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I saw someone arguing recently that

A) Barbarians should always RA. and B) Assuming A) is true ... then it's a waste of time for a Barbarian to try to boost their AC.

I assume that the case for B) is supposed to be "since they've got advantage ... they're so likely to hit you that the AC boost isn't worth what effort / expense you put into gaining it."

i.e. that giving your opponents Advantage on attacks against you means that AC increases are less valuable.

Is this argument for point B) valid?


Let us assume that point A) is in fact true. I'm not convinced it is ... but that's not what I'm interested in, for this question!

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    \$\begingroup\$ Worth is heavily opinion based, I would suggest asking for a comparison to some sort of baseline, such as 'how effective is each point of AC against advantage and normal attacks' \$\endgroup\$
    – SeriousBri
    Nov 28 '21 at 19:42
  • \$\begingroup\$ I'm flagging this as a dupe, because you can easily deduce the answer to this question from the answer to this one: How does rolling two d20 and taking the higher affect the average outcome? Reckless attack renders AC boost less effective. \$\endgroup\$
    – Lovell
    Nov 28 '21 at 20:54
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    \$\begingroup\$ @Lovell there probably is more to talk about that that answer. For ex, do monster have 50/50 to hit character at any point on the leveling curve? So what if the barbarian tries to be a tank vs try to be a killer? The question should probably tell more about the kind of encounter or level range they expect. \$\endgroup\$
    – 3C273
    Nov 28 '21 at 22:21
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    \$\begingroup\$ There are other less obvious considerations, like what if the parties enemies try a high damage sneak attack against the barbarian before they see them? They won't be raging and will take full damage, and the effective loss of AC won't be in effect. \$\endgroup\$
    – Turksarama
    Nov 29 '21 at 6:20
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Enemies attacking with advantage lowers the value of AC boosts, especially in the mid-ranges. It does not remove the value entirely.

Zerohitpoints made a helpful graph of what the probabilities look like when rolling 2d20 and taking the highest.

advantage probability graph

Most enemies have a decent attack bonus, with even CR 1 foes having +3 or more to hit. At higher levels AC does not scale quite as well vs attack bonus, especially with special attacks or means for enemies to have advantage.

However if you go off the graph, there's only a 24% or so chance that an enemy with advantage will roll below 11. For a barbarian with 15 AC say, that means you are going to be hit around 85% of the time vs someone with +5 to hit, and around 80% vs someone with +3. A standard plot of 5%/number roll means you are hit around 50% of the time vs someone with +5 to hit, and only 35% vs someone with +3. Add to that (by having good medium armour and a +2 dex modifier, say, or wielding a shield on top) and you are dodging the vast majority of attacks at lower levels.

So yes, broadly - if you always use Reckless Attack, and your AC isn't particularly high to start with (10 dex, cheap medium armour, no magic items or buffs, no shield) then the odds of you getting hit increase significantly, to the point it may reach 90%+ vs higher to-hit enemies. At that point, increases to AC (from Rings of Protection or buffs or so on) may be less valuable than vs say, giving that AC boost to the paladin with a shield who enemies have to roll against without advantage.

However, even if you were always using reckless attack - if you stacked AC as much as possible you are still seeing a significant benefit, not THAT much lower than someone who isn't being attacked with advantage. The difference is mostly egregious at lower AC levels.

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  • \$\begingroup\$ I think this answer lacks one aspect of discussing AC boosts: AC boosts are always worth more with an already high AC. Or more specifically, AC boosts are better when the attacker already has a lower chance of hitting. In the absence of Advantage, a foe hitting on a 11+ will lose 10% of hit chance if they now have to roll 12, however a foe hitting on a 19+ will lose 50% of hit chance if they now have to roll 20. Advantage skews the percentages a bit, but not that much. \$\endgroup\$ Nov 29 '21 at 11:41
  • \$\begingroup\$ I don't have enough for a full answer, but wanted to agree with this answer and also point out a couple of things (having played a barbarian and DM'd for a barbarian for over a year each). Having a high AC absolutely matters. Just because advantage makes it easier to hit you, doesn't mean they always will, and the higher your AC, the better the chances of them missing on every roll of the dice. Plus, there are a lot of attacks that trigger nasty side effects, so "I'm just soaking damage" can become "now I'm grappled/incapacitated/prone/etc." and really mess up your day. \$\endgroup\$
    – Purplemur
    Nov 29 '21 at 23:37
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TLDR; I kinda screwed up. Yes, it is always true that AC points are less interesting if you're being hit with advantage.

I have posted an earlier answer that I want to correct. While the maths in it are correct, they are not actually taking into account what is most relevant for the game mechanics (I want to remind that I am not a dnd player). I will in this answer correct this with mostly the same maths, but a slightly different point of view on them which will alter the actual practical conclusion. Thanks to Matthieu M., Neil Slater, and Dulkan in the comments of the earlier answer for pointing out the part that I was missing about what actually matters in game

Odds of getting hit with and without advantage:

We will here use odds instead of probabilities to discuss chances of getting hit. That is to say, we will discuss "you have one chance in x to get hit". You could also think of it as "it will take your opponent x attempts to hit you". This x will be much more relevant to your game experience, and will show why AC is indeed less valuable if hit with advantage. From here on, I will refer to it as the required attempts to hit.

Here are the required attempts to hit a given AC with a +4 attack modifier, with and without adavantage:

enter image description here

See how the curves rise more and more sharply? That illustrates what commenters had been pointing out: the higher your AC, the more benefits you get from raising it.

We also see of course that it take significantly less attemtpts to get hit with advantage, especially for high ACs. This is what changes the conclusion.

What benefits do I get from raising my AC:

We now discuss how much gains you can expect from raising your AC by one more point. Of course, as we established above, that depends on your current AC: the higher it already is, the more benefits you will get out of one more point.

Here is a graph showing how many more attempts it will require for your enemies to hit you if you raise your AC by a single point, depending on your current AC:

enter image description here

Let me walk you through it: see for example that the value at 21 AC is almost 2 without advantage? That means that going from 21 to 22 AC will require your opponent to take 2 more attempts before they can actually hit you!

We notice two things: in both cases, as already discussed, you get much more benefits from investing in AC for higher AC values. And you can also simply see that the gains you get from investing in AC are always higher without advantage. So the value you get out of AC points is always reduced if you are being hit with advantage

N.B: In also checked the math in terms of relative gains but did not include it here so as to keep it rather short. It does not alter the conclusion. AC remains less attractive if hit with advantage

The math details:

The required attempts of being hit is simply the inverse of the probability of getting the hit. So without advantage, it is given by: $$P_{usual}=\frac{20}{21-AC+m}$$ with advantage (rolling the dice twice and keeping highest result, if i understood correctly): $$P_{adv}=\frac{400}{400-(AC-m-1)^2}$$

The gains can be obtained from derivation of those odds with respect to AC. The derivatives were calculated numerically by finite differences (this is more accurate than theoretically deriving it, since the AC values are discrete and can only increase by increments of 1)

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