# Conditional Probability: Sheldon Ross Example 2h

The following question comes from Example 2h, in Sheldon Ross's textbook A First Course in Probability on page 64. I got the same answer as the author through a different line of reasoning (given at the end of the solution). However, I would really like to understand the reasoning given in this solution, if someone could elaborate that would be great!

The question: An ordinary deck of 52 playing cards is randomly divided in 4 piles 13 cards in each pile. Compute the probability that each pile has an ace.

The solution:

We want to define four events for this problem : \begin{align*} E_{1} &= \{\text{The ace of spades in any one of the piles}\}\\ E_{2} &= \{\text{The ace of spades and the ace of hearts are in different piles}\}\\ E_{3} &= \{\text{The aces of spades, hearts, and diamonds are all in different piles}\}\\ E_{4} &= \{\text{All four of the aces are in different piles}\} \end{align*}

The desired probability is $P(E_{1}E_{2}E_{3}E_{4})$ and by applying the multiplication rule, $$P(E_{1}E_{2}E_{3}E_{4})=P(E_{1})P(E_{2}|E_{1})\cdots P(E_{4}|E_{1}E_{2}E_{3})$$

Now,
$$P(E_{1})=1$$

Since $E_1$ is in the sample space $S$. Also, $$P(E_{2}|E_{1})=\frac{39}{51}$$ Since the pile containing the pile contain the ace of spades will contain 12 of the remaining 51 card, and $$P(E_3 | E_1 E_2)=\frac{26}{50}$$

Since the piles containing the aces of spades and hearts will receive 24 of the remaining 50 cards. Finally,

and finally, $$P(E_4 | E_1 E_2 E_3)=\frac{13}{49}$$

Multiplying them all together we get $$P( E_1 E_2 E_3 E_4) \approx 0.105$$

My Confusion: I don't quite understand the reasoning; it is clear that $P(E_1)$ is 1, because the ace of spades has to end up some where. But the the conditional probability has me confused; given the the ace of spades is in one pile I have 51 cards to choose from to put in any of the piles. There are $\binom{51}{39}$ ways that I can choose the 39 cards for the piles that don't contain the ace of spades. If I want to assure that I get the ace of hearts in one of the 3 piles that doesn't contain the ace of spades, I can set it aside. Then I am left to choose 38 from 50, there are $\binom{50}{38}$ ways to do this. Then I have $$\frac{\binom{50}{38}}{\binom{51}{39}}=\frac{39}{51}$$

And this same reasoning will lead to the same answer. I see benefit from not having to use the binomial coefficient, but I don't understand the soundness of the authors reasoning could someone elaborate on why his method works?

• When it comes time to decide where the Ace of hearts goes, there are $51$ open slots, of which $39$ are "favourable." Oct 16 '13 at 23:44
• @AndréNicolas Thanks! I guess it makes sense, we don't need to consider combinations since we are controlling the experiment. Oct 16 '13 at 23:50
• It still doesn't make sense to my why we don't need to consider the combinations. My reasoning was as follows. Given we've fixed the location of the Ace of Spades, there are 39 favorable locations to put the Ace of Hearts for each of these locations there are 50! ways to arrange the remaining cards. Hence of the 51! arrangements available after placing the Ace of Spades 39*50! have Ace of Hearts in a different pile. And the probability is then 39*50!/51! = 39/51. I have a hard time imagining the sample space when the permutations are not considered... Aug 6 at 14:21

Imagine that each player has $13$ "slots," each of which will ultimately contain one card. We can assume that the Ace of $\spadesuit$, Ace of $\heartsuit$, Ace of $\diamondsuit$, and Ace of $\clubsuit$ are, in that order, the first four cards in the deck.
Once we have decided where the $\spadesuit$ Ace has gone, there are $51$ open slots, of which $39$ are not in the same group as the group that contains the $\spadesuit$ Ace. So the probability that the $\heartsuit$ Ace lands in one of these slots is $\frac{39}{51}$.
Similarly, when it comes to deal the $\diamondsuit$ Ace, there are $50$ open slots left, of which $26$ are "favourable." And given that the first three Aces have landed nicely, the probability the fourth lands nicely is $\frac{13}{49}$.