, I think the falls under the "better/different/more specific" guideline for questions that might be answered elsewhere since someone could say "ability A should be priced significantly higher than ability B" or "ability A is superior at low dice pool size but ability B is better with a larger dice pool". – user5834 Jan 22 '15 at 2:12
• What do you mean by "exploding" a die? – Matthew Najmon Jan 22 '15 at 16:56
• Typically, what it means when you explode a die is that if the die shows a certain number, you first add the die to your result and then reroll it. If it shows that number again, reroll again, and so on and so forth. – user5834 Jan 22 '15 at 17:56

Here's an AnyDice program to simulate it:

function: test N:n against X:n reroll Y:n later Z:n {
if N >= Y { result: 1 + [test d6 against X reroll Z later Z] }
if N >= X { result: 1 }
result: 0
}
output [test d6 against 4 reroll 6 later 6] named "success on 4-6, reroll on 6"
output [test d6 against 3 reroll 6 later 6] named "success on 3-6, reroll on 6"
output [test d6 against 4 reroll 5 later 6] named "success on 4-6, reroll on 5-6 (re-reroll on 6)"


Looking at the summary output, we can see that the baseline roll (success on 4+, explode on 6) yields on average 0.6 successes per die. Lowering the success threshold to 3+ increases this to an average of 0.8 successes per die, whereas lowering the explode threshold to 5+ gives 0.7 successes per die. (I was a bit surprised that the mean success rates would work out to such nice round numbers, but it seems they do.)

Thus, lowering the success threshold is a significantly better investment than lowering the explosion threshold.

In fact, this is easy to see intuitively: lowering the success threshold by one gives you a 1/6 chance of an extra success; lowering the explosion threshold by one gives you a 1/6 chance of an extra die roll. Given that the expected number of successes per die is less than one, it's pretty obvious that, given the choice, you should choose an extra success over an extra die.

Ps. I did some more testing, and it sems that, if you modify the second ability to also allow successive rerolls on 5+, the average number of successes per die increases to 0.75. That's still less than for the first ability, but closer.

Allowing a reroll on any initial success (i.e. on an initial roll of 4+), but only on a 6 for subsequent rolls, turns out to give the same average number of successes, 0.8, as lowering the success threshold to 3. Thus, if you really want the two abilities to be equally good, that might be a good option to consider.

(Or you could keep the improved reroll ability as originally suggested, i.e. only improving the first reroll chance by 1/6, and make it cost half as much as the improved success ability; if you let the player take the improved reroll ability twice, that would be, on average, equivalent to the first ability.)

• Thanks for the informative answer. Reading this, I'm encouraged to take out the restriction on the exploding dice expander and make them cost roughly the same. – user5834 Jan 22 '15 at 5:22
• You could also consider keeping the restriction, but letting the "explosion expander" allow a reroll on an initial roll of 4+; see edit above. – Ilmari Karonen Jan 22 '15 at 5:33
• If you make them just as good, won't the reroll option always be better, because it allows for MORE successes than the lowered success rate, since you roll more dice? – Erik Jan 22 '15 at 8:05
• @Erik That's looking at spread, not just average successes. And that spread comes with the trade-off of being more likely to fail worse, as well, in this case (chance of 0 successes does not decrease with explosions). You are correct that better average successes isn't the only factor however, that's why looking at the AnyDice graphs is so useful: you can see all the relevant statistics of your data easily, often simultaneously, and comprehensibly. – Please stop being evil Jan 22 '15 at 8:34
• @SevenSidedDie: No problem, that happens. Actually, that literal 6 in the code was kind of ugly anyway, so I just added an extra parameter for it to make the code more generic and, hopefully, self-explanatory. – Ilmari Karonen Jan 22 '15 at 16:35

# Derivation of formulas

First I will derivate a formula that shows how to calculate the probaility to get a number of successes for a standard d6 and for d6 with added first or second ability. Lets call $p(X=k)$ the probability to get $k$ successes with one d6. $pa(X=k)$ and $pb(X=k)$ are the probabilities to get $k$ successes with added first or second ability.

