Finding the optimal amount to Power Attack

[Misirlou]

I'm a bored mathematician who isn't playing D&D right now because our DM is out of town. While watching TV, I came up with a formula for the optimal amount by which you should Power Attack in any situation.

It worked out surprisingly simply:

P = (A - T - D + 21)/2

A is your total attack roll modifier.
T is the target's AC
D is your average total damage per hit (not including any Power Attack), ignoring criticals and any overall multipliers to damage (like Spirited Charge).

EDIT: If using a two-handed weapon, divide D by 2 (oddly enough), and everything works out fine.

EDIT: For any iterative attacks, whether from Flurry of Blows or from full attacking, just use the average for the attack rolls.

 

[CRGreathouse]

Welcome to EN World -- or at least to posting! It's always good to see another mathematician on the boards.

Yep, that works pretty well -- and of all the formulas, it's really the only one simple enough to use at the table.

It doesn't take auto-hits and misses into account, though, nor does it work for two-handed weapons (which account for most Power Attacks, at least in 3.5). These limitations are hard to get past, as is the restriction to integers.

It also doesn't handle crits, but I think those shouldn't change the Power Attack number. Additionally, and more importantly, it assumes the goal is to maximize damage: with the possiblity of weak foes, large damage adds, and Cleave, that isn't always a good assumption.

Still, it's a good formula -- especially because of its elegance.

 

[Misirlou]

Thanks

You're right that it doesn't take auto-hits and misses into account. It actually doesn't cause any error, except in the extreme case where you would otherwise miss on a natural 20. I decided to ignore it for the sake of simplicity--most people will never even try to attack something with such a high AC.

Actually, I did take crits into account. Like you said, it cancels out. Like the auto-hit issue, it only causes an error in the formula in a very rare case (when you Power Attack so far down that you begin to miss within your threat range,like on a 19). But that's such an extreme amount that I surmised it would never near the optimal Power Attack anyway.

Similarly, the damage multiplication factor also cancels out. This was interesting, as one of the characters in my party is a Sprited Charging, lance-wielding Paladin.

Finally, most importantly, there is a very simple way to correct for two-handed weapons and the like. Just divide D by 2.
So if you're a Frenzied Berserker, Supreme Power Attacking with a two-handed weapon, divide by 4.

 

[Joshua Randall]

If you would like to delve into the more complex version of this formula, I suggest a visit to the following thread on the WotC D&D message boards:

http://boards1.wizards.com/showthread.php?s=&threadid=167952

 

[Misirlou]

Ah, pretty nice. He actually derived it exactly the same way, and got the same result I did, which is good.

I never bothered to include iterative attacks, which I guess I could do. I don't feel like reading his quite yet.

My intution says that iterative attacks will result in a small negative correction to the optimal amount. (Because the iterative attack is identical to the original attack, but at a lower attack roll, which disfavors Power Attacking. However, since it doesn't hit nearly as often as the first attack, it will only be small correction).

In fact, by the same logic, Flurries of Blows won't affect anything (the part of the flurry that's at std. BAB).

 

[Conaill]

What I would consider a more interesting question is... how would a character in-game know how much to Power Attack? You could use a simple rule such as "if you hit last time, increase PA by one, otherwise decrease PA by 2", but we can probably come up with something better than that.

A similar question comes up when you have lots of low CR opponents attacking a single higher level opponents: where lies the tradeoff between using Aid Another or attacking directly and hoping for a crit threat? Let's say 24 goblins ganging up against a single Ftr8 or so... how do these guys figure out how much Aid Another to use? They sure won't be doing any math calculations in the midst of combat...

 

[Particle_Man]

I'd say instinct. Just like a baseball player does not work out the complex physics involved in catching a baseball, but just does it. So warriors just know when to pull back a little, when to pull back a lot, and when to go all out, by pure fighting instinct, backed by experience.

 

[Li Shenron]

It worked out surprisingly simply:

P = (A - T - D + 21)/2

I never attempted before, but now I tried to see if I could demonstrate and it wasn't as difficult as I thought. But how did you instead get that criticals "cancel out"? I take it that it means they don't have an effect on the optimal P... but since -P applies to the confirmation roll, doesn't that affect the chance of critical?

 

[Plane Sailing]

P = (A - T - D + 21)/2

Nice formula, I can tell you are a mathematician. Us ordinary mortals would have written P = (21 + A - T -D)/2 to make calculations in our head easier

 

[Thanee]

Optimum power attack is always max...



Bye
Thanee