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.
My attempt:
to randomly divide the cards into two halves $A$ and $B$, I followed this process.
Put each ace, in either $A$ or $B$, with equal probability.(ex. the ace of spades goes into set A with probability $\frac{1}{2}$, and into set B with probability $\frac{1}{2})$
Then, divide the remaining 48 cards between $A$ and $B$, such that both sets of cards contain $26$ cards each.
Question1: Does this process, randomly divide the deck of 52 cards into 2 halves?
Without loss of generality assume, $A$ is the first half, and $B$ is the second half.
define $AL = \text{first half contains at least one ace}$
and
$i = \text{first half contains exactly i aces}$, $i\in {1,2,3,4}$
$=>P(AL) = 1 - P(\text{second half contains all aces}) = 1-(0.5)^4 = \frac{15}{16}$
and
$P(0) = \dbinom{4}{0}*(0.5)^0(0.5)^4 = \dfrac{1}{16}$ and $P(0|AL)=\dfrac{P(0\cap AL)}{P(AL)} = 0$
$P(1) = \dbinom{4}{1}*(0.5)^1(0.5)^3 = \dfrac{1}{4}$ and $P(1|AL)=\dfrac{P(1\cap AL)}{P(AL)} = \dfrac{P(1)}{P(AL)} = \dfrac{4}{15}$
$P(2) = \dbinom{4}{2}*(0.5)^2(0.5)^2 = \dfrac{3}{8}$ and $P(2|AL)=\dfrac{P(2\cap AL)}{P(AL)} = \dfrac{P(2)}{P(AL)} = \dfrac{2}{5} = \dfrac{6}{15}$
$P(3) = \dbinom{4}{3}*(0.5)^3(0.5)^1 = \dfrac{1}{4}$ and $P(3|AL)=\dfrac{P(3\cap AL)}{P(AL)} = \dfrac{P(3)}{P(AL)} = \dfrac{4}{15}$
$P(4) = \dbinom{4}{4}*(0.5)^4(0.5)^0 = \dfrac{1}{16}$ and $P(4|AL)=\dfrac{P(4\cap AL)}{P(AL)} = \dfrac{P(4)}{P(AL)} = \dfrac{1}{15}$
now, if there are $i$ aces in $A$, then after an ace is selected from $A$ and transferred to $B$, $B$ will contain $4-i+1=5-i$ aces, and a total of $27$ cards.
define $C = \text{card selected from B is an ace}$
$=>P(C|i) = \dfrac{5-i}{27}$
Question2: Are the quantities $P(AL)$ and P(i) calculated correctly?
$$P(C) = \sum_{i=1}^4 P(i|AL) \cdot P(C|i,AL) = \sum_{i=1}^4 P(i|AL) \cdot P(C|i) = \frac{4}{15}\frac{4}{27} + \frac{6}{15}\frac{3}{27} + \frac{4}{15}\frac{2}{27} + \frac{1}{15}\frac{1}{27} = \frac{43}{15 \cdot 27} = \frac{129}{45 \cdot 27}$$
The given answer is $\frac{129}{51 \cdot 27}$, what have I done wrong?
edit: correct ways for solving the problem are here and here, I am not asking for these
It is not appropriate to use the binomial distribution for the aces, since the location of each ace is not independent.
The probability that the Ace of Hearts is in deck A is... $$ P(H)= \frac{\binom{51}{25} }{ \binom{52}{26} } =\frac 12$$
But the probability that the ace of spades is in deck A given that the ace of hearts is in deck A is
$$ P(S|H)= \frac{\binom{50}{24} }{ \binom{51}{25} } =\frac{25}{51} <\frac 12$$