Skill System Woes

[iago]

Good idea, but there's a problem Fred. Already we can see that the odds of 1 v 1 is not 50%. That's OK if it's an IGO UGO system, but the effect does decrease with higher levels. So in a fight between two guys with level 100, you have more than a 98% chance of failure. Talk about your whiff factor.

Damn, you're right.  I hadn't extended it out on the other axis.  My bad.

[Walt Freitag]

Hi Fred,

Go with Mike's suggestion if what you want is for the probability of success to depend on the ratio of the two skill levels, so a 2 vs. a 1 would have the same chance (75%) as a 200 vs. a 100, while a 101 vs. a 100 would be very close to even.

On the other hand, you might prefer the probability of success to depend on the difference between the two skill levels -- while still staying below 1.0 no matter how high the difference becomes. So, a 2 vs. a 1 would have the same chance as a 101 vs. a 100, while a 200 vs. a 100 would have the same (extremely high, but not quite guaranteed) chance as a 101 vs. a 1. In that case, you want (using Mike's notation):

C = 100% - (50% / (x ^ ((H-L)/y)))

This gives a 50% chance when H=L, and the chance of failure for H decreases by a factor of x for every y points by which H exceeds L. For instance,

C = 100% - (50% / (2.0 ^ ((H-L)/4.0)))

means that H has a 75% chance of success when H is 4 greater than L, an 87.5% chance of success (that is, one eighth chance to fail) when H is 8 greater than L, a 93.75% chance of success (that is, one sixteenth chance to fail) when H is 12 greater than L, and so forth.

See my old Symmetry thread for a discussion of the implications of this. You can ignore the parts about the actual dicing mechanism, and concentrate on the discussion of the underlying math and the ramifications in play.

- Walt

PS Mike's formula for code use should be C=.5+((H-L)/H*.5). Mine would be C = 1.0 - .5  / (x ^ ((H-L)/y)).