Probability of an even number of sixes 
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.

I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:


*

*The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes. 

*Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$. 
For $n=k+1$, there are two ways the number of sixes are even:
a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n]$
b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$
I have two questions


*

*How do I get from $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$

*Other than that, is my use of induction correct? Is it rigorous enough to prove the formula? 

Your method is correct but the induction hypothesis should be
$$\text{Probabaility of an even number of sixes when you roll } k \text{ dice} = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^k\right]$$
(you have $n$ instead of $k$) and the thing you want to prove is 
$$\text{Probabaility of an even number of sixes when you roll } k+1 \text{ dice} = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^{k+1}\right]$$
and so, using your reasoning, you want to prove that
$$\frac{5}{6}\cdot \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^k\right] + 
\frac{1}{6}\left(1- \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^{k}\right]\right) = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^{k+1}\right]$$
[Also, two footnotes: odds isn't the same as probability, so the words shouldn't really be used interchangeably. And $0$ is even! Whoever wrote the question is probably a bit shaky in number theory...]