### Probabilities $p(X=k)$:

• $p(X=0)=\frac{3}{6}=\frac{1}{2}$ as we have to roll 1,2 or 3 on the d6.
• $p(X=1)=\frac{2}{6}+\frac{1}{6} \cdot \frac{1}{2}=\frac{5}{12}$ as we can roll a standard success (4 or 5) or get an exploding 6 with a follow-up miss (1,2 or 3).
• $p(X=2)=\frac{1}{6} \cdot p(X=1)=\frac{5}{72}$ as we need the first dice to explode for sure, after that we are in the situation that we want to roll another success with one d6, so the multiply with $p(X=1)$.
• $p(X=n+1)=\frac{1}{6} \cdot p(X=n)$ as we can use the same argument for all higher number of successes.

If we use this formula to calculate the probabilities we get following values:

\begin{array}{r|ccccc} X & 0 & 1 & 2 & 3 & 4\\ p(X) & 50.0\% & 41.7\% & 6.9\% & 1.2\% & 0.2\%\\ \end{array}

### Probabilities $pa(X=k)$:

• $pa(X=0)=\frac{2}{6}=\frac{1}{3}$ as we have to roll 1 or 2 on the d6.
• $pa(X=1)=\frac{3}{6}+\frac{1}{6} \cdot \frac{2}{6}=\frac{5}{9}$ as we can roll a standard success (3, 4 or 5) or get an exploding 6 with a follow-up miss (1 or 2).
• $pa(X=2)=\frac{1}{6} \cdot p(X=1)=$ as we need the first dice to explode for sure, after that we are in the situation that we want to roll another success with one d6, so the multiply with $pa(X=1)$.
• $pa(X=n+1)=\frac{1}{6} \cdot pa(X=n)$ as we can use the same argument for all higher number of successes.

If we use this formula to calculate the probabilities we get following values:

\begin{array}{r|ccccc} X & 0 & 1 & 2 & 3 & 4\\ pa(X) & 33.3\% & 55.6\% & 9.3\% & 1.5\% & 0.3\%\\ \end{array}

### Probabilities $pb(X=k)$:

• $pb(X=0)=\frac{3}{6}=\frac{1}{2}$ as we have to roll 1,2 or 3 on the d6.
• $pb(X=1)=\frac{1}{6}+\frac{2}{6} \cdot \frac{3}{6}=\frac{1}{3}$ as we can roll a standard success (only 4) or get an exploding 5 or 6 with a follow-up miss (1,2 or 3).
• $pb(X=2)=\frac{2}{6} \cdot \frac{2}{6}+\frac{2}{6} \cdot \frac{1}{6} \cdot \frac{1}{2}=\frac{5}{36}$ as we need the first dice to explode for sure, after that we are in the situation that we have to roll another success with one d6, so the multiply with $p(X=1)$.
• $pb(X=n+1)=pb(X=2) \cdot p(X=n-1)$ as we can use the same argument for all higher number of successes.

If we use this formula to calculate the probabilities we get following values:

\begin{array}{r|ccccc} X & 0 & 1 & 2 & 3 & 4\\ pb(X) & 50.0\% & 33.3\% & 13.9\% & 2.3\% & 0.4\%\\ \end{array}

# Expected Value and Standard Deviation

If we calculate the expected value or mean of hits we get

• $E[p(X=k)]=0.6$
• $E[pa(X=k)]=0.8$
• $E[pb(X=k)]=0.7$

The standard deviation or distribution of hits is

• $S[p(X=k)]=0.04$
• $S[pa(X=k)]=0.06$
• $S[pb(X=k)]=0.04$

# Dicepools

Here is a barchart for a dicepool of six dice:

Now you clearly see that the maximum for $pa$ is for $k=5$ successes - near $4.8=0.8 \cdot 6$ - where as the maximum for $pb$ is for $k=4$ successes - near $4.2=0.7 \cdot 6$. The first ability has always a lower probability for a lower number of successes and always a higher probability for more successes as the second ability. If you go to extrem numbers of successes like 10+ the second ability has slightly better chances as the second ability but as the probabilities are below 1% it might not matter anymore.

