# Forum > Gaming > Roleplaying Games > D&D 5e/Next >  Bracer of Defense vs Cloak of Displacement

## Skrum

Trying to figure out which of these two are the better defensive item for a spell caster. Base AC of 15 (mage armor + 14 Dex) and a shield spell. Obviously the cloak is far better against one-off attacks, but what is the caster gets focus-fired. 

A CR 8 monster has a +7 chance to hit and deals 53 DPR. If that is broken up over 3 attacks, and there are 2 of them, that adds up to 6 attacks dealing 17.66 damage each with a +7 attack roll. 

With a cloak they'll have 20 AC and the disadvantage thing. 

Bracer instead, and they have 22 but no other defense. 

Expected damage with the bracer is 31.8 damage over the 6 attacks. 

With the cloak -
If the cloak applies to one hit but not the others, expected damage is 38.12

2 hits, it's 33.9
3 hits, it's 29.66

So, if the cloak can get you to 3 misses, it's superior to the bracers. 

With a 16% chance to hit, there's a 48% chance one of the first 3 attacks will be a hit. 

So.... The cloak is better 52% of the time. Basically the same. 

Of course, this is a pretty extreme 6 attacks. A more reasonable 3 attacks, and the cloak is better 68% of the time.

Is my math correct? Has anyone done a more detailed analysis of this?

----------


## Segev

It's worth noting that the cloak also dramatically decreases the chances of a critical hit. The natural 20 has to happen on two dice simultaneously. So, with an AC of 20 and _shield_, I think you may find yourself being frustrated by the fact that _shield_ does nothing if they roll that natural 20 more often than you will be hit by a final attack roll of 21/22 or 26/27. Or is the "20 AC" accounting for the _shield_ spell?

If so, getting base AC up to 17 or so would actually not be a bad move. There are other ways to force disadvantage, including your own spells (e.g. _blur_) or using other means of obscuring your presence/location, so I think the Bracers are a better bet in that case.

----------


## Skrum

> It's worth noting that the cloak also dramatically decreases the chances of a critical hit. The natural 20 has to happen on two dice simultaneously. So, with an AC of 20 and _shield_, I think you may find yourself being frustrated by the fact that _shield_ does nothing if they roll that natural 20 more often than you will be hit by a final attack roll of 21/22 or 26/27. Or is the "20 AC" accounting for the _shield_ spell?
> 
> If so, getting base AC up to 17 or so would actually not be a bad move. There are other ways to force disadvantage, including your own spells (e.g. _blur_) or using other means of obscuring your presence/location, so I think the Bracers are a better bet in that case.


Yeah I was counting that they'd cast shield. Mage armor + 14 Dex + shield spell. And then the respective items. 

Good point about the crits though

----------


## Frogreaver

> With a 16% chance to hit, there's a 48% chance one of the first 3 attacks will be a hit.


This isn't correct.  The only way to get a hit in the first 3 attacks is if all 3 attacks don't miss.  The chance all 3 attacks miss is (84%)^3.  Thus the chance at least 1 attack is 1 - (84%)^3 = 40.73%.




> Is my math correct? Has anyone done a more detailed analysis of this?


I would approach it this way.

With the Displacement Cloak
If you are hit on attack 1 then you will take 1 + .3*5 = 2.5 DPA.  The chance of this occurring is 16%
If you are hit on attack 2 then you will take 1 + .3*4 = 2.2 DPA.  The chance of this occurring is (84%) * 16%
If you are hit on attack 3 then you will take 1 + .3*3 = 1.9 DPA.  The chance of this occurring is (84%)^2 * 16%
If you are hit on attack 4 then you will take 1 + .3*2 = 1.6 DPA.  The chance of this occurring is (84%)^3 * 16%
If you are hit on attack 5 then you will take 1 + .3*1 = 1.3 DPA.  The chance of this occurring is (84%)^4 * 16%
If you are hit on attack 6 then you will take 1 + .3*0 = 1 DPA.  The chance of this occurring is (84%)^5 * 16%
If you are not hit with any attacks then you will take 0 DPA.  The chance of this occurring is (84%)^6

On average this will total to you taking 0.98*DPA.  Lower than the 1.2*DPA you would take when using the Bracers.

*DPA=Damage per attack

----------


## Skrum

> This isn't correct.  The only way to get a hit in the first 3 attacks is if all 3 attacks don't miss.  The chance all 3 attacks miss is (84%)^3.  Thus the chance at least 1 attack is 1 - (84%)^3 = 40.73%.


Right, thanks.





> I would approach it this way.
> 
> With the Displacement Cloak
> If you are hit on attack 1 then you will take 1 + .3*5 = 2.5 DPA.  The chance of this occurring is 16%
> If you are hit on attack 2 then you will take 1 + .3*4 = 2.2 DPA.  The chance of this occurring is (84%) * 16%
> If you are hit on attack 3 then you will take 1 + .3*3 = 1.9 DPA.  The chance of this occurring is (84%)^2 * 16%
> If you are hit on attack 4 then you will take 1 + .3*2 = 1.6 DPA.  The chance of this occurring is (84%)^3 * 16%
> If you are hit on attack 5 then you will take 1 + .3*1 = 1.3 DPA.  The chance of this occurring is (84%)^4 * 16%
> If you are hit on attack 6 then you will take 1 + .3*0 = 1 DPA.  The chance of this occurring is (84%)^5 * 16%
> ...


Sorry, could you give a little more explanation for what's happening here? Ideally, I'd like to explain this to someone else :)

It looks like....
I think the .3 should be .4? Since that's what your AC would revert to, once you are hit? Am I reading this correctly?

----------


## Frogreaver

> It looks like....
> I think the .3 should be .4? Since that's what your AC would revert to, once you are hit? Am I reading this correctly?


This is correct.  For whatever reason I was using .3 for the cloak base chance to hit and .2 for the bracers.  The percent I used for advantage was .16 corresponding to the .4 chance to hit.  So there's that.  But the fix is as you said, replace the .3 with a .4

Totals up to 1.43*DPA vs 1.8*DPA for the Bracers (i had 1.2 before but that was based on .2 chance to be hit with bracers when it should have been .3).




> Sorry, could you give a little more explanation for what's happening here? Ideally, I'd like to explain this to someone else :)


*1a.*  Given the first attack hits what is the expected damage?  (1 + .4 + .4 + .4 + .4 + .4)= (1 + .4*5) = 3*DPA
*1b.*  What chance is there the first attack hits and anything else happens?  (.16 * 1) = .16

*2a.*  Given the first attack misses and 2nd attack hits what is the expected damage?  (0 + 1 + .4 + .4 + .4 + .4) = (1 + .4*4) = 2.6*DPA
*2b.*  What chance is there first attack misses, the 2nd attack hits and then anything else happens?  (.84 * .16 * 1) = (.84) * (.16)

Repeat for 3rd to 6th attack hits while all the ones before miss.  Then do a weighted average.

Hopefully that helps?

*Note the probabilities all add up to 1 if you include the chance every attack misses as a possibility.  (always a good double check on probability work).

----------


## Skrum

> This is correct.  For whatever reason I was using .3 for the cloak base chance to hit and .2 for the bracers.  The percent I used for advantage was .16 corresponding to the .4 chance to hit.  So there's that.  But the fix is as you said, replace the .3 with a .4
> 
> Totals up to 1.43*DPA vs 1.8*DPA for the Bracers (i had 1.2 before but that was based on .2 chance to be hit with bracers when it should have been .3).
> 
> 
> 
> *1a.*  Given the first attack hits what is the expected damage?  (1 + .4 + .4 + .4 + .4 + .4)= (1 + .4*5) = 3*DPA
> *1b.*  What chance is there the first attack hits and anything else happens?  (.16 * 1) = .16
> 
> ...


Yes this is excellent, thank you! One more question though 




> What chance is there first attack misses, the 2nd attack hits and then anything else happens?  (.84 * .16 * 1) = (.84) * (.16)


How do I do this part (the odds there will be a hit in X number of attack). Or, more like, I know it's a binomial distribution, but how do I explain it to someone that doesn't know math, without hitting them with a wall of text. Lol.

Edit: would it be 
1st attack: 84% chance to miss (and 16% chance to hit) 
2nd attack: 84% chance to miss first attack, 16% to land 2nd attack, so .84 x .16 = 13.44%
3rd attack: 84% chance to miss first attack, 84% chance to miss second attack, 16% chance to land 3rd attack, so .84 x .84 x .16 = 11.3%
4th attack: (multiply by an additional .84) for 9.5%
etc.

