Probability that the sum of 6 dice rolls is even

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.

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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:

• OOOOOO $$\binom{6}{6}=1 \text{ arrangement}$$
• OOOOEE $$\binom{6}{4}=15 \text{ arrangements}$$
• OOEEEE $$\binom{6}{2}=15 \text{ arrangements}$$
• EEEEEE $$\binom{6}{0}=1 \text{ arrangement}$$

This gives a total of $32$ arrangements with even sum.

Since there are $2^6 = 64$ total possibilities, we see that your intuition of $50\%$ is correct.

The generating function approach:

$$P(x)=(1+x+x^2+x^3+x^4+x^5)^6=\sum a_i x^i$$

Then $a_i$ counts the number of ways of getting a total of $i+6$ from $6$ dice.

Now, to find the even terms, you can compute $$\frac{P(1)+P(-1)}{2}=\sum_i a_{2i}.$$

But $P(1)=6^6$ and $P(-1)=0$. So $$\frac{P(1)+P(-1)}{2}=\frac{6^6}{2},$$ or exactly half, as you conjectured.

For another example, let $N_{i}$ be the number of ways to roll $6$ dice and getting a value $\equiv i\pmod{5}$. Then it turns out that if $z$ is a primitive $5$th root of unity, then the value can be counted by defining:

$$Q_i(x)=x^{6-i}(1+x+x^2+x^3+x^4+x^5)^6$$ then computing $$N_i=\frac{Q_i(1)+Q_i(z)+Q_i(z^2)+Q_i(z^3)+Q_i(z^4)}{5}$$

This gives the result:

$$N_i =\begin{cases}\frac{6^6+4}{5}&i\equiv 1\pmod 5\\ \frac{6^6-1}{5}&\text{otherwise} \end{cases}$$

More generally, if $N_{n,i}$ is the number of ways to get $\equiv i\pmod 5$ when $n$ dice are rolled, you get:

$$N_{n,i} =\begin{cases}\frac{6^n+4}{5}&i\equiv n\pmod 5\\ \frac{6^n-1}{5}&\text{otherwise} \end{cases}$$

It's this simple because of the fact that $6=5+1$.

If each die has $d$ sides, and you ask what are the number of ways to get a total $\equiv i\pmod {d-1}$, then you get:

$$N_{d,n,i} =\begin{cases}\frac{d^n+{d-2}}{d-1}=\frac{d^n-1}{d-1}+1&i\equiv n\pmod {d-1}\\ \frac{d^n-1}{d-1}&\text{otherwise} \end{cases}$$