TITLE: Why is this counting way wrong, aces drawn from two halves of a shuffled deck
QUESTION [0 upvotes]: 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?

REPLY [1 votes]: The probability of drawing the same card (i.e. the ace drawn from the first half) is $\frac{1}{27}$ and the probability of drawing any other card is $\frac{26}{27}$ and each of these cards has probability $\frac{3}{51}=\frac{1}{17}$ to be an ace.
The total probability is $\frac{1}{27}+\frac{26}{27}\times\frac{1}{17}=\frac{43}{459}$
{"set_name": "stack_exchange", "score": 0, "question_id": 3411085}