Question regarding Conditional Probability with a deck of card 
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.

My attempt was,
The second half which contains 27 cards could have:


*

*1 ace

*2 aces

*3 aces

*4 aces


Hence the total probability of drawing an ace from this half is:
$$\dfrac{ 1 + 2 + 3 + 4 }{27}$$
However the solution from the book is $$\dfrac{43}{459},$$ and I have no idea how they came up with this number. I personally think this problem is straightforward, but I might have missed something. Could anyone give me a hint? Thank you.
 A: Your answer is more than $1/3$, which is clearly not reasonable. If the four cases that you enumerate were equally likely, the desired probability would be $$\frac14 \cdot \frac{1+2+3+4}{27} = \frac{5}{54},$$ which is considerably more plausible. However, the four cases aren’t equally likely. 
If there is just one ace in the larger pack, the first half-deck must have contained all four aces. There are $\binom{48}{22}$ ways to pick $22$ cards to go with the four aces, compared with $\binom{52}{26}$ ways to choose the first half-pack without any restrictions, and $\binom{48}{22}/\binom{52}{26}$ is considerably less than $1/4$.
Can you take it from here?
A: The probability of having $i=0,1,2$ or $3$ ace in the second half-deck when dividing the whole deck at the beginning, is multinomial:
$$\frac{{4 \choose i}{48 \choose 26-i}}{{52 \choose 26}}$$
then...
A: The second deck before adding the ace contains 26 cards all of which have a chance of 3/51 to be an ace. The split is irrelevant for the probabiltiy of the other cards being an ace, there are 3 unseen aces remaining out of 51 possible cards. 
Now you add the ace to a deck of 26 cards creating a 27 card deck and draw from that. You then have a probability of 1/27 to draw that ace but 26/27 to draw one of the other cards which still have a 3/51 chance of being an ace, all that is known about them is that they are not the one specific ace.
Thus the result is:
1/27 * 1 + 26/27 * 3/51 = 17/459 + 26/459 = 43/459
It's not neccesary to think about the distribution of aces between the decks happening by the split. Look at it this way, if you draw a card from a deck it is 1/52 chance to be an ace regardless of burning cards first. Splitting the deck can be seen as just burning half the deck.