----------


## Frogreaver

> Yes this is excellent, thank you! One more question though 
> 
> 
> 
> How do I do this part (the odds there will be a hit in X number of attack). Or, more like, I know it's a binomial distribution, but how do I explain it to someone that doesn't know math, without hitting them with a wall of text. Lol.


It's not a binomial distribution.  There are no coefficients and I never increase the power of the chance to hit term  :Small Smile: 

If you want to know the probability that X attacks miss (all with the same probability and then the xth attack hits it's literally just (Chance to miss)^(x-1) * (chance to hit)

If they all had different then it would just be m1*m2*m3*m4*m5*...*hX.  where m is for a chance to miss and h is for a chance to hit.

If that's too complicated I don't know that you can explain it.




> Edit: would it be 
> 1st attack: 84% chance to miss (and 16% chance to hit) 
> 2nd attack: 84% chance to miss first attack, 16% to land 2nd attack, so .84 x .16 = 13.44%
> 3rd attack: 84% chance to miss first attack, 84% chance to miss second attack, 16% chance to land 3rd attack, so .84 x .84 x .16 = 11.3%
> 4th attack: (multiply by an additional .84) for 9.5%
> etc.


Yes, that's it!

----------


## Corran

> but how do I explain it to someone that doesn't know math, without hitting them with a wall of text.


Explain the weigthed average with a simple example, such as using a d6.

3.5 = 1*1/6 + 2*1/6 + 3* 1/6 + 4*1/6 + 5*1/6 + 6 *1/6

Essentially,
X = A*P1 + B*P2 + C*P3 + D*P4 + E*P5 + F*P6,
where P1=P2=P3=P4=P5=P6=1/6 is the probability that you get any one of {A},{B},{C},{D},{E},{F},
where A,B,C,D,RE,F are 1,2,3,4,5,6 respectively, ie the possible outcomes of the die roll.

Replace P1, P2, etc and A,B,C, etc with the values Frogreaver gave you.

A,B,C,D,E,F, are easy enough to explain, as each of them is a sum of 5 numbers, picked from {0, 0.4, 1}, where 0 represents the enemy missing, 1 represents the enemy hitting, and 0.4 represents not predetermining it, thus making it equal to the chance that the enemy has to hit you without sufferering disadvantage.
So, if for example you have A = 0 + 1 + 0.4 + 0.4 + 0.4, that means that the enemy missed you at first turn, hit you at second turn, and anything goes for the rest 3 turns. So on average in that scenario you are taking damage equivalent to that of 2.2 successful attacks from the enemy. So if the enemy's average dpr is 10 points of damage, that means that on average you are taking 2.2*10 points of damage on average under this scenario. Note that A,B,C,etc must represent exhaustively the number of possible scenarios for these numbers to make sense. You can merge the last two scenarios (ie being hit on the final attack and being missed with all attacks into one, just to keep up with the d6 example for convenience when explaining).

P1,P2, etc are the probabilities mapping to each one of the above scenario.
So, IF P1 corresponds to A, and A = 1+0.4*5 (ie being hit in the first attack and after that anything goes), then P1 = 16%. Why? Because 16% is the probability to be hit on the first attack if the enemy has an inital 40% hit chance which is being reduced to 16% when applying disadvantage.
If P2 corresponds to B, and B = 0 + 1 + 0.4*3, well, then first explain what is B. B translates to being missed in the first attack and being hit with the second attack (after that anything goes so your probability is not affected). Being missed on the first attack happens 84% of the time. Being hit on the second attack when being attacked with disadvantage happens 16% of the time. For both of these to happen, the probability is the product of their respective probabiltiies, since they are independent events (ie one does not affect the other; remember we are talking about being missed and being attacked with disadvantage afterwards, that's why they are independent). So the probability of B happening is P2=84%*16%. Etc.

It's worth noting that the cloak of displacement loses steam the more attacks come your way, but more importantly, the higher the base enemy hit chance becomes. The value of bracers of defense is theoritically fixed (think of it like a straight line), but practically they become worse the higher the average unweighted by hit chance damage of your enemy is (or the lower your current hp are). In situations where you can be taken down fast, you want to take the gamble, and since chances are that in such a situation not too many attacks will be coming your way anyway, the cloak wont be a gamble anyway.

----------

