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MATHS! [Archive] - Wizards Community

Post/Author/DateTime Post
tl
07-13-07, 05:58 PM so, I was trying to develop an equation that will make it easy to decide upon power attack usage. while a quite simple thing (and probably unnecessary), it made me spend quite a few language course hours on it.
so, what's the problem?
if we say that:
q .. number of dice that hit on d20 (19>=q>=t; if it's more then 19, PA at least to (q-a)=19)
X .. on-critical-multiplied-only damage before PA
t .. critical threat
m .. critical multiplier

we can write any attack as:
(t/20)(q/20)(m-1)X + (q/20)X
which equals
(q/20)X((t/20)(m-1) + 1)

now, the only things affected by power attack in this equation are q and x, so it seems that the optimal power attacking isn't affected by type of weapon.
WRONG! this seems so wrong.
so, where did I make my mistake? and can we even calculate in this way, saying that in 20 rolls, every number will fall once?
sonofzeal
07-13-07, 10:03 PM q .. number of dice that hit on d20 (19>=q>=t; if it's more then 19, PA at least to (q-a)=19)
x .. on-critical-multiplied-only damage before PA
t .. critical threat
m .. critical multiplier

(t/20)(q/20)(m-1)x + (q/20)x

Add a power attack variable, P, and assume two-handed weapons ('cause powerattacking with anything smaller is just silly).

(t/20)((q-P)/20)(m-1)(x+2*P) + ((q-P)/20)(x + 2*P)

Rearrange.

((q-P)/20)*(x+2*P)*((t/20)(m-1) + 1)

((q-P)/20)*(x+2*P)*(t(m-1)/20 + 1)

(q-P)*(x+2*P)*(t(m-1)/20 + 1)/20 <- what you should be looking at.

Now, for each set of variables, solve for the Expected Value, then adjust and solve again. Any decent programmer could write a script to optimize P if given the other parameters as input; that's probably the easiest approach, IMHO.
tl
07-13-07, 10:12 PM yes, it is the same conclusion and I could (and did) easily script that.
but my problem is, that optimum P isn't affected by t or m, which is what I find very strange and most certainly wrong.
and the question still remains, am I wrong?
as for 2P, there are ways to get more then twice P on damage... that's why I left that part out, and for the reason that it isn't related to my question
Asymptote
07-14-07, 12:01 PM Well... critical threat range is based on the natural die roll. So how much you Power Attack doesn't really affect that, does it? It still matters in terms of rolling to confirm and how much damage you do, but maybe those are covered by the other variables. :thinks:
BillTheManiac
07-15-07, 12:27 PM Computer programs? No, derivatives.
(q-P)*(x+2*P)*Something Constant

x+2P=2q-2P
P=(2q-x)/4
Fwib
07-15-07, 07:45 PM A treatise on power attack (http://boards1.wizards.com/wotc_archive/index.php/t-167952)

Check that out first, save duplication of effort.
BillTheManiac
07-15-07, 09:14 PM Looks exactly like mine. I got it right, then.
Egres
07-19-07, 08:10 AM Try this (http://www.monkeysushi.net/gaming/DnD/math.html#powerattack) too.
BillTheManiac
07-24-07, 08:11 PM Try this (http://www.monkeysushi.net/gaming/DnD/math.html#powerattack) too.

again, same as mine.

I find this simplest:

Let 21-A be the number needed to hit, so A/20 is the probability of hitting
Let B be the average damage, including x1.05 for a x2 weapon, x1.10 for a x3 or 19-20 weapon, and x1.15 for a x4 weapon as well as /2 for a 2-handed weapon.
Then you need A=B.
So PA=(A-B)/2
On a 2-handed weapon that's
A/2-B/4
which is what we have
B is depended on critical threat range, which makes sense.
Maat_Mons
07-24-07, 09:36 PM Now, the only things affected by power attack in this equation are q and x, so it seems that the optimal power attacking isn't affected by type of weapon.

If your power attack penalty prevents you from hitting on some of the numbers that would otherwise threaten a critical, your t variable needs to shrink to reflect that. So, t is a function of your power attack penalty.
Callix
07-25-07, 02:16 AM If your power attack penalty prevents you from hitting on some of the numbers that would otherwise threaten a critical, your t variable needs to shrink to reflect that. So, t is a function of your power attack penalty.

Only in the extreme cases. Most of the time, optimal power attack is only a few points, and a frontliner who needs a 17 to hit is in a bad way already. If you have a keen greatsword, it might just matter, but in most cases it is irrelevant, since the optimal value still has the maximum threat range.

