Conditional probability - sum of dice is even given that at least one is a five Question:

Calculate the conditional probability that the sum of two dice tosses is even given that at least one of the tosses gives a five.

I'm a bit confused by this. Shouldn't the probability just be 1/2, since we know that at least one of the dice tosses gave us a five, thus the other must give us an odd number?

If the dies were colored, say red and blue, we asked the question, what's chance that sum is even, given that the red die rolls a five, that's $1/2$. But, you see, either red or the blue die could roll a five. So, your sample space is slightly larger. 
Fix $X_{i}$ to be the random variable representing the number on die $i$, $i=1,2$. You could draw a nice tree to think of the possible outcomes of this experiment and count them.
(1) When the first die is rolled, there are three possibilities - $X_{1}=5$, $X_{1}=1 \cup X_{1}=3$, $X_{1}\text{ is even}$. 
$P(X_{1}=5)=1/6$. 
$P(X_{1}=1 \cup X_{1}=3)=1/3$. 
$P(X_{1} \text{ is even})=1/2$.
(2) The second die is now rolled, again having three possibilities each - $X_{2}=5$, $X_{2}=1 \cup X_{2}=3$, $X_{2}\text{ is even}$. So, we have $3 \times 3=9$ branches.
$P(X_{1}=5 \cap X_{2}=5)=1/6 \times 1/6=1/36$. 
$P(X_{1}=5 \cap (X_{2}=1 \cup X_{2}=3))=1/6 \times 1/3=2/36$.
$P(X_{1}=5 \cap X_{2}\text{ is even})=1/6 \times 1/2=3/36$.
$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}=5)=1/3 \times 1/6 = 2/36$. 
$P((X_{1}=1 \cup X_{1}=3) \cap (X_{2}=1 \cup X_{2}=3))=1/3 \times 1/3 = 4/36$. 
$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}\text{ is even})=1/3 \times 1/2 = 6/36$. 
$P(X_{1} \text{ is even} \cap X_{2}=5)=1/2 \times 1/6 = 3/36$.
$P(X_{1} \text{ is even} \cap (X_{2}=1 \cup X_{2}=3))=1/2 \times 1/3 = 6/36$.
$P(X_{1} \text{ is even} \cap X_{2}\text{ is even})=1/2 \times 1/2 = 9/36$.
Counting the outcomes where the sum is even and atleast one roll is 5, the numerator $1/36+2/36+2/36=5/36$.
Restricting our attention to events, where atleast one roll is 5, the denominator = $1/36+2/36+3/36+2/36+3/36=11/36$.
So, the chance is $5/11 < 1/2$.
