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Question: How does one convert disadvantage on the attack of a attacker, into a bonus to AC for a target

Assume a monster is attacking with disadvantage, the target is a level 12 PC with an AC of 16. If we need to take the attack bonus of the monsters into account; assume a monster needs to roll a 10 to hit.

Context

One of my players has a Cloak of Displacement, which grants, for every round, monsters have disadvantage attacks against the player until the first attack hits.

What follows from that power, is that I have to ask the player for every attack, if the monster has disadvantage or not. That is an extra effort I don't want to do; the information about whether or not he has a benefit is on his side of the table. (Note: as a DM I always ask "does AC X hit you?")

So, we want to change the Cloak; instead of granting the monster disadvantage on their attack, the Cloak offers him, for every round, a bonus to AC, until he is hit.

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    \$\begingroup\$ I don't understand why do you need to ask that question every single time, shouldn't you know if the PC has been hit or not earlier during the round? Or is this about you needing to remember the information? \$\endgroup\$ – kviiri Nov 5 at 9:50
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    \$\begingroup\$ I agree with the above comments; this seems like an X-Y problem. It might be worth instead asking the real question that you think this will solve, which appears to be something along the lines of "How can I, as the DM, more easily remember whether I need to roll with disadvantage against a player with a Cloak of Displacement?" (There's probably a better way to word that question...) \$\endgroup\$ – NathanS Nov 5 at 10:10
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    \$\begingroup\$ Shifting responsibility to the player is one option, and a perfectly valid one at that, but so is remembering it yourself. What you're implying is that your way is the only way, that it should always be the responsibility of the player solely, and that this magic item is effectively "broken" and you're trying to "fix" it. So what about a Barbarian's Reckless Attack then? Should that become a flat penalty to AC just because you don't want to have to remember? This is why I think this is an X-Y problem; you think you've solved the problem, but in doing so, you refuse any other solutions. \$\endgroup\$ – NathanS Nov 5 at 11:11
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    \$\begingroup\$ @NathanS Yes, I am refusing other solutions. Because it is not the core of my question, it is context WHY I am asking the question. \$\endgroup\$ – eirvandelden Nov 5 at 11:13
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    \$\begingroup\$ @eirvandelden I'm still not sure what makes the cloak special, and why is it important to you. What do you do if your players take the Dodge action, for instance? That, among others, is a situation where PC imposes a disadvantage independent of the monster. \$\endgroup\$ – kviiri Nov 5 at 11:23
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The bonus varies with the number of times you are attacked

(But a +2 cloak of protection should be of similar tier)

As discussed in Sdjz's answer, if you need to roll a 10 to hit normally, rolling with disadvantage is equivalent to a -5 penalty (or a +5 increase to the AC), although if the normal target number is different then this approximation gets worse. This is a useful result to remember as a rule of thumb.

However, the cloak of displacement is in no way equivalent to a +5 AC magic item. Among other things, the highest AC bonus from a single magic item is +3, which should tell you that granting a static +5 bonus to AC is a bad idea. The biggest difference is that the cloak of displacement only provides disadvantage until you get hit. This means the probabilities change depending on how many times in a round you are attacked.

Let us assume that without the cloak, an attacker needs to roll \$H\$ to hit you. The probability of the attacker hitting you with this unmodified attack is \$ p=(21-H)/20\$.

Now, let us assume that the attacker makes \$n\$ attacks in a round against you. The probability \$P^0_n(p,m)\$ of \$m\$ out of \$n\$ unmodified attacks hitting you is given simply by the binomial distribution,
\$ P_n^0(p,m) = \frac{n!}{m!(n-m)!}p^m(1-p)^m\$.

The mean/average number of attacks which hit you will be \$ \bar{P_n^0}(p) = np.\$ The median and mode are within 1 of the mean.

Now, let us consider the case where you do have the cloak of displacement. The probability of hitting someone with disadvantage is \$p^2\$ if the probability to hit without disadvantage is \$p\$. However, as soon as you are hit once, your attacker no longer has disadvantage. The probability of getting hit by \$m\$ out of \$n\$ attacks is given by the recurrence relations
\$P_0(p,0) = 1,\$
\$P_n(p,0) = (1-p^2)^n,\$
\$P_n(p,1) = p^2 P_{n-1}(p,0) + (1-p) P_{n-1}(p,1),\$
\$P_n(p,m) = p P_{n-1}(p,m-1) + (1-p) P_{n-1}(p,m)\$ for \$m>1\$.