If you want, you could define a function z = min(q, t), and remove the clause that q>=t, then sub z for t into the equations.
Khaim
07-26-07, 09:01 AM again, same as mine.

I find this simplest:

Let 21-A be the number needed to hit, so A/20 is the probability of hitting
Let B be the average damage, including x1.05 for a x2 weapon, x1.10 for a x3 or 19-20 weapon, and x1.15 for a x4 weapon as well as /2 for a 2-handed weapon.
Then you need A=B.

What? That doesn't make any sense to me. Maybe I'm not understanding your notation, but why do you want (# of rolls that hit) to equal (average damage)? Those are totally unrelated!
BillTheManiac
07-26-07, 11:44 AM What? That doesn't make any sense to me. Maybe I'm not understanding your notation, but why do you want (# of rolls that hit) to equal (average damage)? Those are totally unrelated!

Because average damage=A*B/20. With Power Attack that's (A-x)*(B+x)/20. Derivative is (B-A)/20-x/10 so x=(B-A)/2 and A-x=B+x. That when A+B is fixed, optimizing AB implies A=B is just a little trick.
Maat_Mons
07-28-07, 01:40 PM Only in the extreme cases.

Of course it’s only in extreme cases. Remember though, tl’s major quandary was that he couldn’t find a way in which the optimal power attack trade-out could be effected by the weapons critical threat range, leading him to think he must have made an error. Given this, it was pertinent to note the circumstances in which there could be an effect.

If you want, you could define a function z = min(q, t), and remove the clause that q>=t, then sub z for t into the equations.

I could define the damage function in terms of the min function. I could also define it piecewise or in terms of the Heaviside unit step function. I don’t know why you would feel it necessary to point out how to modify the equation. Have I said something that made you suspect I do not have basic competence at math?
Callix
07-28-07, 06:25 PM Maat, it's called "being helpful". CharOp is a collaborative, as well as a competitive, excercise. I saw a simple and easy way to address your concerns, so I pointed it out. The min function is a simple and, to the layperson, obvious way of doing this.
Frankly, I don't know what most of the other functions you talked about mean. You know more maths than me. I just wanted to make a contribution. If it's not useful, then you don't have to use it. But I think a public document that can be understood without a PhD in pure maths is more useful than one that can't. So I tried to help it stay as simple as possible for such a complex mathematical problem.

