Dice and Success...

[The-Most-Compact-of-the-Kyles]

I've done some thinking and did some number crunching and I've come to a question I would like to pose to our ST's.

How do we determine the "success pool" of a challenge?

There are two ways I have seen in game and am wondering if one is the proper way we do it, or if we do both. I'll illustrate my point so I'm more clear about what I mean:



A) - (Dice Pool - Dice Pool)/2 = # of Successes

Example:
(Challenger: DEX+BRAWL to hit. (2+2) = 4 Dice)
-
(Target: DEX+DODGE to dodge. (2+1) = 3 Dice)

---

Total Dice for Success Pool is 1 Die. Challenger begins with 1 success.


B) - (Dice Pool/2) - (Dice Pool/2) = # of Successes

Example:
(Challenger: DEX+BRAWL to hit. (2+2) = 4 Dice)
(Challenger has 2 Successes)
-
(Target: DEX+DODGE to dodge. (2+1) = 3 Dice)
(Target has 2 Successes)

---

Total number of Successes in Pool is 0.



As you can see, although the "Stats" are the same, the end result is not. If I could get a clarification on this, it would be great.

-Kyle

[benime]

In instances where there is direct opposition between someone with an even die pool and someone with an odd die pool, I would use the subtraction method to represent the extra level of skill. I like to keep in mind that this game is not Dungeons and Dragons, so every single point counts (unlike in DnD where odd ability scores aren't really worth anything).

[David Price]

We've (Where we I mean I) been running it the first way, so that the even/odd split matters. Otherwise having a dice pool of 6 instead of 5 to dodge would never really come up (since different difficulties in dodge pools don't happen all that often).

[cassidy]

As with myself I would have to agree with BRoney. I have always done it where we subtract dice pools from one another to figure out who has more dice then subtract that. however that does not mean that the person is any closer to a botch.

Example:
A) - (Dice Pool - Dice Pool)/2 = # of Successes

Example:
(Challenger: DEX+BRAWL to hit. (2+2) = 4 Dice)
-
(Target: DEX+DODGE to dodge. (2+1) = 3 Dice)

--------------------------------------------------------------------------------
Total Dice for Success Pool is 1 Die. Challenger begins with 1 success.

just because when the chops are thrown he has '1' success, that does not mean that if he looses twice he has just botched. (1-2=-1, actually it is still 2 (initial success) -2=0. he still threatens to botch, must test more with the ST). Also with higher dice pools, a pool of 8 verses a pool of 7, the man with the pool of 8 must botch down 5 times (4-5=-1) in order to have botched.

Again this is to give the man who has spent the xp (or starting character dots) the advantage with his even die pool.

[The-Most-Compact-of-the-Kyles]

Thanks for the info. guys. I appreciate it. Always good to know how to do the math.

-Kyle

[cassidy]

no prob dude....

If anyone wants to know more on this stuff feel free to come up to me before any game session and i will be glad to open a book and show you some stuff so you can better understand how the game works.