The mean is given by the formula \$\bar{P_n}(p) = \sum_{m=0}^n m P_n(p,m) \$.

In order to convert the cloak of displacement into an effective AC, we can solve
\$\bar{P_n}(p) = \bar{P_n^0}(q)\$
for \$q\$, where \$q\$ is the probability to hit the effective AC. Taking \$H\$ as the original target (used in \$p\$) and \$B\$ as your effective AC bonus, \$q\$ is given by
\$q=(21-H-B)/20\$.

With a little algebra, the bonus \$B\$ is given by
\$B = 21 - H - \frac{20}{n} \bar{P_n}(p)\$.

Note that this is the effective bonus for the whole round. It assumes that you know a priori how many attacks are going to be made against you in the round.

This result would change depending on your values of \$ H\$ and \$n\$. We can start by looking at the \$H=10\$ case which you specify in the question. I have produced a spreadsheet which does this calculation. If you want to investigate the specific probability distributions you can look at that spreadsheet, although I shall note that the distributions for the cloak appear slightly broader than equivalent static AC distributions.

Let us look at the effective AC bonus with \$H=10\$ and see how it varies with the number of times you are attacked.

Plot of the effective AC bonus provided by a cloak of displacement given 10 as the target number, with respect to the number of attacks

Here we see an immediate trend. For a target roll of 10, if there is only one attack in the round then our effective bonus is +5. However, the effective bonus goes down the more times you are attacked in the round. If you face four attacks then it is equivalent to a +3 to AC. It goes down to +2 if you face eight attacks.

However, assuming that \$H=10\$ always is rather poor. Some enemies have lower to-hit bonuses. Some have higher to-hit bonuses. Sometimes you do something which changes your AC. I have the results for all possible target numbers below.

Table of the effective AC bonus of the cloak of displacement for all target numbers and numbers of attacks.

This table is the full answer to the mathematical part of your question.

We see that an effective bonus of +5 is the highest. The effective bonus drops as the target number deviates away from a middle target number, although the middle of the distribution shifts to higher target numbers as you face more attacks.

If you have a high base AC, the benefits of the cloak of displacement are more likely to last through more attacks because you are less likely to get hit. On the flip side, if you have a low base AC, the cloak of displacement is less helpful since you are more likely to get hit and have its benefits go away. However, for much of the parameter space (especially target numbers near the middle, since bounded accuracy makes them more likely), the effective bonus is at least +2, and often greater than +3, attesting to the effectiveness of the cloak of displacement, at least if you aren't facing too many attacks per round.

If you know in advance how many attacks you will be facing per round, and what the target roll for each of those attacks are, and if you assume that the target roll will be the same for each of them, then you can use the above table to estimate your bonus to hit.

However, by now you should have observed that the magnitude of the bonus provided by the cloak is situational, even in this highly idealised scenario. There are too many variables which can change too frequently to make a static bonus properly represent its behaviour.


I note that even this fairly hefty analysis is a gross over-simplification of the cloak of displacement. I will describe the short-comings of the analysis below to attempt to further convince you that setting a fixed AC bonus for the cloak is a bad idea.

  • Often when facing multiple attacks some of those attacks will be made with different bonuses. This can be represented in the above analysis by changing \$p\$ at each \$n\$ in the recursion relations. The probability for the unmodified attacks will also have to be recalculated since the regular binomial distribution would no longer apply.

  • Disadvantage does not stack with disadvantage, while static bonuses do stack with disadvantage. This means if you impose disadvantage on attacks by some other means, such as by taking the Dodge action, then the static AC item is superior to the RAW cloak.

  • Disadvantage is cancelled by advantage, while static bonuses are not. This means an enemy with advantage would face very different statistics depending on whether you had disadvantage or a static AC bonus.

  • Critical hits are far less likely with disadvantage, and sneak attacks are not allowed with disadvantage. This can be an important consideration.

  • Things which deal damage via means other than attacks are unaffected by the cloak and would normally disable it. If the cloak provides a constant AC bonus, then this disabling probably would not be captured.

  • The cloak is also disabled by things which stop you from moving. If you've made this a static bonus, you might forget to turn off the cloak under those circumstances.

Overall, the rules for the cloak of displacement make attempting to replace its effects with a static AC bonus a futile endeavour. If you really cannot deal with the complexity of the cloak of displacement as-written, then replace it with a cloak of protection. Since a cloak of protection is only an uncommon item while the cloak of displacement is a rare item, increasing the AC bonus from +1 to +2 would be acceptable and in line with the numbers above. But if you do so, do not call it a cloak of displacement, because it would bear no resemblance to that magic item.

