Problem
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.
I gave a try on this question, and I know there is exact-same question online but they solved by conditioning on whether or not the interchanged card is selected. But I did it in the other way. This is my attempt.
Let $D_i$ be the event that the original deck (where the first ace was drawn) contains exactly $i$ Aces, for $i\in \{1,2,3,4\}$, and $A$ be the event where ace was drawn again. Then we have:
$$P(A)=\sum_{i=1}^{4}P(A|D_i)P(D_i),$$ where $P(D_i)$ can be computed as follows: $$\frac{{4\choose i} {48\choose{26-i}}}{{52\choose{26}}}.$$ Whereas for $P(A|D_i)$, for example $P(A|D_1)=\frac{4}{27}$, since there is originally one A in the deck (where the first ace was drawn), this means there are 3 Aces in the unshuffled second deck of 26 cards. After adding the Ace into the second deck, there are 4 Aces out of 27 cards. Similarly, $P(A|D_i)=\frac{5-i}{27}$. After substituting all the values, I get $P(A)=\frac{2269}{22491}=0.10088$, but the actual answer is $\frac{43}{459}=0.0937.$
May someone please help on this? All the works are much appreciated. Thank you. (By the way, if anyone thinks that I need to include all the steps, please comment.)