Math Analysis: Counting Successes

[The Crimson Binome]

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]

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]

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