High AC and encounters

jgsugden said:

Seriously. Chapter 3.

Click to expand...

I have my DMG thanks, I’ve read all five or six incarnations of it, including this one.

Dude if you want people to take you seriously, you’re gonna have to contribute a bit more meaningfully than this. I’m guessing it’s your way of saying you have nothing more to say. At least I’m grateful you’ve stopped telling people to go and play board games.

You suggested I read the DMG as you thought it supported your view. I pointed out the entirety of Chapter 3 pretty much opposes your view. Instead of addressing that, you raised another section. I asked how it might support the argument that you should build an encounter to counter the choices a player makes. You did not address that either. And now you say you know the DMG really well.

Seriously. Chapter 3.

Since this has devolved into a piss'n match over who plays D&D better, I thought I would post a more enjoyable form of unhelpful post.

Next time you find the PCs' high AC bumming you out, have an enemy magic user cast "Arlo's Un-Neutron Bomb"

With 150' radius, centered on the caster, structures melt, all weapons and armor disintegrate, "and there is nothing there but flowers growing, butterflies flying around....Even clothes disappear! And there's nothing but naked people everywhere! Now isn't that the kind of [combat] we want to have?

CITE: https://www.youtube.com/watch?v=onf3ZaU9PqU

iserith said:

That makes sense. I'd be interested to see the probabilities and whatnot, just for funsies, if anyone is inclined. I'm not sure how involved that would be to calculate.

Click to expand...

Let's assume, for the sake of argument, that someone would want to get Help on an attack only when they had a low chance of hitting, similar to our scenario. Let's further assume that "low chance of hitting" is 20% or needing a 17+.

In our scenario of 6 attacks needing a 17+ to hit means that, in a random trial of 100,000 attacks you'll hit this often:

0: 26,132
1: 39,250
2: 24,776
3: 8,181
4: 1,497
5: 156
6: 8

Average number of hits: 1.20

If you attack 3 times at advantage then you get the following distribution (using a different set of 100,000 attacks. It should probably be the same set but I don't want to redo things again and the difference is minimal)

0: 26,184
1: 44,158
2: 24,936
3: 4,722
4: 0
5: 0
6: 0

Average number of hits: 1.08

There are only six attacks so it probably wouldn't be too hard to mathematically figure out exactly what your chances of getting X number of hits would be. It's just that I already had a Monte Carlo simulation up in Excel so it was easier doing it the way I did.

EDIT: One commenter who mentioned it wasn't beneficial using the Help action made sure to add the caveat that the attacks needed to be identical in hit chance and damage in order to definitively state it wasn't useful to use Help. As you can see, in a low hit chance environment the Helped attacker only needs to do a little more than 10% more damage for it to be worth Helping.

For example, if you average 1.20 hits that do 1 point of damage each you'll do an average of 1.20 points of damage a round. The Helped attacker needs to do 1.20/1.08 or 1.11 points or more damage per attack to make Helping an equal or better option.

Oofta said:

Well, let's say you're in the (modern day) army. You see two enemies in front of you. One is wearing heavy body armor, crouching to make a smaller target, has a pistol but isn't shooting anyone. The other guy has no body armor, is standing out in the open and has an RPG pointed at you.

Which enemy do you target first? It shouldn't take a tactical genius to realize which one of the two is the greater imminent threat.

I don't go out of my way to reward or punish character builds, I do what makes sense for the monster. An ogre may walk up to the heavily armored guy and futilely bash away because they are incredibly stupid. Of course they may also decide to pick up the guy and see if they can breath under water when hitting isn't working because they're strong and tired of bashing at something that doesn't make pretty blood spurt everywhere.

I know DMs make concessions to convention and assumptions all the time, but at the same time IMHO it's more important to run monsters logically than to metagame.

Click to expand...

Yes... and what exactly was the point of your post. Everything you said is very obvious... it especially does not need any extra preparation from DM side.

guachi said:

Let's assume, for the sake of argument, that someone would want to get Help on an attack only when they had a low chance of hitting, similar to our scenario. Let's further assume that "low chance of hitting" is 20% or needing a 17+.

In our scenario of 6 attacks needing a 17+ to hit means that, in a random trial of 100,000 attacks you'll hit this often:

0: 26,132
1: 39,250
2: 24,776
3: 8,181
4: 1,497
5: 156
6: 8

Average number of hits: 1.20

If you attack 3 times at advantage then you get the following distribution (using a different set of 100,000 attacks. It should probably be the same set but I don't want to redo things again and the difference is minimal)

0: 26,184
1: 44,158
2: 24,936
3: 4,722
4: 0
5: 0
6: 0

Average number of hits: 1.08

There are only six attacks so it probably wouldn't be too hard to mathematically figure out exactly what your chances of getting X number of hits would be. It's just that I already had a Monte Carlo simulation up in Excel so it was easier doing it the way I did.

