3
\$\begingroup\$

Considering a character is using the same weapon in both of the following scenarios, which would result in a higher damage per round?

  1. The character makes a single attack with advantage.

  2. The character makes two attacks: a regular attack and an attack with disadvatange.

\$\endgroup\$
  • 1
    \$\begingroup\$ I finished an approximate analysis, but it assumes that a critical hit doubles the expected damage, which is not quite the case as only the weapon's damage dice is doubled while the ability modifier is added once. If you tell use more about the character in question (level, any attack modifiers, ability scores, etc), I could guarantee greater accuracy. \$\endgroup\$ – David Coffron Nov 3 '18 at 5:42
5
\$\begingroup\$

TL;DR Choose 1 attack with advantage, if the difference between the AC and the attack modifier is at least 11, choose 2 attacks otherwise.


For those interested in math, let's calculate the expected values. The following calculation assumes that the expected damage is positive.


Let

  • \$D\$ be the difference between the AC and the bonus. We assume \$2\leq D \leq 20\$; other differences should simply be treated as the nearest number in that range (For \$D=2\$ you hit automatically except for natural 1s, for \$D=20\$ you need a 20 to hit).
  • \$E\$ be the expected damage on a normal hit
  • \$E_{crit}\$ be the expected damage on a critical hit
  • \$E_{2}\$ be the expected damage of attacking twice
  • \$E_{adv}\$ be the expected damage of attacking with advantage

$$ \begin{align} E_{adv} &=& \left(1-\left(\frac{(D-1)}{20}\right)^2 \right)\cdot E + \frac{39}{20^2} \cdot (E_{crit} - E)\\ &=& \frac{400E-(D^2-2D+1)E + 39 (E_{crit}-E)}{400}\\ &=& \frac{(360 - D^2 + 2D)E + 39 E_{crit}}{400}\\ E_2&=& \frac{21-D}{20}E + (E_{crit} - E)\frac{1}{20} + \left(\frac{21-D}{20}\right)^2E+\frac{1}{20^2}(E_{crit}-E)\\ &=&\frac{420E-20DE+20E_{crit}-20E+(441 - 42D+D^2)E + E_{crit}-E}{400}\\ &=& \frac{(420-20D-20+441-42D+D^2-1)E + (20+1)E_{crit}}{400}\\ &=& \frac{(840-62D+D^2)E + 21E_{crit}}{400}\\ E_{2}-E_{adv} &=& \frac{(480-64D+2D^2)E - 18E_{crit}}{400}\\ &=& \frac{2E}{400} \left(D^2 - 32 D + 240 - 9 \frac{E_{crit}}{E}\right) \end{align} $$ Using the p-q formula we get $$ \begin{align} K & := & \sqrt{16 + 9\frac{E_{crit}}{E}}\\ E_{adv} \geq E_{2}&\Leftrightarrow & 0\geq E_{2}-E_{adv}\\ &\Leftrightarrow & 16 -K \leq D \leq 16+K \end{align} $$

About \$\frac{E_{crit}}{E}\$ we know that it's at least 1 (no dice involved, e.g. unarmed attack) and assuming the damage before adding the dice rolls is not negative, it's at most 2, which means \$\sqrt{15+9} = 5\leq K<6 = \sqrt{16+9\cdot 2 + 2}\$ and therefore

you should choose the one attack with advantage, if the difference between the AC and your modifier is at least 11.

