Can the information available affect the probability structure? I have a homework assignment on probability that goes like this:

Two cards are selected one after the other, at random and without
  replacement, from a well-shuffled, standard deck of 52 playing cards.
  Let A1 be the event that the first card selected is an ace, and let A2
  be the event that the second card selected is an ace. 
  What is the probability that the second card selected is an ace?

The solution argues that because we don't know what we picked first, the probability that the second card selected is an ace is still $\frac{4}{52}$. However, I am very confused by the idea that not knowing what was picked, is equivalent to assume that an ace was not picked.
Also, this leads to the following sort of paradox. Given two observers, one who knows what is on the first card, the other who does not, the probability of picking an ace on the second card changes from conditional to unconditional.
Instead, I would see the experiment as a case of conditional probability in both cases, which cannot be estimated unless we know what is on the first card.
 A: The solution, as you've stated it here, seems to be incorrect. Perhaps you've misunderstood what it was saying?
A nice way to think about problems of conditioning like this is with the law of total probability:
\begin{align}
\Pr(A_2)
  &= \Pr(A_2 \mid A_1) \Pr(A_1) + \Pr(A_2 \mid \overline{A_1}) \Pr(\overline{A_1})
.\end{align}
That is, we need to both consider both the cases that an ace was picked first, and that an ace was not picked; then we need to combine those two by the probability of the respective first events. From this formula, you can easily work out the probability:
\begin{align}
\Pr(A_2)
= \frac{3}{51} \times \frac{4}{52} + \frac{4}{51} \times \frac{48}{52}
= \frac{1}{17 \times 13} + \frac{16 \times 4}{17 \times 13}
= \frac{1}{13}
\end{align}
Another way to think about it is as whuber suggested in the comments: the second card, given that we don't know what the first card was, is simply going to be completely random out of the deck. How could any card be more likely than any other? Thus, more directly, we have
$$\Pr(A_2) = \frac{4}{52} = \frac{1}{13}.$$
Your "paradox," though, is not really a paradox. The observer who saw whether $A_1$ happened now has more information about whether $A_2$ will happen or not. Thus, from their perspective, the probability of $A_2$ has changed.
