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.
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.