Probability that the sum of 6 dice rolls is even

Question:

6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number?

I think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem?

Notice that whatever the sum of the first 5 rolls, whether the outcome is odd or even is totally determined by the last die. It is even or odd with equal probability, so the probability of an even sum is exactly the same as the probability of an odd sum.

The first 5 dice don't matter.

Edit to formalize this and address concerns in the comments:

Statement:

Let $X_1,\ldots,X_n$ be independent, integer-valued random variables, and let some $X_k$ satisfy $P(X_k \mbox{ odd}) = 0.5$. Then $P(\sum X_i \mbox{ odd}) = 0.5$.

Proof:

Without loss of generality let $k=n$, and let $S = \sum_i^{n-1} X_i$. Since $S$ and $X_n$ are independent random variables, $$ \begin{eqnarray*} P\left(\sum X_i \mbox{ odd}\right) = P(S + X_n \mbox{ odd}) &=& P(S \mbox{ odd and } X_n \mbox{ even}) + P(S \mbox { even and } X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ even}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ odd}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& \left(P(S \mbox{ odd}) + P(S \mbox{ even})\right)P(X_n \mbox{ odd}) \\ &=& P(X_n \mbox{ odd}) \\ &=& 0.5 \end{eqnarray*} $$

The only thing about the numbers rolled that matters is their parity - whether they are even or odd. In order to get an even sum, an even number of the six dice must be even. In order to get an odd sum, an odd number of the six dice must be even.

Using O for odd and E for even, we can list out the possibilities.

Even sum: