Probability that each pile of cards has exactly one ace (different solution, different answer) 
user585792, A deck of $52$ cards is divided into four piles of $13$ cards. What is the probability that each pile has one ace?, URL (version: 2018-09-18): A deck of $52$ cards is divided into four piles of $13$ cards. What is the probability that each pile has one ace?

This question is from Introduction to Probability Models 10th Edition by Sheldon Ross Ch1 Exercises 26,27. I have found another way to solve this problem but the calculated probability is about $\frac{1}{3}$ of the given answer. I want to explain my answer and ask which part of my logic is not valid. 
My solution:
$E_1 = \text{{the first pile has exactly 1 ace}}$
$E_2 = \text{{the second pile has exactly 1 ace}}$
$E_3 = \text{{the third pile has exactly 1 ace}}$
$E_4 = \text{{the fourth pile has exactly 1 ace}}$
$P(E_1E_2E_3E_4) = P(E_4|E_1E_2E_3)P(E_3|E_1E_2)P(E_2|E_1)P(E_1)$
$P(E_4|E_1E_2E_3) = 1$ 
$P(E_3|E_1E_2)$ =  probability given $2$ aces left for the third and fourth pile, the probability that the third pile has only one ace. 
Let $(x,y)= (\text{number of ace in the third pile}, \text{number of ace in the fourth pile})$, then given first and second pile each has exactly one ace, possible $(x,y)$ are $\{(2,0),(0,2),(1,1)\}$.
Therefore, $\underline{P(E_3|E_1E_2) = \frac{1}{3}}$
$P(E_2|E_1)$ =  probability given $3$ aces left for the second, third and fourth pile, the probability that the second pile has only one ace. 
$\begin{align*}\text{Let }(x,y,z)= 
(&\text{number of ace in the second pile},\\
&\text{number of ace in the third pile},\\
&\text{number of ace in the fourth pile})\end{align*}$
Then given first and second pile each has exactly one ace, possible $(x,y,z)$ are $\{(1,1,1),(1,2,0),(1,0,2),(0,0,3),(0,3,0),(3,0,0),(0,1,2),(0,2,1),(2,1,0),(2,0,1)\}$.
Therefore, $\underline{P(E_2|E_1) = \frac{3}{10}}$
Likewise, for $P(E_1)$, there are $32$ possible $(x,y,z,l)$ where non-negative integers $x,y,z,l$ add up to $4$ and among them, there are 10 cases where $x=1$. 
Therefore, $\underline{P(E_1) = \frac{10}{32}}$
To obtain the final result: 
$P(E_1E_2E_3E_4) = P(E_4|E_1E_2E_3)P(E_3|E_1E_2)P(E_2|E_1)P(E_1)$
$= 1 \cdot \frac{1}{3} \cdot \frac{3}{10} \cdot \frac{10}{32} = \frac{1}{32}  $.

I get the same answer if I use the same method to calculate $P(E_1E_2E_3E_4)$ when 
$E_1 = \text{{one of the piles contains the ace of spades}}$
$E_2 = \text{{the ace of spades and the ace of heart are in different piles}}$
$E_3 = \text{{the ace of spades, the ace of heart, and the ace of diamonds are in different piles}}$
$E_4 = \text{{all $4$ aces are in different piles}}$.
Please help! 
 A: The probability for selecting 1 from 2 aces and 12 from the remaining 24 cards when selecting 13 from 26 cards is:
$$\begin{align}\mathsf P(E_3\mid E_1,E_2)&=\left.\dbinom{2}1\dbinom{24}{12}\middle/\dbinom{26}{13}\right.\\[1ex] &=\dfrac{13}{25}\end{align}$$


Let $(x,y)= (\text{number of ace in the third pile}, \text{number of ace in the fourth pile})$, then given first and second pile each has exactly one ace, possible $(x,y)$ are $\{(2,0),(0,2),(1,1)\}$.
  Therefore, $\underline{P(E_3|E_1E_2) = \frac{1}{3}}$

When I buy a lotto ticket, the outcomes are either "a winning ticket", xor "a loosing ticket".   Therefore the probability of winning is $1/2$.   Right?
No, you have to weight the probability; or make sure the outcomes you are counting have equal probability. 
The event you call $(2,0)$ is the event of obtaining 2 from the first 13 places for the aces, when you are selecting 2 from 26.  That has a probability of $\tbinom{13}2/\tbinom{26}2$ or $6/25$.  Likewise for event $(0,2)$. And event $(1,1)$ is the event for obtaining 1 from the first 13 and 1 from the later 13 when selecting 2 from 26. That is $\tbinom{13}1^2/\tbinom{26}2$, which is $13/25$.

Take a deck of 4 cards, two of which are aces, and deal the deck into two piles of 2 cards.  We have six equally probable outcomes, with four for the event "an ace in each pile". $~4/6=\tbinom 21^2/\tbinom 42$ $$\rm\{(AA{,}OO)~,(AO{,}AO)~,(AO{,}OA)~,(OA{,}AO)~,(OA{,}OA)~,(OO{,}AA)\}$$
A: Your notation $(x,y)=\{(2,0),(0,2),(1,1)\}$ is fine, but $P(E_3|E_1E_2) = \frac{1}{3}$ is not true. Let the two aces be $A\clubsuit$ and $A\spadesuit$.
Case 1: $(x,y)=(2,0)$. There is only one way to have $2$ aces in $E_3$ (both $A\clubsuit$ and $A\spadesuit$) and ${24\choose 11}$ ways to have non-aces in $E_3$, while overall there are ${26\choose 13}$  ways to have $13$ cards in $E_3$. Hence: $$P((x,y)=(2,0))=\frac{{2\choose 2}{24\choose 11}}{{26\choose 13}}=\frac6{25}.$$
Case 2: $(x,y)=(1,1)$. There are $2$ ways to have $1$ ace in $E_3$ ($A\clubsuit$ or $A\spadesuit$) and ${24\choose 12}$ ways to have non-aces in $E_3$, while overall there are ${26\choose 13}$  ways to have $13$ cards in $E_3$. Hence: 
$$P((x,y)=(1,1))=\frac{{2\choose 1}{24\choose 12}}{{26\choose 13}}=\frac{13}{25}.$$
Case 3: $(x,y)=(0,2)$. It is symmetric with the Case 1: $(x,y)=(2,0)$.
In conclusion, as noted by other responders, by $(x,y)$ you are only considering the distribution of aces while forgetting the non-aces to be selected to $E_3$ and $E_4$ as well, which put different weights to the distributions.
