Probability of having an even number of a specific face value of a die Here is the question: Given a fair die rolled $n$ times, what is the probability of having an even number of $6$s?
My approach: 
$$\Omega = \{(x_{1}, x_{2}, ..., x_{n}) :1 \le  x_{i} \le 6\}$$
So $$|\Omega| = 6^{n}$$
Since the probability is space is a finite uniform one, $$P(A) = \frac{|A|}{|\Omega|}$$
Let $A_{i}$ be the event that there are exactly $i$ sixes in the $n$-tuple. Then, $$A = \bigcup\limits_{i=0}^{\lceil{(n-1)/2}\rceil} A_{2*i}$$ Since these events are disjoint, $$P(A) = P(\bigcup\limits_{i=0}^{\lceil{(n-1)/2}\rceil} A_{2*i}) = \sum\limits_{i=0}^{\lceil{(n-1)/2}\rceil} P(A_{2*i})$$
Now, $|A_{i}| = \binom{n}{i} * 5^{n-i}$. Hence, $$P(A) = \frac{1}{6^n} \sum\limits_{i=0}^{\lceil{(n-1)/2}\rceil} \binom{n}{2i} * 5^{n-2i}$$ As a side note, when I fed this sum into wolfram alpha engine, it couldn't compute it.
The book however says the anwer is $$\frac{1}{2}(1 + (\frac{2}{3})^{n})$$
Any suggestions?
 A: Your approach is fine and correct, but may not yield the most convenient form of the answer.  Instead, perhaps you could approach via a recurrence relation.  Let $p_n$ be the probability of having an even number of sixes after $n$ rolls.
For $n+1$ rolls of the dice, for there to be an even number of sixes the final die must be a six and the previous $n$ dice must have an odd number of sixes or the final die must not be a six while the previous $n$ dice must have an even number of sixes.
This yields the recurrence relation: $p_{n+1} = \frac{1}{6}(1-p_n)+\frac{5}{6}p_n = \frac{1}{6}+\frac{2}{3}p_n$
Using an initial condition of $p_0=1$ (or the initial condition $p_1=\frac{5}{6}$ if you dislike starting at zero for whatever reason), we can solve the recurrence relation for a closed form solution which will agree with the answer given by the book.

$p_{n}=\frac{1}{6}+\frac{2}{3}p_{n-1}$
characteristic polynomial of associated homogeneous equation:  $x-\frac{2}{3}=0$, thus the general part is of the form $c_1 (\frac{2}{3})^n$
non-homogeneous piece is a constant (and no general part is of the same form) so particular piece is also a constant.  Solving for that constant:  $k = \frac{1}{6}+\frac{2}{3}k \implies \frac{1}{3}k=\frac{1}{6}\implies k=\frac{1}{2}$
So, we have $p_n = \frac{1}{2}+c_1(\frac{2}{3})^n$.  Using our initial condition that $p_0=1$ we find $c_1=\frac{1}{2}$
Thus:  $$p_n=\frac{1}{2}+\frac{1}{2}\cdot (\frac{2}{3})^n$$