A +2 to AC can be argued two ways. One is from magic item rarity: an uncommon cloak gives +1, so a rare cloak could give +2. A +3 is out of the question, since only legendary items give that bonus to AC.

The other is from the numbers in the analysis. The calculated effective AC tends to hover around +3 (with a very wide spread), but that calculation over-estimates the cloak's effectiveness in real combat, so we could reduce the bonus by one to heuristically account for that.

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  • \$\begingroup\$ thank you very much for your very complete write-up. This is exactly the kind of answer I was looking for! \$\endgroup\$ – eirvandelden Nov 8 at 10:15
  • \$\begingroup\$ Bravo for taking the time to do an in depth analysis. (Interestingly, there is no Cloak of Protection +2 in the game, though good point on the legendary, etc) \$\endgroup\$ – KorvinStarmast Nov 9 at 16:27
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How to make Cloak of Displacement less fiddly in play

  1. Conversion to AC deprives the player if a significant benefit: the reduction in chance to being hit with a critial hit as noted in @Sdjz's answer. The cloak makes it ~1/400 chance. No cloak is a ~1/20 chance.

  2. You can place the burden on the player to indicate "cloak on" or "cloak off" using a chip or piece of colored papery under their figure's base. (If doing on-line play on a virtual table top (see later), apply a status color/symbol that is "on" or "off".

  3. Roll two d20's as DM whenever attacking that player, preferably two dice of different colors. Call which color is high before the roll.

Why we used 'on-off condition' indications for the cloak

There are multiple reasons for the cloak to turn "off"...

... causing any creature to have disadvantage on attack rolls against you. If you take damage, the property ceases to function until the start of your next turn. This property is suppressed while you are incapacitated, restrained, or otherwise unable to move.

... and you are right: the DM has a lot to keep track of. The player needs to be tasked with indicating "cloak on" or "cloak off" using a visual cue. <== That's important: the visual cue gives you, the DM, what you need to see to know if it's a one or two d20 roll.

What we did: Player Flags the Token/Figure

My Champion switched from a cloak of protection (the Indomitable feature gives second chances for a saving throws) to a cloak of displacement: I was a front liner, I got hit a lot, and I got critical hit - it seemed to happen a lot. Our wizard appreciated the Cloak of Protection for save boosts and AC boost.

I took ownership of this as the player. We were playing on roll20, but in table top games for years we have used a variety of cards, chips, and colored paper to indicate things like hasted characters, slowed characters, invisible characters ... all visual cues to a status of the Character.

I turned the cloak on or off based on what happened.

  1. Cloak starts combat On. (it was a blue dot on roll20).

  2. Attack misses? No change. Attack hits? I turn off blue dot.

  3. My turn starts? I turn on blue dot.

  4. AoE effect, white dragon breathes on me? Blue dot off.

    Rinse and repeat. All the DM had to do was roll d20 or 2d20.

Get your player to take ownership of this.

Once you have this discussion, the tools themselves are simple as can be. The "on / off" flag can be as simple as a pair of poker chips.
White on top of red? Cloak on.
Red on top of white? Cloak off.

If your player won't take ownership of this, you have some things to discuss before the next gaming session.

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    \$\begingroup\$ I appreciate this answer; It's very complete and gives a helpful solution, with an example on how to hand ownership to the player, one of the essences of my question. An this answer does so without turning it into a "This is YOUR problem" answer. I feel obliged to tell you why I won't be accepting this as the correct answer: Because it does not answer how to convert a disadvantage into a static bonus. But again, this was a helpful answer. \$\endgroup\$ – eirvandelden Nov 5 at 14:34
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    \$\begingroup\$ @eirvandelden No worries, if it helps someone else, then your question and this answer does what they are supposed to do. :-) I also like Sdjz's answer in terms of how it illustrates the effect on AC. \$\endgroup\$ – KorvinStarmast Nov 5 at 14:39
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    \$\begingroup\$ @eirvandelden: I feel like the part about converting (dis)advantage to a static bonus is already pretty well answered e.g. here. This answer does a pretty good job of addressing the parts of your question that have not already been asked here. \$\endgroup\$ – Ilmari Karonen Nov 5 at 14:51
  • \$\begingroup\$ @IlmariKaronen It was never clearly stated in that question (in my reading it) whether the same is 1-on-1 applicable for disadvantage, which is why I made a new question. I'd be okay with marking my question as a duplicate if that was clear from the answer you mentioned. \$\endgroup\$ – eirvandelden Nov 5 at 15:12
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Assuming a monster would need to roll a 10 to hit, +5 AC is a close approximation

This can be seen by using this anydice program:

output 1d20 named "normal"
output [lowest 1 of 2d20] named "disadvantage"

When comparing the outputs using the "at least" option, you can see that when having disadvantage, the odds of getting at least a 10 (so a hit) is 30,25%. This is fairly close to the 30% odds of rolling at least a 15 in a single roll, so adjusting the AC by 5 for this particular situation is a decent approximation.

