richard develyn |

I've never seen this very satisfactorily explained anywhere, but from my maths, and without taking into account natural 1s and 20s, which I'm hoping will even out, you should power attack when:

number-to-hit < 21 - ( average-damage / power-attack-extra-damage )

So, for example, a 1st level barbarian with a two handed sword, +6 to hit and 2d6+6 damage: average damage is 13, power-attack-extra-damage is 3 (two handed), so

number-to-hit < 21 - (4 1/3)

number-to-hit < 16 2/3

number-to-hit <= 16

which basically means anything of AC 22 or less.

Which is going to be most things for a 1st level Barbarian

A Frost Giant, however, with a greataxe, +18/+13 for 3d6+13 damage: average damage 31, extra is 3 again so:

number-to-hit < 21 - (10 1/3)

number-to-hit < 10 2/3

number-to-hit <= 10

which basically means anything of AC 28 / 23 or less (go for the average of 25).

Which for a CR 9 monster is going to be pretty rare.

Which makes sense, as anything which does a lot of damage is going to get a very marginal gain from power attack in exchange for losing the chance to do the lots of damage it's already doing.

Most of the time, giants shouldn't power attack.

Richard

SorrySleeping |

I'm not sure how you got this math, and why the Frost Giant is only getting 3 extra damage from power attack and not 9, but I have to point this out.

The Frost Giant has an AC of 21 and a to-hit of 18 before power attack. He needs to roll a 3 to hit himself. CR 9 monsters have an average AC of 22, meaning a 4 to hit.

The Frost Giant would be stupid to not raise his "minimum needed" to hit to 7 for average CR 9 monsters to get that extra power attack damage, as he adds 70% of his static damage. Thats a 40% gain on average damage for 15% less chance to hit.

EDIT: DPR Math for Frost Giant vs CR 9/22 AC

Standard Attack - 18.8

Standard Attack Power Attack - 21.1

Full Attack - 31.7

Full Attack Power Attack - 34.1

MrCharisma |

I'm assuming this is for PF1, in which case...

If you're a full BAB two-handed weapon using character, the answer is pretty much always.

This is pretty much right.

The only thing I'll add is that a level 1 barbarian with a greatsword is probably going to 1-shot most enemies without power attack, so all it's really doing is lowering your to-hit. You can wait till level 2 or 3 to start using it unless you come up against any big beasties.

richard develyn |

I know it feels counter-intuitive but the more damage you do the less you should power attack.

Consider an extreme example: a mega-titan that does 1000 points of damage on average.

Why would such a thing ever power attack?

I'll do a bit more maths if you like, let's assume that it hits on an 11, and ignore criticals. Assume we have a complete average set of 20 rounds in which every number of the dice is rolled once.

Without Power Attack, total damage is 10 hits times 1000 = 10,000

Let's do Power Attack -1 and +3. Now it only hits 9 times but doing 1003 hit points, for a total of 9027. Much less.

Power Attack -2 and +6, and we have 8 hits doing 1006 damage, or 8048.

And so on.

Richard

blahpers |

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I know it feels counter-intuitive but the more damage you do the less you should power attack.

Consider an extreme example: a mega-titan that does 1000 points of damage on average.

Why would such a thing ever power attack?

I'll do a bit more maths if you like, let's assume that it hits on an 11, and ignore criticals. Assume we have a complete average set of 20 rounds in which every number of the dice is rolled once.

Without Power Attack, total damage is 10 hits times 1000 = 10,000

Let's do Power Attack -1 and +3. Now it only hits 9 times but doing 1003 hit points, for a total of 9027. Much less.

Power Attack -2 and +6, and we have 8 hits doing 1006 damage, or 8048.

And so on.

Richard

With respect, your example is useless, as it doesn't exist in practice. Pathfinder mechanics don't really scale well past a certain point. Exactly where that point is is subject to debate, but when you're doing 1,000 points of damage per hit, you're waaaaaaaay past it. That's not even getting into the probability of said mega-titan having proportionally higher Hit Dice (so higher BAB, higher bonus damage from PA, higher penalty to hit from PA, etc.) because all of those would be subject to the same "Pathfinder scales badly" problem.

richard develyn |

With respect, your example is useless, as it doesn't exist in practice. Pathfinder mechanics don't really scale well past a certain point.

The decision to power attack as a means of achieving optimal damage is a mathematical one, not a game-rule one.

There's a straight formula at play here, and it's as relevant to creatures doing 1000 points of damage as it is to creatures doing 10.

But, you know, if you don't believe me, that's fine :-)

Richard

Gray Warden |

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DPA = max(0.05, min(0.95, (21 + atk - AC)/20)) * (1 + thr * (crit - 1)/20) * dmg

DPA = damage per attack

atk = total attack bonus

AC = target armor class

thr = critical threat range (1 for crit on 20, 2 for crit on 19-20, etc.)

crit = critical modifier

dmg = average damage per hit

Calculate the DPA with and without Power Attack for a reasonably wide range of ACs and plot the two curves. When DPA with PA becomes lower than the one without, then it's more advantageous not to use PA. And if you run this test on a variety of situations that are **likely** to happen in actual games, you'll understand why "ALWAYS POWER ATTACK" is a pretty good approximation of an answer.

