Average Damage? [Archive] - Wizards Community

Average Damage? [Archive] - Wizards Community

Post/Author/DateTime Post
elvnsword
03-29-04, 11:40 PM I am confused as are a pair of my friends, on how to figure average damage for a build.
the best we could come up with is a full attack action, with full damage at a crit, and half the damage twice...

What is the formula you all use for Average damage...

Thanks,
Elvnsword
Vipren
03-29-04, 11:49 PM Just use all your standard damage modifiers, and use the average of your die

d1 = 1
d2 = 1.5
d4 = 2.5
d6 = 3.5
d8 = 4.5
d10= 5.5
d12= 6.5
frisbeet
03-30-04, 12:26 AM I like the average full attack damage/round statistic.

Naturally this number is parameterized by the AC of an opponent--you're going to do more average damage against AC 10 than AC 40.

So, start with what an average means. Redefine it 1st. Here's an example I posted today on the D&D general forum.

Consider the set of 5 numbers:

1 3 3 6 7

The average = sum(all numbers)/N, where N = # of #s. That's (1+3+3+6+7)/5 = 4.

Now ask, how many times does each # occur in this set?

For 1, 6, & 7, just once. But 3 occurs twice. Hmm. I could define the probability of pulling one of these numbers out of a hat as = # occurances/N.

The the P of pulling a 1 out of a hat containing a 1, 3, 3, 6, & 7 = 1/5, or 0.2. Or re-read as a 20% chance. Same with pulling a 6. Or a 7. But of getting a 3, that 2/5, or 0.4 = 40%.

Great.

Back to the average formula: (1+3+3+6+7)/5 = 4
Distribute the factor 1/5:
1 x 1/5 + 3 x 1/5 +3 x 1/5 + 6 x 1/5 + 7 x 1/5 = 4
Simplify by factoring out like #s:
1 x 1/5 + 3 x (1/5 + 1/5) + 6 x 1/5 + 7 x 1/5 = 4
then
1 x 1/5 + 3 x 2/5 + 6 x 1/5 + 7 x 1/5 = 4
or
1 x 0.2 + 3 x 0.4 + 6 x 0.2 + 7 x 0.2 = 4
It does? Yep, = 4.
But that's
1 x P(getting a 1) + 3 x P(3) + 6 x P(6) + 7 x P(7).

To generalize: an average can be expressed as taking the product of each outcome with its probability and summing this product for all outcomes.

So, if we know what the probability of hitting a certain AC, the average damage will simply be the product of this probability with the outcome of doing average weapon damage. "Average weapon damage", of course, is NOT the same thing as average damage/round. Think of it in this analysis as a constant--pick the weapon, accompanying feats, and character strength, and you've set this number in stone. The full attack average damage/round is the sum of the products of these two numbers for all attacks.

How do we find the probability of hitting, taking into account critical hits?

Know your weapon and opponent's AC.

Suppose a level 6 fighter with 18 Str, weapon focus & weapon specialization for greatsword, wields a +2 greatsword. His opponent has an AC of 25.

His attack bonus is 6+4+1+2 = 13

His average weapon damage is ave(2d6) + 6 + 2 + 2= 17

He gets two attacks at +13/+8.

The 1st attack:

There are three possibilities. He can miss. He can hit. He can critically hit.

The probability (P) of hitting for normal damage = P(normally hitting without critical threat) + P(threatening for critical) x P(botching the crit confirmation roll)

Against an AC of 25, any d20 roll 12-18 hits without threatening a crit. That 7 out of 20 possible rolls, so P(normally hitting without critical threat) = 7/20. P(threatening for critical) = 2/20, since two of a possible 20 rolls, 19 or 20, will threaten for critical. P(botching the crit confirmation foll) = 11/20; if I rerolled 1-11 after rolling a 19 or 20 to check for crit, I fail to score a crit but I still hit for normal damage. The probability of two things happening, where the 2nd thing is dependent on the 1st happening, is = to the product of their separate probabilities. Thus to account for doing normal damage when a crit is threatened but not confirmed we multiply these probabilities. Then the P(hitting for normal damage) = 7/20 + 2/20 x 11/20.