EDIT: One commenter who mentioned it wasn't beneficial using the Help action made sure to add the caveat that the attacks needed to be identical in hit chance and damage in order to definitively state it wasn't useful to use Help. As you can see, in a low hit chance environment the Helped attacker only needs to do a little more than 10% more damage for it to be worth Helping.

For example, if you average 1.20 hits that do 1 point of damage each you'll do an average of 1.20 points of damage a round. The Helped attacker needs to do 1.20/1.08 or 1.11 points or more damage per attack to make Helping an equal or better option.

Click to expand...

It is even more easy:

Just multiply hit chance with number of attacks and you are good.

At 20% hit chance, 6 attacks you have 6*0.2=1.2

Advantage raises hit chance to 0.36.
0.36*3 = 1.08.

I will point out a second time, this time with math, that using help to negate disadvantage is beneficial.

3 normal attacks:
3 * 0.2 = 0.6

6 attacks with disadvantage:
0.04*6 = 0.24

Really no need to use montecarlo methods here. It is just a binomial distribution (Bernoulli chain?) Since you repeat identical experiments with only 2 outcomes and a set probability.

Just for completeness sake: standard deviation for 3 attacks at 20% is squrt(3*0.2*0.8)

Aside from having a Monte Carlo simulation up and ready I also wanted to do it to see the distribution. Sometimes people find it easier to grasp probabilities using large, whole numbers instead of decimals, fractions, or percentages.

In this case, the math is actually fairly straightforward as you showed, but a distribution also visually shows the nonexistence of getting 4-6 hits and the sharply reduced number of 3 hits in a way that simple average won't. However, since we have it both ways I hope readers get a better understanding of probability.

To be even mathier the calculations for the possibility of achieving any number of hits are as follows. Find the number of possible combinations for each outcome. In our example it's 6!/(X!*(6-X)!). Where the 6 is the total number of attacks and the ! means factorial and that means to multiply 6*5*4*3*2*1 = 720 possible outcomes. The X is the number of hits you achieve. For example, for two hits the formula is (6*5*4*3*2*1)/((2*1)*(4*3*2*1)) = 15. So there are 15 possible ways to get four hits. Specifically, there are this many ways to get each of 0 to 6 hits.
0: 1
1: 6
2: 15
3: 20
4: 15
5: 6
6: 1

Then, calculate the probability of getting a certain number of hits. So zero hits is 0.8^6 and six hits is .2^6. Two hits is 0.2^2*0.8^4 and so forth. Then multiply this probability by the number of combinations. The results:

0: 26.214%
1: 39.322%
2: 24.576%
3: 8.192%
4: 1.536%
5: 0.154%
6: 0.006%

Which is, not surprisingly, quite similar to the results I got above.

If you don't care a lick about what the distribution looks like then UngeheuerLich showed you just how much easier it is to compute the average.

One positive I will say about Monte Carlo is that as the number of variables increases (like, if you wanted to simulate 20 attacks of various types each with different hit probabilities and damages) it doesn't get much harder to simulate the distribution. Just hit a button and let the computer do the math for you.

So [MENTION=6785802]guachi[/MENTION], assuming I'm not adding in CR0 zombie squirrels willy nilly to help my regular zombies, can you tell us if it's worthwhile for a zombie or two out of a pack of six to do a shove/knock prone? Or are there too many variables?

Not that my zombies need to be tactical geniuses, but occasionally the necromancer controlling them is. I guess it would have to be based on percentage chance to knock prone, same as your percent chance to hit.

Oofta said:

So [MENTION=6785802]guachi[/MENTION], assuming I'm not adding in CR0 zombie squirrels willy nilly to help my regular zombies, can you tell us if it's worthwhile for a zombie or two out of a pack of six to do a shove/knock prone? Or are there too many variables?

Click to expand...

Zombies should totally knock their victims prone then mob them and eat their intestines - like a butcher-shop scene from an old spaghetti-horror movie.

OK, maybe that should be ghouls.

But that's a 'yes' from me.

Oofta said:

So [MENTION=6785802]guachi[/MENTION], assuming I'm not adding in CR0 zombie squirrels willy nilly to help my regular zombies, can you tell us if it's worthwhile for a zombie or two out of a pack of six to do a shove/knock prone? Or are there too many variables?

Not that my zombies need to be tactical geniuses, but occasionally the necromancer controlling them is. I guess it would have to be based on percentage chance to knock prone, same as your percent chance to hit.

Click to expand...

Don't forget zombies are clumsy: "Oops, it stumbles and runs into you, possibly knocking you prone! So this is a Shove attack." I don't need no necromancer to justify why my zombies are tactical geniuses! (Or is that putting strategy over story? LOL)
[MENTION=6785802]guachi[/MENTION], [MENTION=59057]UngeheuerLich[/MENTION]: Thanks for doing the maths that I clearly cannot do. Another scenario for consideration. High-AC character. Heavy hitting monster with weaker monsters. Good for weaker monsters to do Help action? Please feel free to put forth whatever hard numbers you want there. My gut says that's a good move, but I trust maths more than my guts.