Conceptual Doubt: Probability of 4 of a kind hand

Question

I have seen people's solutions to calculating probability of getting 4 of a kind in a poker hand, and I know the answer is $\frac {(13\cdot48)} {^{52}C_5}$. Now I understand that out of $13$ sets of rank, $1$ has to be chosen, i.e. $^{13}C_1$. and the 5th card is chosen using $^{48}C_1$. What confuses me is the number of ways to draw the other three cards of that rank is $^4C_4$. Shouldn't this be $^{3}C_3$, since the $^{13}C_1$ is already helping us decide on a rank and helping us choose the first card for the 4 of a kind hand. Then there's 3 more of the same cards left to choose from and they can be chosen out of 3 possible options for the hand to be 4 of a kind, i.e. $^{3}C_3$.

I know that in the end, both $^{4}C_4$ and $^{3}C_3$ amount to 1, so it shouldn't affect the answer but I'm trying to understand the reason behind why it's $^{4}C_4$. I read other answers on StackExchange like the accepted one for this Probability of being dealt four-of-a-kind in a set of $5$ cards? but it isn't clear to me.

Any help would be extremely appreciated!!

Answer 1

I would imagine that after you choose which of the 13 ranks that you will have 4 of a kind of, it is generally accepted that you then choose all 4 of that rank at the same time.

So the pertinent quad's factor has to be $\binom{4}{4}.$

As you describe it, the procedure that you suggested is that once you select the rank, you then choose one of the $4$ cards of that rank, and then choose the remaining $3$ cards of that rank.

That is accurate as far as it goes, but actually results in the quad factor of $\binom{4}{1} \times \binom{3}{3} = 4.$

This begs the question why you get two different quad factors, if both analyses are valid. The answer is that the 2nd analysis, which is $\binom{4}{1} \times \binom{3}{3}$ constitutes over-counting.

For example, suppose that the rank is Aces, and that there are therefore 4 choices of which card will be the 1st Ace chosen.

The choice of Ace-spades and then the other 3 Aces results in the same collection of 4 Aces as (for example) the choice of the Ace-clubs, and then the other 3 Aces, and so on.

So, with respect to the quad factor of $\binom{4}{1} \times \binom{3}{3}$, you have to apply the overcounting adjustment factor of $\frac{1}{4}$.

So, the idea is that with the overcounting adjustment factor applied as needed, both approaches are valid. The 1st approach is generally preferred as simpler, since it does not require any overcounting adjustment factor.

Answer 2

I view $\binom{13}{1}$ as choosing which rank, so you are not actually choosing any cards at this point, hence you then use $\binom{4}{4}$. It makes a bigger difference in other poker hands, e.g. a full house.

For that you choose a rank and then take $3$ cards of it: $\binom{13}{1}\cdot\binom{4}{3}$. Then you choose a second rank from the remaining ones, and take $2$ cards of it: $\binom{12}{1}\cdot\binom{4}{2}$. Multiplying all of those factors gives the total number of full house hands: $3744$. I hope this illustrates the difference between choosing a rank and actually selecting cards.