\$\endgroup\$
  • \$\begingroup\$ @Medix2 For this reason I double checked my answer by writing a little program simply enumerating all the outcomes of the 1 or 2 dice involved in each part of the attack. The results for \$D=10; E = 3.5; E_{crit} = 7\$ were \$E_2 = 3.1675; E_{adv} = 3.1325\$ and changing \$D\$ to 11 resulted in \$E_2 = 2.80875; E_{adv} = 2.96625\$ Which seems to confirm my results (not sure about the damage used by David though). Either I screwed up both approaches, the expected damage is significantly different from mine or there's an error in the other answer. \$\endgroup\$ – fabian Nov 3 '18 at 14:35
  • \$\begingroup\$ @fabian my answer essentially assumed the expected damage is 1 and the crit damage is 2. Under that assumption where does your calculation lie? I also may have done the probability formula for a hit wrong. I solved literally for each non-crit at every AC and to-hit value, and then divided that by the 95% chance to not-crit. This shows the percentage that a non-crit is a hit/miss. Then I added 2* the probability of a crit and added it to the damage multiplier. \$\endgroup\$ – David Coffron Nov 3 '18 at 16:26
  • \$\begingroup\$ @DavidCoffron I don't get how you came up with your formulae, but your result for AC 8, bonus 3 seems wrong: In this case you miss on 1, 2, 3 and 4 i.e. Q=4/20. To hit you must fail the roll twice: 1-Q² Now we've treated a nat 20 as normal damage requiring us increase the damage for 20s by 1. Of the 400 possible combinations of rolls on the dice 20 have a 20 as first roll and 19 have 20 as second roll but some other number as first roll -> the expected value for damage 1 / crit damage 2 should be \$\frac{400-16+39}{400}=\frac{423}{400}=1.0575\neq 1.15067867\$ \$\endgroup\$ – fabian Nov 4 '18 at 0:10
  • \$\begingroup\$ @DavidCoffron In the 2 attacks sheet the results for the normal attack are clearly wrong: We only use integral values for damage and the probability of each possible outcome is \$\frac{1}{20} = 0.05\$ so the results need to be 0.05 multiplied by some a natural number, but this is not the case. \$\endgroup\$ – fabian Nov 4 '18 at 0:16
  • \$\begingroup\$ @fabian. That's because I'm multiplying by 0.95 (the chance of a non-crit). I'll do some tinkering and see what I did wrong \$\endgroup\$ – David Coffron Nov 4 '18 at 0:39
3
\$\begingroup\$

It depends on your to-hit bonus and the target AC

In short, if the difference between your attack roll modifier and the target AC is 9 or greater, use the one advantage attack

This works because the more likely you are to hit, the better using two attacks becomes. This is because the advantage has less of a chance of mattering as the normal and disadvantage attacks have a good enough chance of hitting even without advantage.

However, as you approach very high Armor Classes, this starts to be less impactful. This is because you are mostly relying on critical hits to deal damage and while Advantage is still better (since you have two chances to score a critical hit as opposed to 1 and then a half-chance), the degree to which it is better is lessened.

Here is a simple chart showing what is best at which to hit bonuses and which Armor Classes. "Adv" means the 1 attack with advantage is better, "TWF" means using two attacks is better even though one has disavantage.

\begin{array}{c|l|l|l|l} \text{Armor Class →} & 8-11 & 12 & 13 & 14 & 15 & 16 & etc. \\ \hline \text{To-Hit Bonus ↓} \\ +3 & TWF & Adv & Adv & Adv & Adv & Adv \\ +4 & TWF & TWF & Adv & Adv & Adv & Adv\\ +5 & TWF & TWF & TWF & Adv & Adv & Adv\\ +6 & TWF & TWF & TWF & TWF & Adv & Adv \\ +7 & TWF & TWF & TWF & TWF & TWF & Adv\\ etc\\ \end{array}

The full table and raw data for this analysis can be found in this Google Sheet. You can download a copy to see the formulas. These numbers represent the multiplier applied to the weapon damage dice. However, do note that the critical damage would not be quite as high, which may change the results (without knowing the ability score modifier, it is impossible to know what portion of the damage will be affected by a critical hit).

\$\endgroup\$
  • \$\begingroup\$ The table formatting overlaps the sidebar, as the answer currently stands. Unfortunately, with so many columns, I don't think using MathJax for the table will really work (even if you swap columns and rows, there are so many rows that the same problem will occur). \$\endgroup\$ – V2Blast Nov 3 '18 at 5:57
  • \$\begingroup\$ @V2Blast ugh. I don't know of a good way to include a table then \$\endgroup\$ – David Coffron Nov 3 '18 at 5:58
  • \$\begingroup\$ Might just have to include it as an image, though I know that has accessibility issues. Or link to a table hosted elsewhere. (Or include an image preview that links to an accessible table hosted elsewhere?) \$\endgroup\$ – V2Blast Nov 3 '18 at 5:59
  • \$\begingroup\$ @V2Blast is that better? \$\endgroup\$ – David Coffron Nov 3 '18 at 6:00
  • \$\begingroup\$ Yep, that looks better. I assume people can see the trend and extrapolate from there. \$\endgroup\$ – V2Blast Nov 3 '18 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.