5
\$\begingroup\$

I have a lot of weapons on my own system and I want to split on four categories like, common, uncommon and stuff.

I want to split them by how good these weapons perform damage wise. But I have difficulties with the underlying math, since some weapons have very close maximum and minimum damage numbers.

For Example:

  • 2d12+3: can roll values between 5 and 27
  • 3d8+2 can roll values between 5 and 26

I have many different weapons with different dices in different quantities. The dice range from d4 to d12 and the count from 1 to 3.

Ex: 1d4+3 / 3d6+2 etc.

I am looking for how to compare my weapons with each other damage wise, so not only for the examples but also how to do the comparsion in general.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ How does the minimum and maximum work? Values are clipped to that range? Rerolled? And are you asking for the math, or for the idea how to write a computer program or excel sheet to process that? \$\endgroup\$
    – Mołot
    May 18 at 22:20
  • 2
    \$\begingroup\$ From the values you give, is there a typo or a math error for the 3d8 example? If you're just giving the highest possible value, it should be 5/26 \$\endgroup\$
    – 3C273
    May 19 at 0:03
  • 2
    \$\begingroup\$ Welcome to RPG.SE! Take the tour if you haven't already, and check out the help center for more guidance. \$\endgroup\$
    – V2Blast
    May 19 at 0:44

3 Answers 3

13
\$\begingroup\$

I mean, if you want the best answer,

Take a basic probability and statistics class.

That’s how you figure these things out, with basic probability and statistics. These are taught in most high schools and pretty much all colleges (barring extremely specialized schools, maybe), plus there are a plethora of online courses out there.

For the very most basic things, a single (fair) die has what’s called a “uniform distribution.” That is, when you roll a d6, you have an even (“uniform”) chance of getting 1, 2, 3, 4, 5, or 6. In probability and statistics, we talk about the “expected value” of distributions, which is what you can expect, on average, from that distribution. For a uniform distribution, the expected value is just the average (arithmetic mean) of each possibility in it. For a d6, that’s

$$ \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5 $$

In a probability and statistics class, you’d learn about the expected value for all sorts of things. You’d also learn shortcuts for calculting them. In the case of a regular dX die (numbered 1 through X), the expected value is (X+1)/2, or (equivalently) X/2+½. And so for the standard dice:

Die Expected Value
d2
d3 2
d4
d6
d8
d10
d12
d20 10½

For multiple dice that are added together (e.g. 4d6), you can determine the expected value by just multiplying the expected value by the number of dice (e.g. 4×3½ = 14). This still works when you use dice of different sizes (e.g. 4d6 + 3d4 = 4×3½ + 3×2½ = 21½). Flat numbers have an expected value of themselves, so you can just add those in (so 4d6+3 has an expected value of 14 + 3 = 17).

And that basically covers simple rolls of dice where you just add everything together. Note that there is a lot more to be said here. For instance, while you can just add expected values together to get the expected value from adding dice together, the sum of several dice does not produce a uniform distribution and 3d6 produces very different results from 1d20 even though they are both expected to average out to 10½ over the long run.

As for minimums and maximums, that’s going to depend a lot on how exactly you’re defining those. But if, say, you apply those to individual dice, so for example “d6, minimum 2,” and you’re just rerolling numbers below the minimum or above the maximum, you’re looking at a set of 2, 3, 4, 5, 6. So you can just do (2 + 3 + 4 + 5 + 6)/5, which equals 4. As long as your minimums and maximums maintain that uniform distribution (each number appears no more than once), all of the above will hold true, you just need to figure out the expected values of individual dice.

On the other hand, if you apply the minimums and maximums to the sums of multiple dice, so something like “3d6, minimum 8,” then you’re not dealing with a simple uniform distribution anymore, and that gets more complicated to calculate. Not necessarily very complicated, but you need more than the above. Likewise, if instead of rerolling numbers below minimums or above maximums, you bump numbers up or down as needed, that will also change things (since, for example, “d6 minimum 2” would be twice as likely to have a result of 2 as any other number). For specific, more complex rolls, separate questions specifically about that roll can be a good way to learn, but to be able to do it yourself, again, I recommend finding a decent online course.