However, do note that doing this approximation also changes a number of other things. First, monsters that would not need a 10 to hit may find this approximation to be worse the further you get from a 10 to hit, as can be seen in this other anydice program:

output 1d20-5 named "+5 AC approximation"
output [lowest 1 of 2d20] named "disadvantage"

Which when seen in graphical form for "at least" displays:

enter image description here

You can see that while the approximation is fairly good for needing to roll a 10, 11 or 12 to hit originally, the further you get from these values, the worse it gets.

Caveat: the odds of getting a critical hit change if you do this

Where having disadvantage means you usually have a 0.25% chance of critting, simply adding +5 AC keeps the 5% odds of rolling a critical hit.

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  • \$\begingroup\$ I threw in that edit because I ran into this very feature in a Tier 3 game. I had a cloak of protection, but got hit a lot by giants. The crits hurt real hard. We found a cloak of displacement and I swapped in ... as the front liner, that made a siginficant difference. YMMV \$\endgroup\$ – KorvinStarmast Nov 5 at 14:06
  • \$\begingroup\$ The caveat about disadvantage is a very valid point, as is the effect on monster with a lower or higher attack bonus. \$\endgroup\$ – eirvandelden Nov 5 at 14:44
  • \$\begingroup\$ A flat +5 is extremely risky. If they need to roll 15, you're essentially going from "less likely to get hit" to "can only get hit by crits". This will make the game extremely swingy, because every time they do get hit, they get hit for a lot, but they'll only get hit 5% of the time. \$\endgroup\$ – Theik Nov 5 at 14:48
  • \$\begingroup\$ @Theik Indeed, the further away you get from the desired approximation of needing a 10 to hit, the worse it gets. At the extremes in particular as you point out it gets really bad. In the case of needing a 15 to hit, it goes from a 9% chance to hit with disadvantage to a 5% with the +5 AC as you note. So yes, I'd agree that this is very limited. Like stated in the answer, the approximation is only actually decent when monsters needed a 10, 11 or 12 to hit in the first place \$\endgroup\$ – Sdjz Nov 5 at 14:53
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I believe using a token of sorts to symbolize if the player has the effect would be optimal.

That being said if you absolutely want to change that you could do something similar to the advantage rules on passive checks (PHB, p. 177):

If you have advantage, +5 to the roll. If you have disadvantage, -5 to the roll. If it seems powerful that's because it is, as can be advantage. The major difference is that you're taking out the luck and giving an "average".

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As you go on there will be more and more PC options that grant advantage or impose disadvantage. Always roll 2d20, and ask if you need to apply advantage or disadvantage before you calculate the monster's total.

You will need a system to distinguish which die is used in the "ordinary" case. The easiest is if you have different coloured dice, but even "the leftmost as they fall" is sufficient.

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    \$\begingroup\$ "Always roll 2d20" is a decent solution for IRL games, but it doesn't work online with a dice roller, since you need to choose either "d20" or "2d20kl1" when you type in what to roll. So you do need to know in advance whether the attack has disadvantage. \$\endgroup\$ – Ryan C. Thompson Nov 5 at 11:34
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    \$\begingroup\$ It's worth noting for online play that Roll20's default setting for character sheets is apparently to "always roll advantage" - that is, roll 2d20 and leave it to users to determine when to use the first die and when advantage/disadvantage is actually relevant (basically, your exact solution). \$\endgroup\$ – V2Blast Nov 5 at 11:35
  • \$\begingroup\$ @RyanThompson copy both commands into a text file, and then use the one you need ... all you do then is paste and enter. (that's what I do on discord ...) \$\endgroup\$ – KorvinStarmast Nov 5 at 14:03
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    \$\begingroup\$ @RyanThompson rather than "2d20kl1 + n", you type "d20 + n" "d20 + n" and ignore the higher/lower/second as appropriate \$\endgroup\$ – Caleth Nov 5 at 14:06

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