# How do you calculate average damage for DPR comparisons?

I have asked a few questions about optimization and DPR (damage per round) and I don't understand how those numbers are reached.

Take, for example, a monster with 15 AC and a level 1 fighter with a greatsword and +2 on strength.

I think that die average is just a simple divide by 2.

• d4 average = 2
• d6 average = 3
• d8 average = 4
• d10 average = 5
• d12 average = 6
• d20 average = 10

A greatsword does 2d6, so a average of 6 damage.

Adding prof and str to the roll makes gives it a + 4 making a d20 roll between 5 and 24.

Does this make a AC 15 with +4 the same as an AC 11 with +0?

An AC 11 gives 9 out of 20 chance to hit = 9/20 = 0.45 = 45% to hit.

Then the 6 damage divided by the percentile to hit for 0.45 * 6 = 2.7 damage per round.

Are my calculations correct or am I missing something?

• "Does this make a AC 15 with +0 the same as an AC 11 with +4?" Did you say that backwards? AC 15 with +4 would be the same as AC 11 with +0. – Darth Pseudonym Dec 31 '18 at 13:02
• @DarthPseudonym yes i did, corrected it. also can people who downvote place a comment on why they downvote. – darnok Jan 2 at 13:22
• @darnok: Related metas: Can we require comments on downvotes?, Why is an answer being downvoted without any comments? (mostly applies to questions too). People can choose to leave a comment suggesting improvements if they like, but they're under no obligation to do so. – V2Blast Jan 3 at 4:45

i think that die average is just a simple divide by 2.

No, you need to add 0.5, since the damage is the average of all possible rolls. You could pair up the lowest and the highest, the second lowest and the second highest roll, ect. to get $$\\frac{sides}{2}\$$ pairs with sum $$\sides+1\$$ each. Divide the sum by the number of rolls in the pair (2) to get the average of $$\(sides+1) / 2\$$.

does this make a ac15 with +0 the same as an ac11 with +4?

That's the wrong way of expressing the fact you used, when calculating the number of positive outcomes. For each roll $$\X\$$ comparing $$\X+4\$$ with $$\AC\$$ and comparing $$\X\$$ with $$\AC-4\$$ yields the same result. Of course this means that the combinations of (AC 15, +4 attack bonus) and (AC 11, +0 attack bonus) have the same chance of hitting.

an AC 11 gives 9 out of 20 chance to hit = 9/20 = 0.45 = 45% to hit.

Not exactly. 11, 12, ..., 20 are $$\10 = (20 - (AC - modifier - 1)) = 21 + modifier - AC\$$ numbers.

are my calculations correct or am i missing something?

You're missing something (except for what's already mentioned): critical hits. Also you're not including your strength modifier in the damage calculation.

### Correct calculation

The expected damage without considering the increased damage of a critical hit

$$E_{non-crit} = \frac{10}{20}\left(2\cdot\frac{1}{2}(6+1)+2\right) = \frac{90}{20} = 4.5$$

Now we also need to double the dice, if a critical hit happens (1/20 cases); note that the base damage and the damage dice of a normal hit were already included in the previous calculation:

$$E_{bonus\;crit} = \frac{1}{20} \cdot 2 \cdot \left(\frac{1}{2}(6+1)\right)= \frac{7}{20} = 0.35$$

Summing both expected damage values we get $$\E=4.5+0.35=4.85\$$ expected damage for a single attack.

# The average of a die

The average result of a die is ultimately just that, the average of all the values on the die -- or in other words, the sum of all the faces divided by the number of faces.

If it's a d4, that's (1+2+3+4)/4 = 10/4 = 2.5

Because of a quirk of the way even distributions work, on a die you can actually just average the highest and lowest values and get the same answer: (1+4)/2 = 5/2 = 2.5

Since the lowest value is always 1, you can further reduce the calculation to "half the die size, plus a half": 4/2 + 0.5 = 2.5

# Multiple dice

When you have multiple damage dice, you can add their averages together, but an easy trick is to remember that the average of two same-size dice is just the highest value plus the lowest -- so a greatsword's 2d6 averages 7 damage, or if you have a spell that does 4d8, you can quickly say that's two 9's, or 18.

And this gets into some statistical math, but something you'll often hear people say in DPR discussions is that "more dice is more average" (or the same thought expressed differently). It's a little counter-intuitive, but in brief, the more dice you roll, the more the randomness will cancel itself out and ensure that the actual result comes closer to the average, while fewer dice means more random results.

In technical terms, we call this the "standard deviation". A large standard deviation means the results are 'more random' -- that is, more likely to fall far away from the average, expected result -- while a small standard deviation means the results are clustered tightly near the average and exceptions are very rare.

Consider the classic Greatsword vs Greataxe. The averages are nearly the same -- 7 vs 6.5 -- but the axe has a 1 in 12 chance of rolling 12 damage, the same as any other value; while the sword has only 1 in 36 (the chance of rolling 6 on one die times the chance of rolling 6 on the second die as well). A greataxe has the same 1 in 12 chance of rolling a 7, while the chance of a greatsword getting a 7 is actually 1 in 6. The 2d6 produces reliable, but not spectacular damage; while the 1d12 is a gamble with a good chance of rolling either very high or very low. Which one you prefer often depends on whether you'd rather risk a crappy damage roll for the chance of a good one or if you'd rather have a boring but reliable weapon.

For a more extreme example, consider the chances of rolling maximum damage on a fireball -- or minimum! Just intuitively, both results are extremely unlikely, since they would depend on rolling 6 (or 1) on many dice at the same time. And if you threw 1,000 d6s and summed them up, the result would almost certainly be very close to 3500 (probably to within a hundred or so), even though the range of results is a huge 1000-6000.

# To-hit percentages

Your to-hit values are basically correct. Some people do what you did and figure out the range of the attack die, then compare it to the target; but a more common method is to ask "What do I hit on?" (i.e., what number do you need to roll to match the AC?) Count how many faces on the d20 will result in a hit and then multiply that by 5%. For example, if you have a +6 attack bonus, and the target has AC 14, you hit on an 8 or higher. So that means there are 7 faces that will result in failure, and 13 faces that result in success on this particular roll. 13 * 5% = 65%

The complications come in when you start working with conditionals. If you just have multiple attacks, it's easy enough to multiply the average single-attack DPR by the number of attacks to get your true expected damage per round. But then you get effects that use an "If X then Y" structure, and those require some (admittedly fairly basic) statistical math to figure out, which is too complicated to go into here.

• Just to be explicit: Standard deviation is proportional to square root of the number of dice rolled, so it increases as the number of dice increases. However, standard deviation divided by mean does decrease. With 1000d6 the standard deviation seems to be around 54 (I only used a calculator and did not verify the result). – Tommi Dec 31 '18 at 21:39