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Suppose you throw a six face dice three times, how many times will be the sum of the faces even?

I approached it this way: You either get all three times even face, or twice odd and once even.

As there are only 3 faces that are even, you have $3^3$ possibilities.

Then, for the second situation, you have $3$ choices for then face, twice $3$ choices for the odd one. Thus giving again $3^3$ possibilities.

Overall, there will be $3^3 + 3^3$ possibilities, yet my textbook shows $3^3 + 3^4$ possibilities. What's wrong with my reasoning?

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    In the second case you must multiply by 3 since your first, second, or third roll can be even. In other words take $3 \choose 1$ to choose which roll will be even. Then your logic holds. – 2017-01-24

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For the second case we have only one even number that can be on any place. So we have 3 places. Then 3 ways.

So in second case $3 × 3^3 = 3^4$

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P(all even)$=\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{1}{8}$

P(one even and two odd)$=\dbinom{3}{2}\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{3}{8}$

Required probability $=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{1}{2}$