The P(critically hitting) = 2/20 x 9/20. Again, first you've got to threaten, and that P is 2/20...then multiply by the chance of confirming the crit. A roll of 12-20 is a successful confirmation, so that's 9 out of 20 possible rolls. Ergo 2/20 x 9/20.

The average damage for the 1st hit is the sum of the product of each outcome (hit or crit hit) multiplied by the damage done when hitting or crit hitting.

That's (7/20 + 2/20 x 11/20) x 17 + (2/20 x 9/20) x 17 x 2 (the crit factor) = 8.415.

But he's full attacking, so do the same thing but with an AB of +8 for the 2nd attack.

That's (2/20 + 2/20 x 16/20) x 17 + (2/20 x 4/20) x 17 x 2 = 3.74

The sum of these numbers is 12.155. That's your answer, for said fighter against an opponent with an AC of 25.

At least, that's how I do it. There's lots of other ways to look at the problem, and average damage/round isn't always the best statistic to consider.
Thane of Eurmal
03-30-04, 12:39 AM Somewhere along the lines of:

Average round = (average # hits) x (average damage)

Average damage = (min dam + max dam)/2

Average hits = 21 + (your Attack bonus) - (opponent's AC) , times 5%

This ignores crits and the automatic miss on 1, hit on 20 rules... With an example, this becomes:

Longsword (1D8 + 4 (str and such)) = (5+12)/2 = 8.5

Hitting an orc, AC14 with your +8 attack = 29-14 = 15 = 75%

So this character does 75% of 8.5 damage a round to AC 14, or 6.4 damage per round.

If he has Power Attack and gives himself a -2 attack for +2 damage, he has a 65% of 10.5 or 6.8 damage a round.

If he has a second hit (at +3), he does 75% of 8.5 plus 50% (24-14= 10 = 50%) of 8.5 for a total of 10.6 damage.

If the orc has a shield (AC 16), he has a 65% of 8.5 for 5.5 damage.

Crits actually only add a 5% chance of +100% damage for 20/ x2 weapons, so I wouldn't worry about it much... If you want more for crits, check out the Sword and Fist explanation.

