Probability Refresher - Help I have a suspicion an algorithm on a video game I play assigns an equal chance of "rolling" a stat value between 90 to 146, i.e. (57 - edit) total possibilities with a probability of 1/(57)th each. However, I want to know how many different ways there are to combine any two numbers between the values of (45-73) to reach each integer from 90 to 146.
This is basically a two-dice probability problem writ large, and I can't for the life of me remember the formulas from high school to do it myself. I'd like to see the distribution of probabilities given this large range.
I can obviously start off by saying 
90 can only be made once using integers from 45-73, (45, 45)
91 can only be made twice using integers from 45-73, (45, 46) (46, 45)
92 can only be made three ways using integers from 45-73, (45, 47) (47, 45) (46, 46)
93 can only be made four ways using integers from 45-73, (45, 48) (48, 45) (46, 47) (47, 46)
94 can only be made five ways using integers from 45-73, (45, 49) (49, 45) (46, 48) (48, 46) (47, 47)
95 can only be made six ways using integers from 45-73, (45, 50) (50, 45) (46, 49) (49, 46) (47, 48) (48, 47)
And so on...but obviously these numbers become much greater as we reach cumulatives of 100+. There must be a way of making a table using an excel program or using a simple formula to find out how many ways each increasing integer from 90 can be made using the subset (45-73). Ideally, I'd also like to calculate the probability of rolling a given total (90-146), given an equal probability of either die rolling (45-73). 
So I know to start: 90: 1/784 chance, 91: 1/392 chance, 92: 3/784, 93: 1/196 - but I'd want to know the rest of the table to 146 without thinking of literally every combination.
This is incredibly redundant, no doubt, but i'm just making sure my bases are covered - it's been 7 years since I've really done academic math. In conclusion, a way to easily calculate and display the probability of rolling certain outcomes on two "(29 - edit) sided dice" so to speak.
Thanks very much.

Note: Each of your two 'dice' with faces $45, 46,\dots, 73$ has $29$ possibilities; so your denominator is $29^2=841$ (rather than $28^2=784$).
You get a 'triangular distribution' of probability with probability $\frac{1}{841}$ for each of the sums $90$ and $146$; $\frac{2}{841}$ for each of the sums $91$ and $145$; ...; up to a maximum probability of $\frac{29}{841}$ for the sum of $118$.
So the probability of getting a sum of $n$ is given by $P(n)=\frac{29-|118-n|}{841}$ for $n=90,91,92,\dots,146$.
