Probability that each pile of cards has exactly one ace (different solution, different answer)

Question

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!

Answer 1

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". $~4/6=\tbinom 21^2/\tbinom 42$ $$\rm\{(AA{,}OO)~,(AO{,}AO)~,(AO{,}OA)~,(OA{,}AO)~,(OA{,}OA)~,(OO{,}AA)\}$$

Answer 2

Another way to compute $P(E_3\mid E_1,E_2)$ (from your first attempt) is to consider the third and fourth piles as an array of $26$ locations where cards can be put, one card in each location. (Top of the third pile, card next to the top of the third pile, second card from the top of the third pile, etc.) There are two aces that can be put in two of these $26$ locations. There is nothing that makes one pair of locations more likely than another. So the probability space can be partitioned into $\binom{26}{2}$ equally likely events, each event identified with two of the $26$ locations.

The event that there is exactly one ace in each pile is the union of $13^2$ of these events, one event for each way you can choose a location from the $13$ locations in the third pile and a location from the $13$ locations in the fourth pile. So $$ P(E_3\mid E_1,E_2) = \frac{13^2}{\binom{26}{2}} = \frac{13^2}{(26\cdot25)/2} = \frac{13}{25} = 0.52.$$

For $ P(E_2\mid E_1)$ you have $39$ locations in the second, third, fourth piles, of which exactly three locations will be occupied by aces, which makes $\binom{39}{3}$ equally likely events. The number of those events that have one ace in the second pile is $13\cdot\binom{26}{2}$ ($13$ locations in the second pile, $\binom{26}{2}$ pairs of locations where the other two aces might go). So $$ P(E_2\mid E_1) = \frac{13\cdot\binom{26}{2}}{\binom{39}{3}} = \frac{13\cdot(26\cdot25)/2}{(39\cdot38\cdot37)/6} = \frac{13\cdot25}{19\cdot37} \approx 0.462.$$

For $P(E_1)$ you have $52$ locations with aces in four locations, $13$ places to put an ace in the first pile, and $\binom{39}{3}$ sets of places to put the other three aces. So $$ P(E_1) = \frac{13\cdot\binom{39}{3}}{\binom{52}{4}} = \frac{13\cdot(39\cdot38\cdot37)/6}{(52\cdot51\cdot50\cdot49)/24} = \frac{13\cdot19\cdot37}{17\cdot25\cdot49}\approx 0.439.$$

Multiplied together, \begin{align} P(E_3\mid E_1,E_2) P(E_2\mid E_1) P(E_1) &= \frac{13^2}{\binom{26}{2}} \cdot \frac{13\cdot\binom{26}{2}}{\binom{39}{3}} \cdot \frac{13\cdot\binom{39}{3}}{\binom{52}{4}} \\ &= \frac{13^4}{\binom{52}{4}} \\ &= \frac{13^4}{(52\cdot51\cdot50\cdot49)/24} \\ &= \frac{13^3}{17\cdot25\cdot49} \\[.4ex] &\approx 0.1055. \end{align}

It is also possible to find that the probability is $13^4/\binom{52}{4}$ through direct counting; see this answer.

Answer 3

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.