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Here is the 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.

My solution:

$$\frac { {4 \choose 1} {48 \choose 12} {3 \choose 1} {36 \choose 12} {2 \choose 1} {24 \choose 12} } {{52 \choose 13} {39 \choose 13} {26 \choose 13} }.$$

I'm not sure if I did it right because the book gives the following solution.

Solution from the book:

E1 = {the ace of spades is in any one of the piles}.

E2 = {the ace of spades and the ace of hearts are in different piles}.

E3 = {the aces of spades, hearts, and diamonds are all in different piles}.

E4 = {all 4 aces are in different piles}.

The desired probability is $P(E1E2E3E4)$, and by the multiplication rule, $P(E1E2E3E4) = P(E1)P(E2|E1)P(E3|E1E2)P(E4|E1E2E3)$:

$$ P(E1 E2 E3 E4) = \frac {39 \cdot 26 \cdot 13} {51 \cdot 50 \cdot 49} .$$

I do understand that the book gave a neat solution, but I wonder if my logic was correct for this problem, because I did not realize how to use the multiplication rule here.

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  • 2
    $\begingroup$ Your logic is correct. Both solutions are equivalent. $\endgroup$ – Graham Kemp Oct 25 '14 at 23:48

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