If your wild shaped form has damage resistance and you revert to your normal form, does your normal form take the halved damage?

Question

Let's say we had a druid with the Circle of the Moon archetype. They can use their wild shape feature to turn into an elemental as per their Elemental Wild Shape feature (PHB, p. 69):

Elemental Wild Shape
At 10th level, you can expend two uses of Wild Shape at the same time to transform into an air elemental, an earth elemental, a fire elemental, or a water elemental.

Those elemental creatures all have resistance to nonmagical bludgeoning, piercing and slashing damage, and some forms have other damage resistances too. Let's say your elemental form was on 5 HP and you took 20 nonmagical slashing damage. Your elemental form would halve that damage to 10 damage. This exceeds the total HP the elemental form was on, and so you would revert back to your normal form, as per wild shape (PHB, p. 67):

... if you revert as a result of dropping to 0 hit points, any excess damage carries over to your normal form. For example, if you take 10 damage in animal form and have only 1 hit point left, you revert and take 9 damage.

Let's assume that your normal form has no damage resistances. Does your normal form really take just 5 slashing damage, even though your normal form isn't resistant to slashing damage, or would your normal form take 10 damage, since that's what it would have been if it wasn't halved by a resistance that no longer applies to your reverted form? Intuitively I think the normal form should take 5 damage, but the below related question's answers (even through it's the opposite scenario) would imply that it would take 10 damage.

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Related (but the other way around, since that Q&A is talking about the normal form having the resistance, whereas I'm asking about the wild shaped form having the resistance): How does resistance/vulnerability/immunity interact with carryover damage after reducing Polymorphed (or Wild Shaped) form to 0 HP?

Answer 1

As far as I can tell, the damage taken is only calculated once. The same applies to a spell-caster with stoneskin up who loses concentration--they gain the benefit of the resistance for the triggering damage, but lose it for future ones.

The order of operations goes

1. Calculate raw damage (the result of the damage roll)
2. Apply resistance, immunity, and vulnerability to determine the damage taken.
3. Reduce the wildshaped form's hit points by min(remaining HP, damage taken)
4. If the wildshaped form hits zero hit points as a result of the damage with some of that pool remaining, the regular form takes the remainder.

I am not aware of any effect that has you recalculate damage after it's already been partially applied--that would go against the simplicity which is a major design goal of 5e for one thing and leave lots of nasty edge cases to be worked out, all without any guidance from the rules.

Answer 2

It depends on how you conceptualize damage

Answers to the following question show that the understanding of vulnerability and resistance is not a wholly agreed upon thing:

• How does resistance/vulnerability/immunity interact with carryover damage after reducing Polymorphed (or Wild Shaped) form to 0 HP?

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The rules on "Damage Resistance and Vulnerability state:

[...] If a creature or an object has resistance to a damage type, damage of that type is halved against it. If a creature or an object has vulnerability to a damage type, damage of that type is doubled against it. [...]

However, people don't agree whether such modifications apply to the total damage or whether they affect the rate of damage. These might not sound particularly different but let's look at how each interpretation applies to a scenario where your altered form has 10 HP remaining, and has resistance to an incoming 40 damage:

1. Resistance applies to the total damage

   • You are dealt 40 damage, your resistance immediately cuts this to 20; there is 20 damage left to take.
   • Your altered form takes the first 10 damage and you revert to your normal form; there is 10 damage left to take.
   • Your normal form takes the remaining 10 damage.

2. Resistance is a change in rate of hit point deduction

   • You are dealt 40 damage, each hit point you lose deducts 2 hit points from this total; there is 40 damage left to take.
   • Your altered form reaches 0 hit points, deducting 20 damage from the total and you revert to your normal form; there is 20 damage left to take.
   • Your normal form takes the remaining 20 damage.

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To understand these differences further let's see what happens if your altered form is vulnerable to the incoming 40 damage (and the altered form is neutral):

1. Vulnerability applies to the total damage

   • You are dealt 40 damage, your vulnerability immediately enhances this to 80; there is 80 damage left to take.
   • Your altered form takes 10 damage and you revert to your normal form; there is 70 damage left to take.
   • Your normal form takes the remaining 70 damage.

2. Vulnerability is a change in rate of hit point deduction

   • You are dealt 40 damage, every 2 hit points you lose deducts 1 hit point from this total; there is 40 damage left to take.
   • Your altered form reaches 0 hit points, deducting 5 damage from the total and you revert to your normal form; there is 35 damage left to take.
   • Your normal form takes the remaining 35 damage.

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The first interpretation makes resistance stronger and vulnerability worse compared to the second. This is just a ramification of the interpretations.

That said, the scores on the answers to the question I linked at the top show that the general agreement is that the first interpretation is correct.