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: By the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Let $A$ be the event of the two rolls adding to an even number and let
 $B$ be the event of rolling at least one five. 
Note that $A \cap B$ = the event of the two rolls adding to an even number AND rolling at least one five. Now we need to find $P(A \cap B)$ and $P(B)$:
There are three ways we can roll at least one five.
1) You roll a five on the first roll but not the second. The probability of this is $(1/6)*(5/6) = 5/36.$ 
2) Not rolling a five on the first roll, but rolling  a five on the second: Probability of this is $(5/6)*(1/6) = 5/36.$ 
3) Or you can roll two fives: Probability $= (1/6)*(1/6) = 1/36.$ 
These three events are disjoint, so to find the probability of getting $\textit{at least}$ one five, we add them: $5/36 + 5/36 + 1/36 = 11/36$. 
Now to calculate $P(A \cap B)$: If you roll two fives, the sum is even. Otherwise, the roll that is not a five must be an odd number (ODD + ODD = EVEN). In each of cases 2) and 3), the only other rolls that give you an even sum are 1 and 3. That's a total of 5 scenarios out of a possible 36. So $P(A \cap B) = 5/36.$  Finally,
$$P(A|B) = \frac{5/36}{11/36} = \frac{5}{11}.$$
A: As suggested above, the Bayes th. can help too:
Let $S$ be the event that the sum is even, and $A5$ the event that at least one $5$ shows up, and $S^c$ is the event that the sum is odd.
Then, 
$$P(S|A5) = \frac{P(A5|S)P(S)}{P(A5|S)P(S)+P(A5|S^c)P(S^c)}
$$
$P(S)=P(S^c)=0.5$, $P(A5|S)= 5/18$, as there are $5$ in $A5$ for even sum. Also, $P(A5|S^c)= 6/18$ when the sum is odd.
Thus, $P(S|A5) = 5/11$.
A: 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 $5/11 < 1/2$.
A: 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
A: It depends upon how you found out that at least one of the dice is 5.
Are you familiar with the children’s game Guess Who? Given a set of 24 characters, you have to guess which one is mine by asking yes/no questions. Suppose you ask me “Does your person have glasses?” And I answer yes. Since most of the characters do not have glasses, there’s a lot of information in my answer. Out of 24 characters, only 6 remain. 
It’s a similar principle with the dice. If you ask, “is at least one of them 5”, and I answer yes, you eliminate 25 possibilities and leave 11. Whereas the original 36 possibilities have an equal distribution of odds and evens, the remaining 11 do not. 
By asking a question, you learned something about the result of rolling the dice. You’ve narrowed the possibilities, and gotten a little closer to knowing for sure whether the sum is even. 
If you didn’t ask about a particular number, but I just looked at one of the dice and said “at least one of them is X”, then I haven’t given you any information. I don’t even know the answer myself, since I only looked at one of the dice. In that case, the answer is 1/2. 
A: A visual explanation. Given that a pair is in the right half, what’s the probability that it’s in the top half?
11    13                           15      
   22    24    26        
31    33                           35
   42    44    46
                       51    53    55
   62    64    66


   12    14    16      
21    23                           25
   32    34    36
41    43                           45
                          52    54    56
61    63                           65

```

A: 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 $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$.
