Problem Statement
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.
The solution is as follows:
Let I denote the event that the interchanged card is selected. Let S be the event that the two values are the same. Then
$P(S)=P(I)P(S|I)+P(I^c)P(S|I^c)=\frac{1}{27}\cdot 1+\frac{26}{27}\frac{3}{51}$
Solution by different user:
Let A be the event that we draw an ace the second time, and let P(i) be the probability that there are i aces originally in the second pile and P(A|i) be the probability that we draw an ace conditional on the second pile starting with i aces. We have
$P(A)=\sum_{i=0}^{3} P(A|i)P(i)$
It is easy to compute $P(A|i)=\frac{i+1}{27}$. Computing $P(i)$ is slightly more involved. We need to compute the number of possible half-decks with i aces, which we will denote $N_i$, and then $P(i)=\frac{N_i}{\sum_{j=0}^{3}N_j}$.
If we have i aces and $26-i$ other cards (out of the 48 non-aces), then there are $\binom{4}{i}\binom{48}{26-i}$ possible hands.
Computing the answer is now a matter of plugging into the formula. ... when we computed our probabilities, we divided by $N_0+N_1+N_2+N_3$, and not by $\binom{52}{26}=N_0+N_1+N_2+N_3+N_4$....
My Question:
I did the calculation in excel and yet I still cannot get the right answer. The second solution is similar to the same argument I followed. Is this 2nd solution incorrect or perhaps the way I executed it? Below is a snippet of the excel sheet:
