Problem
A deck of card is shuffled 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.
This is how I count based on the hint given in this thread Question regarding Conditional Probability with a deck of card.
Assuming the 27-half deck will contains from 1, 2, 3, 4 aces, then I conditioned on how we can distribute the a 52 cards into 2 decks, each has 26 cards. Let $A$ denote the event that the card chosen from this 27 half-deck is ace, and $D_i$ denotes this second half deck (before adding an ace) contains $i$ aces.
$$P(A \mid D_0)P(D_0) + P(A \mid D_1)P(D_1) + P(A \mid D_2)P(D_2) + P(A \mid D_3)P(D_3)$$
where
$$P(A \mid D_0)P(D_0) = \dfrac{1}{27} \cdot \dfrac{\binom{48}{26}}{\binom{52}{26}} = \dfrac{46}{22491}$$
$$P(A \mid D_1)P(D_1) = \dfrac{2}{27} \cdot \dfrac{\binom{4}{1}\binom{48}{25}}{\binom{52}{26}} = \dfrac{416}{22491}$$
$$P(A \mid D_2)P(D_2) = \dfrac{3}{27} \cdot \dfrac{\binom{4}{2}\binom{48}{24}}{\binom{52}{26}} = \dfrac{325}{7497}$$
$$P(A \mid D_3)P(D_3) = \dfrac{4}{27} \cdot \dfrac{\binom{4}{3}\binom{48}{23}}{\binom{52}{26}} = \dfrac{832}{22491}$$
Thus, $$P(A) = \dfrac{46}{22491} + \dfrac{416}{22491} + \dfrac{325}{7497} + \dfrac{832}{22491} = \dfrac{2269}{22491}$$
The correct answer is $\dfrac{43}{459}$. My answer is off by 0.007 percent, but I couldn't figure out why it was wrong. I understand the way in the book, but I just want to see how can I count it in a different way. Any idea?
Update
Following the hints given by Didier Piau and TonyK, I got:
$$\dfrac{1}{27} \cdot \dfrac{\binom{48}{26}}{\binom{51}{26}} = \dfrac{92}{22491}$$
$$\dfrac{2}{27} \cdot \dfrac{\binom{3}{1}\binom{48}{25}}{\binom{51}{26}} = \dfrac{208}{7497}$$
$$\dfrac{3}{27} \cdot \dfrac{\binom{3}{2}\binom{48}{24}}{\binom{51}{26}} = \dfrac{325}{7497}$$
$$\dfrac{4}{27} \cdot \dfrac{\binom{3}{3}\binom{48}{23}}{\binom{51}{26}} = \dfrac{416}{22491}$$
$$P(A) = \dfrac{92}{22491} + \dfrac{208}{7497} + \dfrac{325}{7497} + \dfrac{416}{22491} = \dfrac{43}{459}$$