Post/Author/DateTime	Post
tl 07-13-07, 05:58 PM	so, I was trying to develop an equation that will make it easy to decide upon power attack usage. while a quite simple thing (and probably unnecessary), it made me spend quite a few language course hours on it. so, what's the problem? if we say that: q .. number of dice that hit on d20 (19>=q>=t; if it's more then 19, PA at least to (q-a)=19) X .. on-critical-multiplied-only damage before PA t .. critical threat m .. critical multiplier we can write any attack as: (t/20)(q/20)(m-1)X + (q/20)X which equals (q/20)X((t/20)(m-1) + 1) now, the only things affected by power attack in this equation are q and x, so it seems that the optimal power attacking isn't affected by type of weapon. WRONG! this seems so wrong. so, where did I make my mistake? and can we even calculate in this way, saying that in 20 rolls, every number will fall once?
sonofzeal 07-13-07, 10:03 PM	q .. number of dice that hit on d20 (19>=q>=t; if it's more then 19, PA at least to (q-a)=19) x .. on-critical-multiplied-only damage before PA t .. critical threat m .. critical multiplier (t/20)(q/20)(m-1)x + (q/20)x Add a power attack variable, P, and assume two-handed weapons ('cause powerattacking with anything smaller is just silly). (t/20)((q-P)/20)(m-1)(x+2P) + ((q-P)/20)(x + 2P) Rearrange. ((q-P)/20)(x+2P)((t/20)(m-1) + 1) ((q-P)/20)(x+2P)(t(m-1)/20 + 1) (q-P)(x+2P)*(t(m-1)/20 + 1)/20 <- what you should be looking at. Now, for each set of variables, solve for the Expected Value, then adjust and solve again. Any decent programmer could write a script to optimize P if given the other parameters as input; that's probably the easiest approach, IMHO.
tl 07-13-07, 10:12 PM	yes, it is the same conclusion and I could (and did) easily script that. but my problem is, that optimum P isn't affected by t or m, which is what I find very strange and most certainly wrong. and the question still remains, am I wrong? as for 2P, there are ways to get more then twice P on damage... that's why I left that part out, and for the reason that it isn't related to my question
Asymptote 07-14-07, 12:01 PM	Well... critical threat range is based on the natural die roll. So how much you Power Attack doesn't really affect that, does it? It still matters in terms of rolling to confirm and how much damage you do, but maybe those are covered by the other variables. :thinks:
BillTheManiac 07-15-07, 12:27 PM	Computer programs? No, derivatives. (q-P)(x+2P)*Something Constant x+2P=2q-2P P=(2q-x)/4
Fwib 07-15-07, 07:45 PM	A treatise on power attack (http://boards1.wizards.com/wotc_archive/index.php/t-167952) Check that out first, save duplication of effort.
BillTheManiac 07-15-07, 09:14 PM	Looks exactly like mine. I got it right, then.
Egres 07-19-07, 08:10 AM	Try this (http://www.monkeysushi.net/gaming/DnD/math.html#powerattack) too.
BillTheManiac 07-24-07, 08:11 PM	Try this (http://www.monkeysushi.net/gaming/DnD/math.html#powerattack) too. again, same as mine. I find this simplest: Let 21-A be the number needed to hit, so A/20 is the probability of hitting Let B be the average damage, including x1.05 for a x2 weapon, x1.10 for a x3 or 19-20 weapon, and x1.15 for a x4 weapon as well as /2 for a 2-handed weapon. Then you need A=B. So PA=(A-B)/2 On a 2-handed weapon that's A/2-B/4 which is what we have B is depended on critical threat range, which makes sense.
Maat_Mons 07-24-07, 09:36 PM	Now, the only things affected by power attack in this equation are q and x, so it seems that the optimal power attacking isn't affected by type of weapon. If your power attack penalty prevents you from hitting on some of the numbers that would otherwise threaten a critical, your t variable needs to shrink to reflect that. So, t is a function of your power attack penalty.
Callix 07-25-07, 02:16 AM	If your power attack penalty prevents you from hitting on some of the numbers that would otherwise threaten a critical, your t variable needs to shrink to reflect that. So, t is a function of your power attack penalty. Only in the extreme cases. Most of the time, optimal power attack is only a few points, and a frontliner who needs a 17 to hit is in a bad way already. If you have a keen greatsword, it might just matter, but in most cases it is irrelevant, since the optimal value still has the maximum threat range. If you want, you could define a function z = min(q, t), and remove the clause that q>=t, then sub z for t into the equations.
Khaim 07-26-07, 09:01 AM	again, same as mine. I find this simplest: Let 21-A be the number needed to hit, so A/20 is the probability of hitting Let B be the average damage, including x1.05 for a x2 weapon, x1.10 for a x3 or 19-20 weapon, and x1.15 for a x4 weapon as well as /2 for a 2-handed weapon. Then you need A=B. What? That doesn't make any sense to me. Maybe I'm not understanding your notation, but why do you want (# of rolls that hit) to equal (average damage)? Those are totally unrelated!
BillTheManiac 07-26-07, 11:44 AM	What? That doesn't make any sense to me. Maybe I'm not understanding your notation, but why do you want (# of rolls that hit) to equal (average damage)? Those are totally unrelated! Because average damage=AB/20. With Power Attack that's (A-x)(B+x)/20. Derivative is (B-A)/20-x/10 so x=(B-A)/2 and A-x=B+x. That when A+B is fixed, optimizing AB implies A=B is just a little trick.
Maat_Mons 07-28-07, 01:40 PM	Only in the extreme cases. Of course it’s only in extreme cases. Remember though, tl’s major quandary was that he couldn’t find a way in which the optimal power attack trade-out could be effected by the weapons critical threat range, leading him to think he must have made an error. Given this, it was pertinent to note the circumstances in which there could be an effect. If you want, you could define a function z = min(q, t), and remove the clause that q>=t, then sub z for t into the equations. I could define the damage function in terms of the min function. I could also define it piecewise or in terms of the Heaviside unit step function. I don’t know why you would feel it necessary to point out how to modify the equation. Have I said something that made you suspect I do not have basic competence at math?
Callix 07-28-07, 06:25 PM	Maat, it's called "being helpful". CharOp is a collaborative, as well as a competitive, excercise. I saw a simple and easy way to address your concerns, so I pointed it out. The min function is a simple and, to the layperson, obvious way of doing this. Frankly, I don't know what most of the other functions you talked about mean. You know more maths than me. I just wanted to make a contribution. If it's not useful, then you don't have to use it. But I think a public document that can be understood without a PhD in pure maths is more useful than one that can't. So I tried to help it stay as simple as possible for such a complex mathematical problem.