## Friday, January 30, 2015

### On the Average Results of Savage Worlds Dice

Something occurred to me the other day, regarding the average results of dice in Savage Worlds.  For those who don't know, in Savage Worlds, each die (with the possible exception of damage dice - I forget) can explode.  For those who don't know what that means, if a die rolls its highest result, that result is kept and then added to a new roll of that die.
Obviously, the minimum result of any roll (except 1d4-2) is a 1, and the maximum result is an arbitrarily large number.  However, what's the average result of any die roll?

Consider a 1d6 roll, the most common roll in Savage Worlds.
Let x be the average result, which we seek.

$x&space;=&space;\frac{1+2+3+4+5+(6+x)}{6}$

We're re-using x on the right, since the average roll being added to 6 in the best case is, itself, the average case of a new die roll.  Thus, we can use basic algebra to solve for x:
$6x=1+2+3+4+5+6+x$

$5x=1+2+3+4+5+6$

$x=\frac{\sum_{n=1}^{6}n}{5}$

In this specific case, the result is x = 21/5 = 4.2.  However, I've left the sum to reveal the generalized form.  For a die of size k, the average result is
$x_{k}=\frac{\sum_{n=1}^{k}n}{k-1}$
While I don't think that this is new knowledge, I also suspect that tables are commonly referenced, so I thought that I'd put this generalized form out there.

EDIT:  I had a botched example result; I was thinking of the d8 case.
EDIT:  I discovered that the fancy math stuff is rendering weirdly on Blogger.  I'll fix it later; for now, you can click on it if it's illegible:  The page where you land will have it (correctly) in about the middle.