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Re: [DL] Dice Rollin'



> No problem what-so-ever.  If my math was incorrect, I want to know ... one
> of the reasons I posted it.  Obviously, I didn't then do a proof.  :-)  My
> numbers might be off when going above the max for a die as well ... I based
> it on the chance of getting an Ace, and may not have taken into account the
> chance of getting more than one Ace.  If so, that would mean that the
> percentage chances are all slightly higher ... but not by much.

I assume that you use:
pd = chance of dice succeeding
p = chance of roll succeeding
n = number of dice
p = 1 - (1 - pd)^n
Or, in words, the chance of not all dice failing
That takes care of everything. If not all dice fail,
then at least one succeed, possibly more but that
doesn't matter (in DL/HoE, it matters in other
games, that uses successes, like Vampire or Shadowrun). 
Your numbers are correct as far
as I can tell (no, I didn't check them all... ;-),
with the exception of bustin', but as you clearly
stated that you didn't handle that.

The above formula is appealing since it is simple
and takes care of everything except a bustin'. 
You need to do the combinatorial approach to 
solve that one, at least that is the only solution 
that I have found. The problem is that you need
to compute the number of results that are busts
but beat the TN.

Let's do the 3D10 vs 7 case.

There are 28 bust results.
A "succeeding bust" result is:
1-1-(7-10)       4 combinations
And there are 3 dice that can end up at 7+,
4 * 3 = 12
The chance of getting a succeding bust is 1.2%
Now, the chance of rolling 7+ on a single D10
is 40%, on 3D10 it is 78.4% (1 - 0.6^3) but
you also need to avoid the bust chance so the
final probability ofs success is: 
0.784 * (1 - 0.012) = 0.774592 ~= 77.46%

> I understand this, but I only see 28 combinations.  What am I missing?

Absolutely nothing, I somehow managed to get 3 * 9 to 29...
No matter how much math you have studied, you can always
screw up the basics... ;-(
Yes, it should be 28, and the bust chance is 2.8%.

/Johnny