Probability that the total of three rolls of a fair die is odd A fair die is thrown three times.  Events A, B, C are defined as follows:
A:The total score is an odd number. 
B:A six appears as the first throw
C:The total score is 13.
My question is how to find the probability of event A?  
 A: Here is how to do it for two dice. Consider the following table:
$$
\begin{align*}
2 &= 1+1 \\
3 &= 1+2=2+1 \\
4 &= 1+3=2+2=3+1 \\
5 &= 1+4=2+3=3+2=4+1 \\
6 &= 1+5=2+4=3+3=4+2=5+1 \\
7 &= 1+6=2+5=3+4=4+3=5+2=6+1 \\
8 &= 2+6=3+5=4+4=5+3=6+2 \\
9 &= 3+6=4+5=5+4=6+3 \\
10 &= 4+6=5+5=6+4 \\
11 &= 5+6=6+5 \\
12 &= 6+6
\end{align*}
$$
You can diligently count that there are $18$ ways of obtaining odd sums, so the probability is $18/36 = 1/2$.
This was the brute-force computation, but the fact that the answer is nice prompts us to look for a simple reason that the answer is half. Indeed, let $X,Y$ be the values of the two dice; I claim that no matter what the value of $X$ is, the probability that $X+Y$ is odd is exactly $1/2$. Why? If $X$ is odd, then $X+Y$ is odd if $Y=2,4,6$; while if $X$ is even, then $X+Y$ is odd if $Y=1,3,5$. Can you generalize this to three dice?
A perhaps more striking way is to consider the dice modulo $2$, which turns them into fair coins. Now the question, if $X,Y$ are chosen randomly from $\{0,1\}$, what is the probability that $X \oplus Y = 1$? It's not hard to check that the probability is $1/2$.
A: Whatever the parity of the sum of the first two throws, the probability that the last throw has the opposite parity is obviously $\frac12$.
