How to handle dice probability? ie, how much more likely would 3 six sided dice give a higher sum than 3 four sided dice? I am playing at making my own table-top gaming system/rules and I wanted to have a better handle on how likely different dice combinations will give a higher result than one another. I know that a six sided die roll averages to 3.5, and an eight sided die roll averages 4.5, but I still don't quite have a grasp on just how likely it is an 8 sided die comes up with a higher result than a 6 sided one. 
I would also like to know how adding integers to die results effects their comparative advantage as well, like how often would the sum of 3 six-sided dice with a 1 added to the final result give a higher outcome than just 3 six-sided dice? 
Thanks in advanced for any advice, I'm just not really sure where to start with this, I focused mostly on algebra/calc/trig in school and never really did any probability/stat.
 A: For your first question, six-sided die vs. eight-sided, make a $6$ by $8$ table, with values $1, 2, 3, 4, 5, 6$ in one direction and $1, 2, 3, 4, 5, 6, 7, 8$ in the other direction.  
The resulting $48$ small squares in the table determine $48$ possible outcomes of the two dice.  You can then see for how many squares does six-sided beat eight-sided; how many ties; and how many times eight-sided beats six-sided.  Assuming the dice are fair, you can then divide by $48$ to get the desired probabilities.
A: There are online utilities that can do the calculations for you. This one for example looks good to me: you can combine dice rolls (subtract 1d6 from 1d8 for example) to get a probability curve for all the possible results.
A: Because you're making a gaming system, I guess you might be
comfortable with some simulations that require little
programming. Here is a program with an approximate answer to your
question about 3 rolls of a six-sided die with and without
a bonus point. (The third decimal place may be off by 1 or 2.)
The probability is about 0.546
that three rolls with a bonus point beats three rolls without.
(About 0.36 of being greater than or equal.)
m = 10^6;  x1 = x2 = numeric(m)
for(i in 1:m) {
  x1[i] = sum(sample(1:6, 3, rep=T))
  x2[i] = sum(sample(1:6, 3, rep=T)) + 1 }
mean(x2 > x1);  mean(x2 >= x1)
## 0.546143
## 0.636343
mean(x2 - x1)
## 1.000961  # Reality check: theoretical value 1

This kind of simulation might make it very easy for you
to investigate various options quickly and informally.
Of course, you'd have to do the math for exact probabilities
to many places.
[The small program above is written in R, excellent statistical
software available free at r-project.org.]
