For each of $k = 0, 1, 2, 3, 4$, find the probability that a poker hand $($five cards$)$ contains just $k$ aces. Since we're dealing with a hand of cards, our sample space would be $$\binom{52}{5}$$ since we want to pick $5$ cards out of $52$. The reason the sample space's order doesn't count, and there's no replacement is because generally when you pick a card, you don't put it back, and the order the cards appear do not matter, right?
So for $k = 0$, that means our hand has no aces, and $5$ random cards. So there are $\binom{4}{0}$ ways to pick no aces out of a total of $4$ aces, which is $1$ way. For each of these ways, there are $\binom{48}{5}$ ways to pick $5$ other cards from the remaining of the deck. Is this correct? I think it is because since our hand can have no aces, we take the total cards $(52)$ and subtract $4$ from it which gives you $48$ total cards left to choose from. So that means there are a total of $$\binom{4}{0} \binom{48}{5} = 1712304 $$ ways to pick a hand with no aces? My next question is, do I need to divide by the sample space? This is something I don't quite understand. Why do we need to divide by the sample space? I'm assuming since our probability we just found was only an event, then that event is a possibility that could occur from the sample space. The probability of the sample space is always $1$, and never greater than $1$. So $$\frac{1712304}{2598960} = 0.65884199833$$ which is about a $65$% chance that you will get a hand of $5$ cards with no aces? This seems realistic in the real world.
 A: You've got it.
The reason why you divide by the size of the sample space is because that tells us the proportion of outcomes in which the desired outcome occurs. It may help to think of a simpler example, such as throwing a fair 6-sided die. Say you want to know the probability of rolling an odd number. There are 3 outcomes in which an odd number is rolled: 1, 3, or 5. There are six total outcomes. They are all equally likely, so the probability is $\frac{3}{6}=\frac{1}{2}$.
In your problem, you found there are $2598960$ ways to draw a hand of five cards from a deck, and there are $1712304$ ways to draw a hand of five cards such that the hand contains no aces. All hands are equally likely, so the probability of no aces is the proportion of outcomes in which there is no ace, i.e., $\frac{1712304}{2598960}$.
A: Yes, you do need to divide by ${52\choose5}.$  The ultra-classical definition of probability is, "the number of ways of success, divided by the number of outcomes, assuming that each outcome is equally likely."  Probabilities are always between $0$ and $1$.
A: Dividing by the size of the sample space turns a count into a probability.  $1,712,304$ is the number of ways to get the hand that you're looking for (which is calculated correctly) out of the total $2,598,960$ different possible hands.
Without dividing by the number of possibilities, we don't know where $1,712,304$ events is a large proportion of the events or a small proportion of them.
