If you roll two six-sided dice, what is the probability that the dice add to 10 or higher? When answering these sort of questions people mostly resort to diagrams and I'm wondering if there is a way to calculate the probability without going through each outcome, just solely on the given variables.
 A: Sure you may do that. Call the result of the first die $X$ and the second die $Y$ (presuming you can identify the die).
$$\begin{align}\mathsf P(X+Y\geq 10) ~=&~ \sum_{x=1}^6\mathsf P(X=x)\mathsf P(Y\geq 10-x) \\[1ex]=&~\sum_{x=4}^6\;\sum_{y=10-x}^6 \frac 1{36} \\[1ex]=&~\sum_{k=1}^3\;\sum_{h=1}^{h}\frac 1 {36} \\[1ex]=&~\frac{1+2+3}{36}\\[1ex]=&~\frac{1}{6}\end{align}$$
But for this exercise, listing outcomes really is easiest.
A: Comments:
This is a very perceptive question. And the answer of @GrahamKemp illustrates an important convolution
formula that works quite generally for sums of independent
random variables. 
For example, suppose you put a lead weight into each die just beneath the corner
where faces 1, 2, and 3 meet in order to bias the die in
favor of 4, 5, and 6. Perhaps the result is that faces
1, 2, and 3 each have probability 5/36 and faces 4, 6, and 6
each have the probability 7/36. You can
use the convolution formula to answer the same question.
(For biased dice, not all cells in a $6 \times 6$ diagram are equally likely.)
Also, simulation is sometimes useful to get good approximations---
and even to check analytic solutions. This can be especially
useful for messier problems. Here is an example of a
simulation (using R statistical software) based on a million rolls of two fair dice,
colored red and blue. The answer should be correct to three places.
red = sample(1:6, 10^6, rep=T)
blue = sample(1:6, 10^6, rep=T)
total = red + blue
mean(total >= 10)
## 0.16684

For dice biased as above, one would use 
bias = c(5,5,5,7,7,7)/36
red = sample(1:6, 10^6, rep=T, prob = bias)
...

to get the approximation 0.226999 (or on another run 0.22651).
Exact answer seems to be $6(7/36)^2 = 0.22685.$
