# Thread: How is base damage calculated?

1. ## How is base damage calculated?

On weapons you see:
A[BdC]+D
But the formula isn't:
[(A[Lowest Outcome]+D)+(A[Highest Outcome]+D)]/2

Let's take Celestia
2.5[1d10]+7
Lowest Damage: 9
Highest Damage: 32
Average Damage: 20.5
Listed Base Damage: 22.55

I have no idea how they came to that number.
Pinion
2.5[2d6]+7
Lowest Damage: 12
Highest Damage: 37
Average Damage: 24.5
Listed Base Damage: 29.4

I'm not sure how crits effect this, but if anybody knows, I'd like to find out how they get that number.

2. Originally Posted by Maelodic
On weapons you see:
A[BdC]+D
But the formula isn't:
[(A[Lowest Outcome]+D)+(A[Highest Outcome]+D)]/2

Let's take Celestia
2.5[1d10]+7
Lowest Damage: 9
Highest Damage: 32
Average Damage: 20.5
Listed Base Damage: 22.55

I have no idea how they came to that number.
Pinion
2.5[2d6]+7
Lowest Damage: 12
Highest Damage: 37
Average Damage: 24.5
Listed Base Damage: 29.4

I'm not sure how crits effect this, but if anybody knows, I'd like to find out how they get that number.
If you multiply by 1+critrate*(critmultiplier-1) you get those numbers.

I don't fully understand why 2.5[1d10]+7 gives 20.5 average damage and not 2.5*5.5+7 = 20.75, but that is probably due to the outcomes being integers.

3. Originally Posted by Forzah
If you multiply by 1+critrate*(critmultiplier-1) you get those numbers.

I don't fully understand why 2.5[1d10]+7 gives 20.5 average damage and not 2.5*5.5+7 = 20.75, but that is probably due to the outcomes being integers.
Thanks, you're awesome.

Ninja edit:
It's just Lowest damage roll+highest damage roll over 2.
So (9+32)/2

4. Originally Posted by Maelodic
Thanks, you're awesome.

Ninja edit:
It's just Lowest damage roll+highest damage roll over 2.
So (9+32)/2
Ok... it seems like they changed the definition of average then . Example: you roll a 1 with probabily 0.9 and a 10 with probability 0.1, then the average roll will be 1.9.... not 5.5.

5. Originally Posted by Forzah
Ok... it seems like they changed the definition of average then . Example: you roll a 1 with probabily 0.9 and a 10 with probability 0.1, then the average roll will be 1.9.... not 5.5.
Ah wait, that 20.5 is due to rounding down.

Outcomes:
9.5 12 14.5 17 19.5 22 24.5 27 29.5 32

Rounded outcomes:
9 12 14 17 19 22 24 27 29 32

Average: 20.5

6. Originally Posted by Maelodic
On weapons you see:
A[BdC]+D
But the formula isn't:
[(A[Lowest Outcome]+D)+(A[Highest Outcome]+D)]/2

Let's take Celestia
2.5[1d10]+7
Lowest Damage: 9
Highest Damage: 32
Average Damage: 20.5
Listed Base Damage: 22.55

I have no idea how they came to that number.
Pinion
2.5[2d6]+7
Lowest Damage: 12
Highest Damage: 37
Average Damage: 24.5
Listed Base Damage: 29.4

I'm not sure how crits effect this, but if anybody knows, I'd like to find out how they get that number.
It appears to be including crits, but not automisses on a 1. For a 19-20/x2 crit profile, take your average * 22 / 20; for 19-20/x3, it's average * 24/20.

7. Originally Posted by Forzah
Ok... it seems like they changed the definition of average then . Example: you roll a 1 with probabily 0.9 and a 10 with probability 0.1, then the average roll will be 1.9.... not 5.5.
So if you have a 10 sided dice, and you roll it 10 times, probability suggests that that you'll hit each number at least once.
1-10

1+2+3+4+5+6+7+8+9+10=55
Then divide it by the number of rolls for the most average number: 5.5
1+10/2 works because you're taking the highest number and the lowest number and finding the most average roll between them assuming every other possibility is the same.

In Celestia's case, the lowest roll is 9.5, but since we cannot have a half point of damage, the devs have said round down.
1.1[(9+32)/2]=22.55
I got that part, should have got the crit part a bit easier though.

8. Originally Posted by Nibor
It appears to be including crits, but not automisses on a 1. For a 19-20/x2 crit profile, take your average * 22 / 20; for 19-20/x3, it's average * 24/20.
Easiest way to think about for me, anyway:

If you have a 10% chance to crit, it's 1.1
If you have a 20% chance to crit, it's 1.2

Times the decimal by your critmod-1
So Carnifex for example
3-14 Damage, 20% chance to crit, 3x crit rate
1.4[(14+3)/2]=11.90

With weapons with improved die or extra crit profiles, you just need the highest and lowest roll for comparison.

For example a fully geared unarmed DPS monk with an antipod in earth stance will have:
10[1d8]+8
10% , 3x

So the base damage on that sucker would be:
1.2[(18+88)/2]=63.6

Monks OP.

9. Originally Posted by Maelodic
So if you have a 10 sided dice, and you roll it 10 times, probability suggests that that you'll hit each number at least once.
1-10 <snip>
Pedantic rant:
Please don't say that probability suggests that you'll hit each number at least once.
Probability suggest nothing like that. Assuming 10 roll of a 10 sided fair die, the chances that you get each number at least once (which btw is the same as exactly once)
is 10!/10^10 which is roughly 0.00036288 so less then 4 in 10000.

What you are invoking is that assuming a very large number of rolls the average number of 1's 2's ... 10's will be about the same 1/10*number of rolls. So the next bit you do is just an expected value.

End of pedantic rant.

The other issue of course is that base damage is an extremely misleading number. It ignores grazes, crit confirmations, damage from strength and many other effects which can and often do completely change the numbers when you are playing. And that's ignoring DR, susceptibility and lot's of other stuff.
Just my 2 copper,
Rawel

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