Probability of Getting a 3 on a die on the 2nd throw knowing that 3 only appears on an even throw The question in the title (A die is repeatedly thrown until a 3 appears. Knowing that 3 only appears on an even roll, what is the probability that a 3 appears on the 2nd roll?) was a question in an exam.
Intuitively, I thought that it is only $1/6$ because it is already impossible to get the $3$ on the 1st roll, so it is just equal to the probability that we get a 3 when we roll the die the 2nd time, which is $1/6$ (Another way to see it is that this case is equivalent to a case where we are throwing the die repeatedly to get a 3 while denoting the rank of each throw as $({2,4,6,...})$instead of $({1,2,3,...})$ and then asking what the probability of getting a $3$ at rank $2$ is, which is only $1/6$).
However, after the exam, I thought of just applying the conditional probability formula and got a completely different answer (which was the right one). The probability of getting 3 at the 2nd throw and at an even roll is nothing bu the probability of getting $3$ on the 2nd roll, which is $5/36,$ and the probability of getting 3 at an even number (calculated by geometric series) is $5/11,$ so dividing these numbers gives us the final (correct) answer of $11/36.$ Any idea where my logic was flawed to give a wrong answer?  
 A: There is a difference between the following two questions and probabilities:
When you keep throwing dice until you get a 3, and knowing that you end up stopping after an even number of throws, what is the probability that you got the 3 that ended this process on the second throw?
When you keep throwing dice until you get a 3, what is the probability of getting a 3 on the second throw, given that the first throw is not a three?
In the second case the probability is clearly just $\frac{1}{6}$. This is how you thought about the problem initially.
But in the second case you know something more: You know that you stopped after an even number of throws. Knowing this will lower the probability of having gone beyond the second throw, since otherwise you could have thrown a 3 in the third throw. Similarly, the probability of having gone beyond the fourt throw gets lowered as that opens the door for throwing a 3 on the fifth throw, etc. Hence, the probability of having ended the process at the second throw gets increased. Thus, the probability asked about in case 1 is higher than in case 2.
A: Let $R$ count the rolls until we obtain the first 3.   The answers you obtained are actually:
(1) $\mathsf P(R=2\mid R\neq 1) = 1/6$
(2) $\mathsf P(R=2\mid R\in 2\Bbb N) = \dfrac{5/36}{5/11}=\dfrac{11}{36}$
Thus the first answer does not exclude the possibility of obtaining a three on odd rolls after the first should the favoured event fail to happen, which we are informed is somehow prohibited.   The second answer correctly evaluates the probability under the required condition.

Any idea where my logic was flawed to give a wrong answer? 

You assumed that on an even roll, the probability for obtaining a three would remain $1/6$ even when you know the event can only happen on an even roll.
However, knowledge that a result of three will occur on some even roll, and is prohibited on all odd rolls, should increase the expectation for obtaining it on any particular even roll.
A: You have a 5 sided dice, with {1, 2, 4, 5, 6} and a 6 sided dice with {1, 2, 3, 4, 5, 6}.
Which dice do you throw on the 2nd turn?  What is the probability of getting a 3 when that dice is thrown? Do the results of other turns affect that probability?
A: You are mixing two questions.
The answer to the question in which 3 on 2nd roll is $\frac 5{36}$
And answer of getting 3 on even roll is  $\frac 5{11}$
