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I'm struggling with the following introductory probability problem:

Bob randomly chooses 8 cards from a 52-card deck. What is the probability that all the aces are contained in his choice?

The answer is given as:

Since there are 4 aces, Bob can only choose 4 more cards out of the remaining 48 cards, so total number of hands Bob can have such that they include the aces is $${4\choose 4} {48\choose 4}$$ The total number of ways Bob can choose any 8 cards is $$52 \choose 8$$ So the probability is $$\frac{{4\choose 4} {48\choose 4}}{52\choose 8}$$

I understand the denominator of this expression, but I don't understand the numerator. To me, the numerator seems to give the number of outcomes where the aces are the first 4 cards selected out of 8, but not those outcomes where the aces are interspersed, so to speak, among the other cards, and thus undercounts the viable outcomes.

I tried to simplify the question to resolve my confusion, but I ran into the same issue. For instance, if you were to select 3 cards from a 4-card deck of $\{A, B, C, D\}$, what is the probability that your selection would contain $A$ and $B$? If you write out all 24 permutations of 3 cards, you find that $1/2$ of them contain $A$ and $B$. However, if you substitute in the problem values to the solution above, you get $$\frac{{2\choose 2}{2\choose 1}}{4\choose 3} = \frac{\frac{2!}{(2-2)!}\frac{2!}{(2-1)!}}{\frac{4!}{(4-3)!}}=\frac{2!\cdot 2!}{4!}=\frac{4}{24}=\frac{1}{6}$$ Thus underestimating the probability.

Could someone please tell me where I am going wrong? Thanks.

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    $\begingroup$ To make a hand of $8$ cards with all four aces is the same as making a hand of $4$ cards from the set of non-aces. $\endgroup$
    – lulu
    Jul 22 at 1:20
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    $\begingroup$ Directly, your computations of $\binom22,\binom43$ are wrong. (By coincidence, $\binom21=2$ is right) $\endgroup$
    – Brian Moehring
    Jul 22 at 1:33

1 Answer 1

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You are confusing combinations with permutations, when they are actually two distinct combinatorial objects. Combinations deal with the number of ways to choose items, while permutations deal with the number of ways to order items. We don't care about the number of ways to pick 8 cards with 4 aces and then order them, we just care about the number of ways to pick 8 cards with 4 aces. In other words, in your shorter example ABC, ACB, BAC, BCA, CAB, and CAB are all the same combination, because the order does not matter.

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  • $\begingroup$ Ok, I see. Thank you. How do we know to use combinations instead of permutations here? Is it as if the 8 cards are selected all at once instead of sequentially, and if it were sequentially permutations would be appropriate? $\endgroup$
    – user601846
    Jul 22 at 1:44
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    $\begingroup$ Theoretically, with probability questions, you can use either permutations or combinations, but you have to do so consistently in the numerator and denominator. One of them may be easier. If you say that the denominator is $\binom{52}{8}$, then you are using combinations in the denominator and you must also use combinations in the numerator. If you use permutations in the denominator, you get $52 \cdot 51 \cdot \ldots \cdot 45$, but then you have a much more complicated calculation in the numerator where you have to look at the all the way the aces could be interspersed, as you mentioned. $\endgroup$
    – Ted
    Jul 22 at 1:56
  • $\begingroup$ @user601846 As long as the deck is sufficiently shuffled so the order is random, why would it matter how many cards are dealt at a time? $\endgroup$
    – Brian Moehring
    Jul 22 at 1:59
  • $\begingroup$ @Ted appreciate the response. $\endgroup$
    – user601846
    Jul 22 at 2:11

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