Math Analysis: Counting Successes

Saelorn

Hero
Hi everyone,

I've been thinking about system mechanics recently, and I realized that I have difficulty intuiting the mathematical implications of a count-the-successes type system, such as used in World of Darkness or Shadowrun. In something like D&D, it's easy to see how a +1 to hit compares to +1 AC and get a rough estimate for probabilities. Likewise with GURPS, where both the attack roll and the defense roll are independent of each other and have a direct percentage chance of succeeding, it's easy to intuit the combined probabilities.

How do probabilities compare when you're rolling a bunch of dice and counting successes, if both the attacker and the defender get to roll? If you need a 7 or higher on a d10 to count as a success, and the attacker is rolling nine dice against the defender's five dice, what is the chance that the defender will roll more successes? Has anyone seen a big chart somewhere?

Altamont Ravenard

Explorer
I've been playing around with that mechanic for a homebrew in which I wanted to mix dice and playing cards...

I would refer you to the Binomial distribution, which calculates the probability of k successes in n tries with a p probability of success as such :
(n!/(k!(n-k!)) * p^k * (1-p)^(n-k) where n! = n * (n-1) * (n-2) * ... * 1 and a^2 = a²

But that's just the probability for, if we look at your example, the attacker rolling exaclty k successes.

The next step will be to calculate the cumulative probability (ie getting k successes or more).

Then you'll have to do the same thing for the defender.

Then you'll have to pit those two probability sets against each other, which right now I wouldn't know how to do, except by building a nice Excel sheet.

As you can tell, it's not an easy calculation, and there are better statisticians than me on this board to help you out

tomBitonti

Explorer
Hi,

For counting successes on independent rolls of a die, that gives you a binomial distribution, which looks more and more like a normal distribution with more dice rolls. A normal distribution is the classical bell shaped curve.

The feel of a normal distribution is that values towards the mid point are more likely than values towards the ends. A good example is comparing 3d6 with d20. 11 on a 3d6 is much more likely than 10 (or 11) on a d20, while 18 on a 3d6 (1 in 216) is much less likelly than a 20 on a d20 (1 in 20, about 10 times more likely).

The basic consequence is that results are much more predictable with a normal distribution.

Thx!
TomB