Player Defense Rolling

[DrSpunj]

Here's the file some were requesting from me. It's the handout I provided to my players about a year ago when we adopted this DMG/UA variant.

Thanks!

DrSpunj

 

[Nyaricus]

I have been playing around with that idea for awhile > i have also noticed that the static +10 was just your character "taking 10" on his AC.

i wouldn't go as far as to let a PC rolling a 1 or 2 (with his opponent having a longsword in hand) mean that he has been critically threatened (which was my interpreattion, correct me if i am wrong) because, he essentually is being threatened with twice as many crit. threats > one for the longsword being wielded by the PCs opponent (a 19-20 on his attack roll), and one for the PCs def roll (a def. roll of 1-2).
Also, i don't like the idea of a 1 = -10, and a 20 = 30. Simply, a 1 is a 1, and a 20 is a 20. i go by the standard critical hit method, and on a one, i generally have an AoO occur (if the character is threatened when he rolls to hit).

Other than those points, i have had one session in an Evil Realms Campaign which i am DMing, and i think i might run it by the guys to see if they'd like to roll for Defense. It may go either way, half are noobs and half arent, so it'll be interesting . I was also thinking of implimenting a Armour as DR for this campaign, but one guy is a complete noob, so i dont want to freak him out (or kill too many of his characters)!!

Peace

 

[Ulorian - Agent of Chaos]

Hi DrSpunj.. not following your math.

d20 + MAB >= 10 + ACM
d20 + MAB - MAB - ACM >= 10 + ACM - MAB - ACM
d20 - ACM >= 10 - MAB

or

d20 + MAB >= 10 + ACM
d20 + MAB + MAB + ACM >= 10 + ACM + MAB + ACM
d20 + ACM + 2MAB >= 10 + MAB + 2ACM
d20 + ACM >= 10 + MAB + 2ACM - 2MAB
d20 + ACM >= 10 + MAB + 2(ACM - MAB)

Since your formula is:

d20 + ACM >= 12 + MAB

It looks like you're approximating ACM - MAB as 1. What's the justification for this? Or am I missing something obvious?

-------------------------------------

Nyaricus, the critical range isn't doubled; the defense roll is used instead of the attack roll. The objective of this variant is to eliminate the monster's attack roll.

 

[DanMcS]

Since your formula is:

d20 + ACM >= 12 + MAB

It looks like you're approximating ACM - MAB as 1. What's the justification for this? Or am I missing something obvious?

Yes. The >= sign.

Seriously, it's easier to explain as probabilities. When rolling to hit in D&D, you hit if you roll >= the DC, which is the armor class. If you want to keep the system the same, then you want a defense roll to succeed when it is >= the attack value.

Consider two characters; the attacker has a +0 BAB and the defender has a +0 defense bonus.

If the attacker is rolling to hit, he wants to hit armor class 10, which he will do on a roll of 10 to 20 on a d20, which is 11/20 rolls.

If the defender is rolling to dodge the exact same attack, you would want him to be hit or missed the same portion of the time, so he should successfully dodge on a roll of 12 to 20, which is 9/20. The attack would still hit 11/20 times.

Thus, when setting up the attack difficulty for a defense roll, the DC equals the attackers attack bonuses +12.

In the Unearthed Arcana version of this rule, they screwed up and used 11 as the base, not 12.

 

[Ulorian - Agent of Chaos]

Consider two characters, the attacker has a BAB of +9 and the defender has an AC mod of +2.

Under the standard rules:

d20 + MAB >= 10 + ACM
d20 + 9 >= 10 + 2
d20 >= 3

Using DrSpunj's formula:

d20 + ACM >= 12 + MAB
d20 + 2 >= 12 + 9
d20 >= 19

Oops.

Check my initial post again.. the math behind DrSpunj's formula doesn't stand up. The mistake you and DrSpunj seem to be making is that you're making your comparisons under very specific conditions (i.e. the attacker and defender have the same BAB/AC) where your formula coincidentally happens to be correct, so aren't seeing the error in your arithmetic.

 

[DanMcS]

Consider two characters, the attacker has a BAB of +9 and the defender has an AC mod of +2.

Under the standard rules:

d20 + MAB >= 10 + ACM
d20 + 9 >= 10 + 2
d20 >= 3

Using DrSpunj's formula:

d20 + ACM >= 12 + MAB
d20 + 2 >= 12 + 9
d20 >= 19

Right, that's perfectly correct. The first guy hits on 18/20 rolls, the second guy successfully dodges on 2/20 rolls. The odds of a hit stay the same.

 

[Ulorian - Agent of Chaos]

Ah, I see the problem: I was assuming that both rolls were to determine the success of an attack (as opposed to the success of a defense). Thanks for the clarification.

Neat system. I'm going to play with it a bit more.

 

[Ulorian - Agent of Chaos]

DanMcS and DrSpunj.. in case you were curious about the math behind your formula.

Formula for calculating odds to hit:

d20 + MAB >= 10 + ACM

To convert this formula to calculate odds to dodge:

d20 + MAB < 10 + ACM

DrSpunj wants a high roll to signify a success, so have the side of the equation with the d20 be greater than the other side. You can do this by multiplying both sides of the equation by -1 (this is the same principle that explains why -4 is bigger than -5, but 5 is bigger than 4).

-d20 - MAB > -10 - ACM

Add ACM to both sides of the equation, then add MAB to both sides.

-d20 - MAB + MAB + ACM > -10 - ACM + ACM + MAB
-d20 + ACM > -10 + MAB

Add 21 to both sides.

21 - d20 + ACM > -10 + 21 + MAB
(21 - d20) + ACM > 11 + MAB

Notice that 21 - d20 gives you a range of numbers between 1 and 20. In other words:

21 - d20 = d20

So:
(21 - d20) + ACM > 11 + MAB
d20 + ACM > 11 + MAB

Change the 'greater than' sign to 'greater than and equal to' by increasing the value of the right hand side by 1:

d20 + ACM >= 12 + MAB

Good on you guys for catching that error in the Unearthed Arcana!