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I was reading A First Course in Probability by Sheldon Ross. I read one of the problems and then tried building logic for it. Then read books solution which was completely different. So was guessing if my logic is wrong.

Problem: An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.

Solution given in the book: Define event:

$E_1$ = {the ace of spades is in any one of the piles}

$E_2$ = {the ace of spades and the ace of hearts are in different piles}

$E_3$ = {the ace of spades, hearts and diamonds are all in different piles}

$E_4$ = {all 4 aces are in different piles}

The desired probability

$P(E_1E_2E_3E_4) = P(E_1)P(E_2|E_1)P(E_3|E_1E_2)P(E_4|E_1E_2E_3)$

Now $P(E_1) = 1$

To find $P(E_2|E_1)$, consider the pile that contains the ace of spades. Probability that the ace of hearts is among the remaining 12 cards is $12/51$. Thus $$P(E_2|E_1) = 1 - \frac{12}{51} = \frac{39}{51}$$

Similarly, given that the ace of spades and hearts are in different ppiles, the set of the remaining 24 cards of these two piles is equally likely to be any set of 24 of the remaining 500 cards. The probability that the ace of diamonds is one of these 24 is $24/50$. Thus, $$P(E_3|E_1E_2) = 1 - \frac{24}{50} = \frac{26}{51}$$ Similarly,

$$P(E_4|E_1E_2E_3) = 1 - \frac{39}{49} = \frac{13}{49}$$ Thus,

$$P(E_1E_2E_3E_4) = \frac{39.26.13}{51.50.49} \approx .105$$

Thus 10.5 precent chnce that each pile will contain an ace.

My logic was below and I feel it more intuitive and natural

  1. First pile will contain 13 cards out of 52. Thus sample space will be $^{52}C_{13}$. The first pile is to contain 1 of 4 aces and rest 12 cards can be any of remaining 48. Thus the event space will be $^4C_1^{48}C_{12}$ Thus the probability that first pile will contain one of the aces is $$\frac{^4C_1*^{48}C_{12}}{^{52}C_{13}}$$
  2. Similarly for second pile, the probability for containing one of the aces is $$\frac{^3C_1*^{36}C_{12}}{^{39}C_{13}}$$
  3. Similarly for third pile, the probability for containing one of the aces is $$\frac{^2C_1*^{24}C_{12}}{^{26}C_{13}}$$
  4. Lastly for forth pile, the probability for containing the last of the aces is $$\frac{^1C_1*^{12}C_{12}}{^{13}C_{13}}$$

So the desired probability is:

$$\frac{^4C_1*^{48}C_{12}}{^{52}C_{13}} * \frac{^3C_1*^{36}C_{12}}{^{39}C_{13}} * \frac{^2C_1*^{24}C_{12}}{^{26}C_{13}} * \frac{^1C_1*^{12}C_{12}}{^{13}C_{13}}$$

What I dont understand is

  1. In book's solution, it says

    $E_4$ = {all 4 aces are in different piles}

    But I feel this is what we need to find. So the desired probability is $P(E_4)$. But book says its $P(E_1E_2E_3E_4)$. Is it like the probability $P(E_4)$ cannot be standalone? and we cannot specify/calculate $P(E_4)$ independently. Or $P(E_4) = P(E_1E_2E_3E_4)$

  2. The book's solution says:

    To find $P(E_2|E_1)$, consider the pile that contains the ace of spades. Probability that the ace of hearts is among the remaining 12 cards is $12/51$.

    I am not completely clear about how $\frac{12}{52}$ came.

  3. Is there any logical correspondence in the two approaches? Do they align each other in terms of thinking they follow? If yes can you please put it in words how this correspondence exist? Also may be finding and putting correspondence in terms of similarities in the algebraic equation will help me get it more clearly.

  4. Is the book's approach more sort of defining the events recursively? If yes, does this means that problem involving any sequence of dependent event can be solved in the way given in the book thus effectively defining the desired probability recursively and solving to the probability of base condition (in this particular problem base condition might be first pile contains only one ace that $E_1$). I am just trying to generalize the pattern of the question, so that I will know this method is applicable when I am confronted with some question.

  5. What are the difference between two approaches? and Can both be equally used to solve problems?

  6. All I am trying here is to make the book's approach logically more intuitive and structured to me, the way the other approach feels intuitive to me, so that I can form equations as in books' approach very easily and without confusion. So anything in that direction will be appreciable. The book follows very formula/equation-oriented way at explaining the solutions, not the logical way.

PS: May be I am thinking unnecessarily. Or the question may sound very vague or rambling. But please, probability stuff always confused me and am giving it a last try.

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  • $\begingroup$ What is the numerical value of your solution? The logic of the books solution is clearer if you draw the contingency tree for the problem. Also simulation gives a result close to the book answer, not proof that it is right but it is a reality check. Your answer if it differs from the book should at least be close to it if you are right and the book wrong. $\endgroup$ – Conrad Turner Apr 20 '15 at 8:21
  • $\begingroup$ The answer comes 0.105498199279712 as can be seen in the output at the bottom of this C# code. $\endgroup$ – anir123 Apr 20 '15 at 17:13
  • $\begingroup$ Which agrees with the book answer to 15 significant figures ... $\endgroup$ – Conrad Turner Apr 21 '15 at 9:13
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But I feel this is what we need to find. So the desired probability is $\mathsf P(E_4)$. But book says its $\mathsf P(E_1E_2E_3E_4)$. Is it like the probability $\mathsf P(E_4)$ cannot be standalone? and we cannot specify/calculate $\mathsf P(E_4)$ independently. Or $\mathsf P(E_4)=\mathsf P(E_1E_2E_3E_4)$

The event is the conjunction of the four events, $E_4 = E_1 E_2 E_3 E_4$ , because each subsequent event in the list is a subset of its previous.   $E_4\subseteq E_3\subseteq E_2\subseteq E_1$.

Did the textbook skip this justification? It uses the conjunction to make it clear where the product comes from.

$$\mathsf P(E_4)= \mathsf P(E_1E_2E_3E_4) = \mathsf P(E_1)\;\mathsf P(E_2\mid E_1)\;\mathsf P(E_3\mid E_1E_2)\;\mathsf P(E_4\mid E_1E_2E_3)$$

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I am not completely clear about how $12/52$ came.

Given that the ace of spades is in one of the four piles of $13$ cards, then there are $51$ places the ace of hearts could be, but $12$ of these are in a pile you don't want it to be (that's the other cards in the same pile as the ace of spades).

Thus the probability that the ace of hearts is not in the same pile as the ace of spades is $1-\tfrac{12}{51}$.

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  • $\begingroup$ hi Graham, sorry for super late reply, I came across this problem during revision and ended up looking here. I think the book did not specify the first point you made. So is this the case with all probabilities specified by the multiplication rule? That is it always the case that: $P(E_1E_2E_3...E_n)=P(E_1)P(E_2|E_1)P(E_3|E_1E_2)...P(E_n|E_1...E_{n-1})=\bf{E_n}$ That is in the multiplication rule, does $E_n$ is always the subset of other events? $\endgroup$ – anir123 Aug 3 '15 at 20:28
  • $\begingroup$ @Mahesha999 Not always. You have to set it up. Its the case whenever $E_4\subseteq E_3\subseteq E_2\subseteq E_1$; that is, if you are seeking to find $P(E_n)$ and $E_n$ is a subset of event $E_{n-1}$ it might be that $P(E_{n-1})$ and $P(E_n\mid E_{n-1})$ are easier to find. And so on. $\endgroup$ – Graham Kemp Aug 3 '15 at 22:19

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