Overcounting in drawing cards problem (combinations)

Question

The problem

In how many ways can a gambler draw $5$ cards from a standard deck ($52$ cards) and get $3$ of the same kind?

My solution (wrong): $\dbinom 4 3 \dbinom {13} 1 \dbinom {48} 2 = 58,656$ .

Solutions from my book and questions

1. $\dbinom 4 3 \dbinom {13} 1 \dbinom {12} 2 \dbinom 4 1 \dbinom 4 1 = 54,912$

I get the logic behind this. However, comparing it with my solution, I cannot understand what have I overcounted. Where did these extra $3744$ cases come from?

2. $\dbinom 4 3 \dbinom {13} 1 \dbinom {48} 1 \dbinom {44} 1 / 2$.

This solution is provided with the remark that division by $2$ is needed since no distinction is made for the order in which the other two cards are drawn. I understand the point of division, but I'm not sure why $\dbinom{44}{1}$ is being used in this context.

I realize the problem is trivial, but I'm a beginner who is studying independently, and it's been on my "come back later" list of problems for a couple of months now. So, any help would be appreciated.

Answer 1

Without understanding with complete certainty how you arrived at your expression, I can provide some hints based on what I think you are trying to do.

I'm assuming $\binom{4}{3}$ represents picking the "three-of-a-kind" from $4$ available, and of course there are $13$ card values. So that makes sense. We are left with the task of picking the remaining $2$ cards to complete the hand.

If there's no restriction on the remaining cards (other than of course the fact that neither of them can be the remaining card with the value we have already chosen above), then indeed there are $\binom{48}{2}$ possibilities, as you indicated.

However, it appears that there is an additional (implied?) restriction that the two remaining cards cannot be equal in value. In this case, one completes the hand by choosing the two other values, which is done in $\binom{12}{2}$ ways, and then picking the suit for each, which is done in $4^2$ ways.

Of course, another way to satisfy the restriction would be to pick one card in $52 - 4 = 48$ remaining cards, then the last one from $48 - 4 = 44$ available. In counting this way, we have implicitly assumed that the order matters: picking the ace of hearts and the $2$ of spades was counted twice, for instance. So, we need to divide the product by $2$ to correct.

Answer 2

I'd rather put the correct computation in a different order

$\binom{13}1\binom43 \binom{12}2\binom41\binom41$ [Choose a rank]$\times$[choose 3 from that rank]
$\times$[Choose another two ranks]$\times$[Choose one each from these ranks]

That is the definition of three of a kind in poker hands

Answer 3

Your count includes a "full house" - three of a kind and two of another kind. The number of these can be found with ${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}= 13\times 4\times 12\times 6=3744$ ways.

The book excludes this possibility and requires the two cards which are not part of the three-of-a-kind to be different values from each other.

In calculation (2) the $/2$ comes from the point that these two cards can come in either order but are treated as the same case.