What you are doing with your mega-titan example is treating atk and dmg as two independent variables while, guess what, they are not independent at all, as they are both related, to some extent, to the creature CR. Therefore, you can't put a prior on one of the two without looking at the implications it has on the other.

**THIS** is how you do the math, not with random examples and anecdotes that never apply. Thread closed.

MrCharisma |

blahpers wrote:With respect, your example is useless, as it doesn't exist in practice. Pathfinder mechanics don't really scale well past a certain point.The decision to power attack as a means of achieving optimal damage is a mathematical one, not a game-rule one.

There's a straight formula at play here, and it's as relevant to creatures doing 1000 points of damage as it is to creatures doing 10.

But, you know, if you don't believe me, that's fine :-)

Richard

While that's true it's probably not a particularly useful example. It might help someone heavily theory-crafting, but someone who's asking: "When to Power Attack" is probably not up to that level.

For most people who aren't doing a Goliath-Druid Vital-Strike character it's basically a case of "Always Power Attack" (and for the Goliath, grab Furious-Focus and "Always Power Attack").

To the OP, people have done the math, and for most 2-handed builds "Always Power Attack" really is the best advice.

Slim Jim |

Are you under 3rd level? ...you don't need Power Attack yet.

Do you rage? ...well consider doing that w/Extra Rage before Power Attack.

Do you use a polearm? ...Combat Reflexes is more important than Power Attack.

4th-level and playing "up" in PFS? ...now's an excellent time to start Power Attacking.

Do you *need* Furious Focus to make Power Attack viable? ...take two other feats.

SorrySleeping |

Towards higher levels, monsters become so easy to hit you should always be power attacking.

A 20th level Barbarian with starting 18 Str ends with 37 Str (5 from level up, 6 from belt, 8 from rage) and a +5 weapon. That Barbarian has a +38 to hit and can take a -6 from power attack. If he faces a CR 20 monster, he barbarian can roll a -2 and hit.

Also again, why are you only using the -1/+3 power attack? Higher level creatures have to take the bigger penalty.

Talonflash |

blahpers wrote:With respect, your example is useless, as it doesn't exist in practice. Pathfinder mechanics don't really scale well past a certain point.The decision to power attack as a means of achieving optimal damage is a mathematical one, not a game-rule one.

There's a straight formula at play here, and it's as relevant to creatures doing 1000 points of damage as it is to creatures doing 10.

But, you know, if you don't believe me, that's fine :-)

Richard

Doing your example from the other end:

10 hits @ 10 dam/hit = 1009 hits @ 13 = 117

8 @ 16 = 128

You may want to think your examples through a little further.

This is really just the "how much is +x to hit worth vs +y to damage" question, and it all depends on how well you already hit AND what your damage/hit already is due to the multiplication of the two.

The answer essentially comes down to (current per hit damage)/20 * (change in AB), i.e. what you're gaining or losing by changing AB, compared against the change in per hit damage.

Differential calculus has its uses. :-)

richard develyn |

Turns out my formula was slightly wrong, it should be:

number-to-hit < 21 - ( average-damage / power-attack-extra-damage ) - power-attack-bab-multiplier

where the latter starts at 1, goes to 2 on bab 4, etc

Like I said, this doesn't take into account the 1s and 20s (I'll leave a better mathematician for that one).

Here's my workings:

DPA = max(0.05, min(0.95, (21 + atk - AC)/20)) * (1 + thr * (crit - 1)/20) * dmg

Remove the 1s and 20s bit, and let Z = (1 + thr * (crit - 1)/20)

DPA = ( (21 + atk - AC) / 20 ) * Z * dmg

Factor out the / 20:

DPA = (21 + atk - AC) * dmg * ( Z / 20 )

Power Attack has two components, a bab based one (1, 2 when BAB = 4, and so on), which we'll call B, and a component based on whether you are using a one or two handed weapon, i.e. either 2 or 3, which we'll call T.

So, when you power attack, your atk goes down by B and your damage goes up by BT, giving us:

DPA(pow) = (21 + atk - B - AC) * ( dmg + BT ) * ( Z / 20 )

We should power attack when this is better than not doing so, i.e. Power Attack when:

DPA(pow) > DPA, i.e.

(21 + atk - B - AC) * ( dmg + BT ) * ( Z / 20 ) > (21 + atk - AC) * dmg * ( Z / 20 )

Cancel the Z / 20 on both sides:

(21 + atk - B - AC) * ( dmg + BT ) > (21 + atk - AC) * dmg

Expand the left hand side a bit:

(21 + atk - AC) * ( dmg + BT ) - B * ( dmg + BT ) > (21 + atk - AC) * dmg

And a bit more:

(21 + atk - AC) * dmg + (21 + atk - AC) * BT - B * ( dmg + BT ) > (21 + atk - AC) * dmg

Subtract (21 + atk - AC ) * dmg from both sides:

(21 + atk - AC) * BT - B * ( dmg + BT ) > 0

Move the negative bit across to the other side:

(21 + atk - AC) * BT > B * ( dmg + BT )

Cancel out one B and divide by T

(21 + atk - AC) > ( dmg + BT ) / T

Divide the T on the right hand side into the brackets:

(21 + atk - AC) > ( dmg / T ) + B

Subtract each side from 21 (which reverses the inequality)

AC - atk < 21 - ( dmg / T ) - B

Which is basically my (now slightly corrected) formula.

Richard