Show there are $\frac{6^n}{2}$ possible even results from the sum of $n$ dice rolls. I was requested to show there are $\frac{6^n}{2}$ possible even results from the sum of $n$ dice rolls. I did it via induction and was wondering whether my demonstration is correct.
$I$. Let $a_n=\{a_1, a_2,..., a_n\}$ be the sequence of outcomes out of $n$ dice rolls. Let $s_n=\sum_{i=1}^n a_i$ be the sum of all $n$ elements of $a_n$.
$II.$ If $n=1$ we have $s_1=a_1$, a single result whose possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. It is trivial to see $3=\frac{6^1}{2}$ possible outcomes are even and the property holds for the base case.
Let us assume the number of possible even outcomes of $s_k$, the sum of $k$ dice rolls, is $\frac{6^k}{2}$. Then notice that
$$s_{k+1}=s_k+a_{k+1}$$
By inductive hypothesis, there are $\frac{6^k}{2}$ even outcomes for $s_k$. There are $\frac{6^1}{2}$ even outcomes for $a_{k+1}$, a single dice roll. For the sum of both terms to be even, either both terms are odd or both terms are even. There are
$$\frac{6^k}{2}\times \frac{6^1}{2}+\frac{6^k}{2}\times \frac{6^1}{2}=\frac{6^{k+1}}{2}$$
possible ways for this to occurr. Therefore there are $\frac{6^{k+1}}{2}$ possible even outcomes of $s_{k+1}$ and the proof by induction concludes.
Two questions. $A)$ Is this proof correct? $B)$ What is an equivalent proof using probability theory instead of discrete mathematics/induction?
 A: The question seems ambiguous to me.
After all, isn't the result $6,4,3,1,5,3=22$ the same thing as $6,2,4,6,3,1=22$?
Well, never mind.  If we interpret as it is clearly intended then note there $6^n$ possible rolls and "clearly" an equal number of them are even as odd so there are $\frac {6^n}2$ even results.
If you don't trust that "clearly" (then you are a discerning mathematician with a future ahead of you) you can note there are $6^{n-1}$ ways to roll the dice $n-1$ times.  If the result after $n-1$ rolls is odd then there are $3$ possible outcomes for then $n$th roll, namely $1,3,5$ to make the sum of $n$ rolls even.  If the result after $n-1$ rolls is even then there are $3$ possible outcomes, $2,4,6$, for then $n$th roll to make the result even.
Either way, there are $6^{n-1}\times 3 = \frac {6^n}2$ ways for the result to be even.
Or alternative, you can note if the $k$th dice roll is $a_k$ and the $n$ dice rolls are $a_1, ...., a_n$.  Now for any odd roll $(a_1, ...., a_{n-1}, a_n)$ we can map that uniquely to an even roll via$(a_1, ...., a_{n-1}, a_n)\mapsto(a_1, ....., a_{n-1}, (a_n + 1)\%6)$.  And for every even roll $(b_1, ....., b_{n-1}, b_n)$ that will be uniquely mapped from $(b_1, ....., b_{n-1}, (b_n -1)\%6)\mapsto (b_1, ....., b_{n-1}, b_n)$.  So each odd roll is unique matched one-to-one to each even roll.
So the "clearly" has been justified.
And our argument that there are $6^n$ total rolls and $\frac 12$ of them are even so there are $\frac{6^n}2$ even rolls is now airtight.
