Needless comparison of EWP, WF & Imp Crit

[Infiniti2000]

Yes.

The reason you want to include that in the formula now if you change the weapon to +1 blessed (or something) or use Power Critical or whatever. It depends on how far you want to take the analysis.

 

[Nail]

The correct formula to account for confirmation would be A = P{D[1+PPc(Mc-1)] + Db}..

Nope.

A = P{D[1+PPc(Mc-1)] + Db}

A = PD[1+PPc(Mc-1)] + PDb

A = PD+(P^2)DPc(Mc-1) + PDb <= "P^2"??????

You've got too many pees, bub.

 

[Infiniti2000]

You've got too many pees, bub.

Blame it on old age.

Then the original formula is plain incorrect. Without my change:

A = P{D[1+Pc(Mc-1)] + Db}
A = PD[1+Pc(Mc-1)] + PDb
A = PD+PDPc(Mc-1) + PDb

The problem with this formula then is that it doesn't actually represent the two seperate rolls. You shouldn't use the same P unless you expect the results to be identical, which is a huge, false assumption. Instead, we should have something like:

A = PpD+PsDPc(Mc-1) + PDb

where
Pp - Primary roll
Ps - Secondary roll

 

[Nail]

Then the original formula is plain incorrect. Without my change:

A = P{D[1+Pc(Mc-1)] + Db}
A = PD[1+Pc(Mc-1)] + PDb
A = PD+PDPc(Mc-1) + PDb

The problem with this formula then is that it doesn't actually represent the two seperate rolls. You shouldn't use the same P unless you expect the results to be identical, which is a huge, false assumption. Instead, we should have something like:

A = PpD+PsDPc(Mc-1) + PDb

where
Pp - Primary roll
Ps - Secondary roll

Almost. Still a bit wrong, I'm afraid.

I know this can be a bit hard. Probability can be a hard subject to wrap your brain around; it catches even my best students sometimes.

Look at it this way: "Pp" and "Ps" are the same chance. You have to hit the same AC - twice. (Sure, you're rolling the dice twice - but the eqn isn't recording the result of each die roll. Remember, we're looking at averages.) If Pp = Ps, then there's no reason to distinguish them in the formula. (Putting aside, for a moment, the CW feat: Power Critcal.)

How's that?

 

[Infiniti2000]

Well, it's not wrong. If you set Pp = Ps, you get the same result as before. Maybe you just mean that the original equation is not wrong for Pp == Ps. Okay, then, I'll buy that. I still think you should adopt my formula, though, so you can evaluate Power Critical, blessed, etc.

 

[delericho]

A = PpD+PsDPc(Mc-1) + PDb

where
Pp - Primary roll
Ps - Secondary roll

That last term should be PpDb - the probability on the second roll is irrelevant to this term, as this represents the damage that is not multiplied on a critical hit.

Additionally, Ps is a variable used to represent a number somewhere between 0.05 and 0.95 which in every case matches the numerical value of the number represented by Pp. We're not assuming the results are the same here, but the probabilities are.

The representation of two rolls comes from that Pc term, which is the probability of a critical threat, which is the smaller of Pp and the probability generated from the weapon's threat range.

So, the formula:

A = PD + PDPc(M-1) + PDb

where

A is the average damage scored
P is the probability of a hit
Pc is the probability of the die landing in the threat range and the attack being a hit
M is the critical multiplier of the weapon
D is the damage caused that is multiplied on a critical hit
Db is the damage caused that is not multiplied on a critical hit

will give numerically correct results.

Of course, this can be reduced to:

A = P{D[1+Pc(M-1)]+Db}

Getting back to the analysis of the feats for the moment, what if the probabilities on the attack and confirmation roll were not the same? What if, instead of increasing the threat range, the Improved Critical feat gave a +4 bonus to the confirmation roll (or change the bonus to suit)? This could then be allowed to stack with a Keen weapon without making critical threats so common as to be mundane, and yet prevent the disappointment that comes from rolling a threat only to fail to confirm.

Edit: of course, what I've just described in the paragraph above is the Power Critical feat from Complete Warrior. Oh, well.

 

[Nail]

I still think you should adopt my formula, though, so you can evaluate Power Critical, blessed, etc.

Sure, should you need to do that. (shrug) Most of the time, you don't need to, as your comparison interests lie else where.......such as in this thread.........(hint)........

A more interesting way to derive the formula to get at Power Critcal, etc, might be this:

A = PD+PtDPc(Mc-1) + PDb

A = P{D+(Pt/P)DPc(Mc-1) + Db}

A = P{D[1+(Pt/P)Pc(M-1)]+Db}

Where
A = average damage per attack
P = Probability to hit, as a fraction
D = average weapon damage plus Str, Magic, etc
Pc = Probability to Threaten, as a fraction
Pt = Probability to Confirm the threat, as a fraction.
Mc= Critical Multiplier
Db = Bonus Damage dice that are not multiplied by a confirmed critical

The quasi-interesting bit: It's the ratio of Pt to P that controls the effectiveness of the crit, and that ratio has a median of 1.4. Meaning: the feat Power Critical can make your "criticals" 40% more effective, on average.

(shrug)

That is, I found it interesting.

 

[Nail]

Edit: of course, what I've just described in the paragraph above is the Power Critical feat from Complete Warrior. Oh, well.

Yep.

Great minds think alike, eh? You and the designers, that is?

 

[Infiniti2000]

That last term should be PpDb - the probability on the second roll is irrelevant to this term, as this represents the damage that is not multiplied on a critical hit.

Yes, thanks for the correction. Now, you can modify Ps with Power Crit, set to 1 with blessed and an evil opponent (or to the probability that you'll encounter an evil opponent), etc. Oh, and now include a term for extra crit damage, like a flaming burst weapon.

A = PpD+PsPc[(Mc-1)D + Dc] + PpDb

where
Dc = extra crit-based damage

Of course, most burst weapons have extra damage for multipliers, so it could be

A = PpD+PsPc(Mc-1)(D + Dc) + PpDb

We could continue with additional terms, but then we're just getting crazy.

 

[Nail]

A = PpD+PsPc(Mc-1)(D + Dc) + PpDb

Sure, but you can clean that up a bit more......although it's still messy:

A= P(D+Db) + PtPc(Mc-1)(D+Dr)

Where:
A = average damage per attack
P = Probability to hit, as a fraction
D = average weapon damage plus Str, Magic, etc
Pc = Probability to Threaten, as a fraction
Pt = Probability to Confirm the threat, as a fraction.
Mc= Critical Multiplier
Db = Bonus Damage dice that are not multiplied by a confirmed critical
Dr = Ave of one die of "Burst"=type damage, typically a d10, therefore ave = 5.5.