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?
A = event when one of the tosses gives a five. (Sample space for the conditional probability)
Let (n1, n2) represent the outcomes of die1 and die2
A = { (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), 
      (5,1), (5,2), (5,3), (5,4), (5,6) } // (5,5) must be counted once only 

Thus n(A) = 11

B = sum of two dice tosses is even
n(B|A) = { (1,5), (3,5), (5,5), (5,1), (5,3)|

P(B|A) = n(B|A)/n(A) = 5/11
Examine this sample space

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),6,6)}

If you did not know anything about the two die, then probability of having even sums when you have at least a 5 is 5/11. Your intuition for 1/2 however requires you to have some information about which die has turned out a 5. If you knew for sure everything about column 5 (or row 5) of this matrix, i.e. {(...,5), (...,5), (...,5), (...,5), (...,5), (...,5)}, this means you know which die has a five (first or second die). Then you only have row 5 (or column 5) to deal with. And in that case you probability is 1/2, i.e. your outcomes are: 

$\{(5,1), (5,3), (5,5)\}$ 

In a sample space of $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$.