Can you tell I majored in math? :D

Post/Author/DateTime	Post
elvnsword 03-29-04, 11:40 PM	I am confused as are a pair of my friends, on how to figure average damage for a build. the best we could come up with is a full attack action, with full damage at a crit, and half the damage twice... What is the formula you all use for Average damage... Thanks, Elvnsword
Vipren 03-29-04, 11:49 PM	Just use all your standard damage modifiers, and use the average of your die d1 = 1 d2 = 1.5 d4 = 2.5 d6 = 3.5 d8 = 4.5 d10= 5.5 d12= 6.5
frisbeet 03-30-04, 12:26 AM	I like the average full attack damage/round statistic. Naturally this number is parameterized by the AC of an opponent--you're going to do more average damage against AC 10 than AC 40. So, start with what an average means. Redefine it 1st. Here's an example I posted today on the D&D general forum. Consider the set of 5 numbers: 1 3 3 6 7 The average = sum(all numbers)/N, where N = # of #s. That's (1+3+3+6+7)/5 = 4. Now ask, how many times does each # occur in this set? For 1, 6, & 7, just once. But 3 occurs twice. Hmm. I could define the probability of pulling one of these numbers out of a hat as = # occurances/N. The the P of pulling a 1 out of a hat containing a 1, 3, 3, 6, & 7 = 1/5, or 0.2. Or re-read as a 20% chance. Same with pulling a 6. Or a 7. But of getting a 3, that 2/5, or 0.4 = 40%. Great. Back to the average formula: (1+3+3+6+7)/5 = 4 Distribute the factor 1/5: 1 x 1/5 + 3 x 1/5 +3 x 1/5 + 6 x 1/5 + 7 x 1/5 = 4 Simplify by factoring out like #s: 1 x 1/5 + 3 x (1/5 + 1/5) + 6 x 1/5 + 7 x 1/5 = 4 then 1 x 1/5 + 3 x 2/5 + 6 x 1/5 + 7 x 1/5 = 4 or 1 x 0.2 + 3 x 0.4 + 6 x 0.2 + 7 x 0.2 = 4 It does? Yep, = 4. But that's 1 x P(getting a 1) + 3 x P(3) + 6 x P(6) + 7 x P(7). To generalize: an average can be expressed as taking the product of each outcome with its probability and summing this product for all outcomes. So, if we know what the probability of hitting a certain AC, the average damage will simply be the product of this probability with the outcome of doing average weapon damage. "Average weapon damage", of course, is NOT the same thing as average damage/round. Think of it in this analysis as a constant--pick the weapon, accompanying feats, and character strength, and you've set this number in stone. The full attack average damage/round is the sum of the products of these two numbers for all attacks. How do we find the probability of hitting, taking into account critical hits? Know your weapon and opponent's AC. Suppose a level 6 fighter with 18 Str, weapon focus & weapon specialization for greatsword, wields a +2 greatsword. His opponent has an AC of 25. His attack bonus is 6+4+1+2 = 13 His average weapon damage is ave(2d6) + 6 + 2 + 2= 17 He gets two attacks at +13/+8. The 1st attack: There are three possibilities. He can miss. He can hit. He can critically hit. The probability (P) of hitting for normal damage = P(normally hitting without critical threat) + P(threatening for critical) x P(botching the crit confirmation roll) Against an AC of 25, any d20 roll 12-18 hits without threatening a crit. That 7 out of 20 possible rolls, so P(normally hitting without critical threat) = 7/20. P(threatening for critical) = 2/20, since two of a possible 20 rolls, 19 or 20, will threaten for critical. P(botching the crit confirmation foll) = 11/20; if I rerolled 1-11 after rolling a 19 or 20 to check for crit, I fail to score a crit but I still hit for normal damage. The probability of two things happening, where the 2nd thing is dependent on the 1st happening, is = to the product of their separate probabilities. Thus to account for doing normal damage when a crit is threatened but not confirmed we multiply these probabilities. Then the P(hitting for normal damage) = 7/20 + 2/20 x 11/20. The P(critically hitting) = 2/20 x 9/20. Again, first you've got to threaten, and that P is 2/20...then multiply by the chance of confirming the crit. A roll of 12-20 is a successful confirmation, so that's 9 out of 20 possible rolls. Ergo 2/20 x 9/20. The average damage for the 1st hit is the sum of the product of each outcome (hit or crit hit) multiplied by the damage done when hitting or crit hitting. That's (7/20 + 2/20 x 11/20) x 17 + (2/20 x 9/20) x 17 x 2 (the crit factor) = 8.415. But he's full attacking, so do the same thing but with an AB of +8 for the 2nd attack. That's (2/20 + 2/20 x 16/20) x 17 + (2/20 x 4/20) x 17 x 2 = 3.74 The sum of these numbers is 12.155. That's your answer, for said fighter against an opponent with an AC of 25. At least, that's how I do it. There's lots of other ways to look at the problem, and average damage/round isn't always the best statistic to consider.
Thane of Eurmal 03-30-04, 12:39 AM	Somewhere along the lines of: Average round = (average # hits) x (average damage) Average damage = (min dam + max dam)/2 Average hits = 21 + (your Attack bonus) - (opponent's AC) , times 5% This ignores crits and the automatic miss on 1, hit on 20 rules... With an example, this becomes: Longsword (1D8 + 4 (str and such)) = (5+12)/2 = 8.5 Hitting an orc, AC14 with your +8 attack = 29-14 = 15 = 75% So this character does 75% of 8.5 damage a round to AC 14, or 6.4 damage per round. If he has Power Attack and gives himself a -2 attack for +2 damage, he has a 65% of 10.5 or 6.8 damage a round. If he has a second hit (at +3), he does 75% of 8.5 plus 50% (24-14= 10 = 50%) of 8.5 for a total of 10.6 damage. If the orc has a shield (AC 16), he has a 65% of 8.5 for 5.5 damage. Crits actually only add a 5% chance of +100% damage for 20/ x2 weapons, so I wouldn't worry about it much... If you want more for crits, check out the Sword and Fist explanation. Can you tell I majored in math? :D