I've got confused about the question. There are several same questions, still I got confused.
A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.
My thought is: As the distribution of the initial condition is equally likely, we have probability $ \frac{\left( \begin{array}{c} 26\\ 1\\ \end{array} \right) \left( \begin{array}{c} 26\\ 3\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{1}{26} $ of dividing it into "1,3", and take 1 outo $\frac{\left( \begin{array}{c} 26\\ 2\\ \end{array} \right) \left( \begin{array}{c} 26\\ 2\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{2}{26},\frac{\left( \begin{array}{c} 26\\ 3\\ \end{array} \right) \left( \begin{array}{c} 26\\ 1\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{3}{26},\frac{\left( \begin{array}{c} 26\\ 4\\ \end{array} \right) \left( \begin{array}{c} 26\\ 0\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{4}{26} $ of other three. The denominator should be the sum of them. Thus, by conditional probability, the result is $ \frac{\frac{\left( \begin{array}{c} 26\\ 1\\ \end{array} \right) \left( \begin{array}{c} 26\\ 3\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{1}{26}*\frac{4}{27}+\frac{\left( \begin{array}{c} 26\\ 2\\ \end{array} \right) \left( \begin{array}{c} 26\\ 2\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{2}{26}*\frac{3}{27}+\frac{\left( \begin{array}{c} 26\\ 3\\ \end{array} \right) \left( \begin{array}{c} 26\\ 1\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{3}{26}*\frac{2}{27}+\frac{\left( \begin{array}{c} 26\\ 4\\ \end{array} \right) \left( \begin{array}{c} 26\\ 0\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{4}{26}*\frac{1}{27}}{\frac{\left( \begin{array}{c} 26\\ 1\\ \end{array} \right) \left( \begin{array}{c} 26\\ 3\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{1}{26}+\frac{\left( \begin{array}{c} 26\\ 2\\ \end{array} \right) \left( \begin{array}{c} 26\\ 2\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{2}{26}+\frac{\left( \begin{array}{c} 26\\ 3\\ \end{array} \right) \left( \begin{array}{c} 26\\ 1\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{3}{26}+\frac{\left( \begin{array}{c} 26\\ 4\\ \end{array} \right) \left( \begin{array}{c} 26\\ 0\\ \end{array} \right)}{\left( \begin{array}{c} 52\\ 4\\ \end{array} \right)}*\frac{4}{26}} $
It seems that this answer is not equal to the correct probability. Where do I go wrong?
I seems to me that your answer is correct.
For $k=0,1,2,3,4$ let $A_{k}$ denote the event that there are $k$ aces among the $26$ cards from which a card is drawn at first hand.
Let $E$ be the event that this first drawn card is an ace.
Let $F$ denote the event that the second card drawn is an ace.
Then $P\left(E\right)=\frac{1}{13}$ and to be found is $P\left(F\mid E\right)=P\left(F\cap E\right)/P\left(E\right)=13P\left(F\cap E\right)$.
This can be solved by:$$P\left(F\mid E\right)=13P\left(F\cap E\right)=$$$$13\sum_{k=0}^{4}P\left(F\mid E\cap A_{k}\right)P\left(E\mid A_{k}\right)P\left(A_{k}\right)=13\sum_{k=0}^{4}\frac{5-k}{27}\frac{k}{26}\frac{\binom{26}{k}\binom{26}{4-k}}{\binom{52}{4}}=$$$$\frac1{54}\binom{52}{4}^{-1}\sum_{k=1}^{4}(5-k)k\binom{26}{k}\binom{26}{4-k}$$
This answer agrees with the (nice!) answer of Daniel Mathias (check it e.g. in Excel) and only differs from yours by the fact that you took a more complicated route by calculation of $P(E)$.
There you went for: $$P(E)=\sum_{k=1}^4P(E\cap A_k)=\sum_{k=1}^4P(A_k)P(E\mid A_k)$$ which also results in $P(E)=\frac1{13}$