I think the formula for the expected successes is this:

\begin{align} E &= n \cdot \frac{3d - t - 2e + 1}{e-1}, &\text{where } & 1 ≤ t ≤ e ≤ d \end{align}

While the variance could be this (not tested):

\begin{align} V = n \cdot \left(\frac{d-t+1}{d-1} - \frac{(e-t)^2-(d-e+1)^2}{(d-1)^2}\right) \end{align} Here is what all the variables mean:

• $d$ ... number of sides a single die has (in Shadowrun $d = 6$ - we roll plain old six-sided dice)

• $n$ ... number of such dice in the pool (usually $n = Attribute + Skill$ in Shadowrun)

• $e$ ... minimum roll for a die to explode ($e = 6$ in Shadowrun - only the 6 explodes)

• $t$ ... minimum roll for a success ($t = 5$ in Shadowrun - 5 and 6 are successes)

• $h$ ... number of hits, i.e. dice with a result $≥t$ in the roll (not needed here)

Knowing the average spread (from the variance) is nice too, because you'll also want to know if it is still a frequent occurrence to get, I don't know, 12 successes on a roll of just 16 dice, or if 8 hits is already very unlikely. I.e. with a lower explosion threshold, higher hit counts become more likely. However, the expectation value might be very similar to that of a lower hit-threshold $t$ at higher explosion-threshold $e$.

# The Math behind

## Exploding on 6 only:

If you want formulae, I thought I might give a brief summary of my question about exploding die pools and its answers. You can show the formulae below to be true for probabilities of exactly $h$ hits, the expectation values of hits $E$ and their variances $V$:

\begin{align} p^\text{non-exp}_{d,n,t,h} &= \binom{n}{h}\left(\frac{d-t+1}{d}\right)^h\left(1-\frac{d-t+1}{d}\right)^{n-h}\\ E^\text{non-exp}_{d,n,t} &= n\ \frac{d-t+1}{d}\\ V^\text{non-exp}_{d,n,t} &= n\ \frac{(d-1)(d-t+1)}{d^2}\\ % p^\text{exp}_{d,n,t,h} &= \frac{(t-1)^n}{d^{n+h}} \sum_{k=0}^{\max(h,n)} \binom{n}{k}\binom{n+h-k-1}{h-k}\left[\frac{d(d-t)}{t-1}\right]^k\\ E^\text{exp}_{d,n,t} &= n\ \frac{d-t+1}{d-1}\\ V^\text{exp}_{d,n,t} &= n\ \frac{t\,(d-t+1)}{(d-1)^2}\\ \end{align}

The ideas for proofs can be found on math stackexchange. Now this assumes, that dice only explode at the maximum roll of 6 in your case. So it can't tell you anything about rolls where dice explode e.g. on 5 and 6. Except, that it stands to reason that a a roll of a six sided where 1 and 2 are no successes, 3 and 4 are successes without re-rolls and 5 and 6 are successes with explosion is equal to a roll of three-sided dice where 1 is not a success, 2 is a success without re-roll and 6 is an exploding success.

I've put together a small web-page (useful for Shadowrun or the oWoD) for this and tested it with a simulation:

## Arbitary explosion thresholds:

The formulae should be fairly easy to modify for arbitrary explosion thresholds with the same reasoning used in my link. Let's call the explosion threshold $e$. So if the roll explodes on 5 and 6, then $e = 5$ in this case (for Shadowrun we'd have $e = d = 6$). The expectation value $E_1$ of a single die has to fulfill this equation:

$$E_1 = 0 \cdot \frac{t-1}{d} + 1 \cdot \frac{2d-t-e}{d} + (E_1+1) \cdot \frac{d-e-1}{d}$$

Zero successes with a probability $\frac{t-1}{d}$, on success and no explosions with a probability of $\frac{2d-t-e}{d}$ and in case of exploding dice we have a probability of $\frac{d-e-1}{d}$ to get $E_1$ more successes.

This can be solves for $E_1$. Now the expectation value for $n$ dice is just $n$ times that for one dice ($E = n E_1$):

\begin{align} E &= n \cdot \frac{3d - t - 2e + 1}{e-1}, &\text{where }& 1 ≤ t ≤ e ≤ d \end{align}

Note, that while the formulae for exploding on the highest value are thoroughly tested, I did not test the above formula.

• This would be a great post on math.SE. Unfortunately I think it's a bit too dense for RPG.SE, and doesn't really answer the question in a way that makes it useful to most RPG.SE readers. – LegendaryDude Feb 23 '16 at 14:28
• I edited the post to give an easy to digest answer. – con-f-use Feb 23 '16 at 14:45