\$\endgroup\$
5
  • \$\begingroup\$ I think for a d6 minimum result being a 2 it’d be (2 + 2 + 3 + 4 + 5 + 6)/6, as you have two chances to roll a 2. \$\endgroup\$ May 19 at 15:43
  • \$\begingroup\$ Related: while I do not want to discourage basic math education (with many benefits beyond our hobby), Anydice will calculate expected value, distribution, min and max, and my other things for you, just enter for example “output 3d8+2” and hit calculate \$\endgroup\$ May 19 at 15:49
  • 2
    \$\begingroup\$ @GroodytheHobgoblin Re: d6 min 2: good point—depends on the definition of minimum! I was assuming 1s were just rerolled, but that assumption wasn’t clear in my answer. As for Anydice, yes, it’s great, but without some background, I think you’d be hard pressed to know what you’re looking at. \$\endgroup\$
    – KRyan
    May 19 at 18:06
  • \$\begingroup\$ Khan Academy has a statistics and probability unit that's completely free and at 7th grade level, so fairly accessible. (That's basically early high school, or middle school for societies that have those.) \$\endgroup\$ May 27 at 17:28
  • \$\begingroup\$ Alluded to, but not explicit: the more dice you roll, the more likely you are to roll closer to the average value. A single d4 can roll 1, 2, 3, 4 with equal probability. 2d4 have a 25% chance of rolling a 5, and only 6.25% chance of rolling a 2 or 8 (minimum and maximum). anydice.com/program/15cd3 shows that tendency for 1 through 9d4. Rolling a single die will generally feel more "swingy" than rolling multiple dice, since each value will (theoretically) come up equally often; rolling more dice will generally feel more consistent-but-average. \$\endgroup\$
    – minnmass
    May 27 at 21:23
3
\$\begingroup\$

While I agree with KRyan's answer, I'm going to suggest something simpler:

Look at probability distributions.

AnyDice is a good (and free!) tool for this. Put output 3d6 into the text box and hit "Calculate", and you'll see something like this:

probability distribution of 3d6

The length of each bar indicates the probability of rolling that result. You can see that the minimum here is 3 and the maximum is 18.

Now replace that with output 1d16+2 and hit "Calculate" again.

probability distribution of 1d16+2

Once again you have a minimum of 3 and a maximum of 18, but the distribution is very different! This is formally called a uniform distribution, with every result being equally probable, while the 3d6 is closer to a normal distribution, with the middle values being much more likely than the extremes.

If you don't have much experience with probability or statistics, I think looking at the distributions in nice bar graphs like this is a good way to get an intuition for them. AnyDice can also tell you statistical measures like the mean and standard deviation, but if you don't have experience working with those, the bars show you the same thing in an easier-to-eyeball way.

P.S. AnyDice in particular can also handle all sorts of complicated dice systems beyond standard D&D XdY+Z, which is one reason I recommend it in particular.

\$\endgroup\$
1
  • \$\begingroup\$ The limit of AnyDice is about 10-ish dice though \$\endgroup\$
    – Trish
    May 30 at 5:41
1
\$\begingroup\$

As the other answers already explored the mathematical and a tool supported approach I want to give a less "mathy" rule-of-thumb approach.

Using dice pools with sameish maximum and minimum values both are on avarage equally strong. However, you can compare a weapon with less high value dice (e.g. 1d20, 2d12) with a weapon with more low value dice (e.g. 4d6, 8d3) this way:

The weapon with more dice will more likely roll results close to the avarage, while the weapon with less dice will (in comparsion) have a higher probability to roll the extrem values close to maximum and minimum. Note however that all dice-pools with more than 1 die will always have a higher probability to roll the avarage value, than to roll min or max.

So to look at the weapons you gave for an example:

  • 3d8+2
    • more reliably rolls close to avarage results (53% chance to roll 13 to 18)
    • less likely to roll high damage results (4% chance to roll 23 to 26)
    • less likely to roll low damage results (4% chance to roll 5 to 8)
  • 2d12+3:
    • less reliably in the "avarage zone" (44% chance to roll 13 to 18)
    • more likely to roll high damage results (10% chance to roll 23 to 27)
    • more likely to roll low damage results (7% chance to roll 5 to 8)

This effect gets much more extreme when comparing dice-pools that differ even more in the dice-count (for example 1d20+4 vs 6d4).

Which of the two is better is highly situational and also depends on the system you are playing.

For example the 3d8+2 weapon would be better than the 2d12+3 to finish off an opponent with 11 hp left (89% vs 85%), while the 2d12+3 weapon has better chances in one-hitting an opponent with 23 hp left (4% vs 10%).

\$\endgroup\$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .