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At Level 12, a Pact of the Blade Warlock can take the Lifedrinker Eldritch Invocation, which states the following:

When you hit a creature with your pact weapon, the creature takes extra necrotic damage equal to your Charisma modifier (minimum 1).

Assuming a Hexblade Warlock with a 20 in Charisma, this translates to 10 bonus damage per attack. This enormous bonus means that every missed attack results in a significant reduction in damage.

The Sharpshooter and Great Weapon Master feats grant options of the following pattern:

Before you make [an attack] with [a weapon] with which you are proficient, you can choose to take a -5 penalty to the attack roll. If you do so and the attack hits, it deals +10 damage.

At what point is the Warlock's expected damage per attack reduced by using the optional attack modifier of either the Sharpshooter or Great Weapon Master feat?

Assume a Hexblade, Pact of the Blade Warlock with the following characteristics:

  • At least 12th level
  • Lifedrinker Eldritch Invocation
  • 20 in Charisma
  • Using a pact weapon, either:
    • A longbow
    • A greatsword
  • ... with the respective weapon feat:
    • Sharpshooter
    • Great Weapon Master

With the following variables:

  • Enemy AC
  • Attack roll and damage roll bonuses (due to spells, proficiency bonus, magic weapons, etc.)
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  • \$\begingroup\$ Could you clarify how a 20 Cha translates into +10 damage? Is that also including 20 Str? Or is it a miscalculation (20 Cha is a +5 Mod) \$\endgroup\$
    – Ifusaso
    Jul 17, 2020 at 19:28
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    \$\begingroup\$ @Ifusaso Hexblades can use Cha for attack and damage with their pact weapon. \$\endgroup\$
    – Szega
    Jul 17, 2020 at 19:29
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    \$\begingroup\$ Related: Are features that allow −5 to attack to get +10 to damage mathematically sound? \$\endgroup\$
    – Sdjz
    Jul 17, 2020 at 19:59
  • \$\begingroup\$ That does not clear up why you think a 20 Cha is +10 (or if its a typo) \$\endgroup\$
    – Ifusaso
    Jul 17, 2020 at 23:24
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    \$\begingroup\$ @Ifusaso +5 from Hex Warrior, +5 from life drinker for +10 damage. \$\endgroup\$ Jul 18, 2020 at 3:10

1 Answer 1

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The break even formula for GWM/SS requires a bit of algebra. First, our variables:

  • \$DAM\$ is average damage per hit before GWM/SS.
  • \$HIT\$ is bonus to hit before GWM/SS.
  • \$AC\$ is target armor class.

This is all we need to get our formula. We will set expected damage without GWM/SS equal to expected damage with it, and then solve for \$AC\$.

In this initial equation, the left is expected damage without GWM/SS, right is with.

$$DAM\left(\frac{21+HIT-AC}{20}\right)=(DAM+10)\left(\frac{16+HIT-AC}{20}\right)$$

The steps for solving for \$AC\$, left as an exercise to the reader, yield a break even AC of:

$$AC=16+HIT-\frac{DAM}{2}$$

Note, expected crit damage would be added as a constant on each side of our initial equation, so it is ignored.

Further, when the target's AC is so high that the only way to hit it is with a natural 20, which is when \$AC-HIT\geq20\$, it becomes best practice again to take GWM/SS on every attack since the penalty is meaningless.

The numbers.

With a mundane longbow, assuming a proficiency bonus of +4, \$DAM=1d8+10=14.5\$ and \$HIT=9\$. Then our break even AC is 17.75. This means on average, we should use SS for target ACs 17 and below.

With a mundane greatsword, assuming proficiency bonus of +4, \$DAM=2d6+10=17\$ and \$HIT=9\$. Then our break even AC is 16.5. This means on average, we should use GWM for target ACs 16 and below.

These are just examples, the formula is simple enough to substitute your own numbers with little issue.

A user named Bacon Bits works this result in greater detail in this answer, go give them some love.

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    \$\begingroup\$ Crit damage doesn’t add “a constant” it adds the same random variable which since it has the same expected value still makes no difference to the outcome but the description is not quite correct. \$\endgroup\$
    – Dale M
    Jul 18, 2020 at 1:06
  • \$\begingroup\$ Correct, but the value of crit damage is a random variable- you imply it’s constant. \$\endgroup\$
    – Dale M
    Jul 18, 2020 at 3:07

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