Computing probabilities involving increasing event outcomes Suppose I have an unfair die with three outcomes: $A,B,C$
\begin{align*}
A &\sim 1/512 \\
B &\sim 211/512 \\
C &\sim 300/512 \\
\end{align*}
The case of one dice is trivial. However, if I roll two of these dice, what is the probability of at least one of the dice having outcome $A$? 
I do not know how to tackle this problem since the outcomes in the sample space do not all have the same probability of occurring.
Thanks!
 A: The previous answers flesh out the standard take quite well. Here's another slant on these kinds of problems.
Treat the die faces as a polynomial:
$(\frac{a}{512}+\frac{211 b}{512}+\frac{300 c}{512})$
Squaring this represents rolling two (or twice), cubing it three, etc.
Squared, we get:
$(\frac{a^2}{262144}+\frac{211 a b}{131072}+\frac{75 a c}{32768}+\frac{44521 b^2}{262144}+\frac{15825 b c}{32768}+\frac{5625 c^2}{16384})$
This "encodes" the possible outcomes. The coefficients (like $\frac{211}{131072}$) are the probabilities of the different possible outcomes, and the variables and exponents encode the two faces seen and their multiplicity (e.g., $a^2$ means both were $a$).
So, adding up all the coefficients for where there is an $a$ component gives you the probability of getting at least one $a$.
Nicely, if your problem were two die (or more) with differing face probabilities, or differing number of faces, you simple multiply the polynomials representing each die as needed for the rolls, and same applies.
A: Laars method in the comments is the most straight-forward method, but if you want another method you could add up the other 5 probabilities that do include at least one A
$$\frac{1}{512}\cdot\frac{211}{512}+\frac{1}{512}\cdot\frac{300}{512}+\frac{1}{512}\cdot\frac{1}{512}+\frac{211}{512}\cdot\frac{1}{512}+\frac{300}{512}\cdot\frac{1}{512}$$
Or, lumping B and C together as "not A"
$$\frac{1}{512}\cdot\frac{511}{512}+\frac{1}{512}\cdot\frac{1}{512}+\frac{511}{512}\cdot\frac{1}{512}$$
In general, when you see the phrase, "at least one" it is a good i deal to just calculate the complement and subtract from one.
A: There are 4 ways your two dice can end up:
1.) First is $A$ and second is $A$, 
2.) First is not $A$ and second is $A$,
3.) First is $A$ and second is not $A$,
4.) First is not $A$ and second is not $A$.
Each die has $P(A)=1/512$ and $P(not A)=511/512$ so
1.) First is $A$ and second is $A$: $P(A)\cdot P(A)=(1/512)(1/512)=\frac{1}{512^2}$.
2.) First is not $A$ and second is $A$: $P(not A)\cdot P(A)=(511/512)(1/512)=\frac{511}{512^2}$.
3.) First is $A$ and second is not $A$: $P(A)\cdot P(notA)=(1/512)(511/512)=\frac{511}{512^2}$.
4.) First is not $A$ and second is not $A$: $P(not A)\cdot P(not A)=(511/512)(511/512)=\frac{511^2}{512^2}$.
The probability of at least one outcome being $A$ is the sum of the probabilities that satisfy the description. Thus:
$$P(\text{at least one }A)=\frac{511}{512^2}+\frac{511}{512^2}+\frac{511^2}{512^2}$$
Or equivalent it is the complement of everything that doesn't satisfy your description:
$$P(\text{at least one }A)=1-P(\text{no }A)=1-\frac{1}{512^2}$$
