Number of ways to get a full house in Poker

combinatoricsdiscrete mathematicspoker

I'm trying to understand how to represent the number of ways to get a full house in poker without using the combinations method. I read that this is true:

$\frac{52\times3\times2\times48\times3}{3!2!}=3744$

I understand where the numerator comes from but I'm not sure why we divide by "$3!2!$". I'm assuming it's because we have 3 cards of the same value and 2 others with the same value? But I don't really understand why the division…

I am also questioning why we couldn't start by doing $52\times51\times…$ because we could really pick any card to start with and then pick any of the remaining 51 as a second card (either same value as the first or any other)? I guess the third number in the calculation would depend on if we picked two same values or two different values but I'm all a bit confused about this.

Thanks for your help!

Best Answer

I guess the third number in the calculation would depend on if we picked two same values or two different values

With that approach, yes, that's something we would have to take into account. And even moreso with the fourth and fifth cards. It's a mess. Which is why that's not the approach used. Instead of picking one card, and then the next, and then the next, and keeping track of how many possible options would keep the possibility of a full house open, they build the full house out of one three-of-a-kind and one pair separately.

First we take the three-of-a-kind. How many ways can that be done? Well, the first card can be anything, so that's $52$ possibilities. The next two cards must have the same value, so this gives ${}\times3\times2$.

However, in a poker hand order doesn't matter. So picking first the ace of spades, then hearts then clubs is considered the exact same three-of-a-kind as ace of hearts, then clubs, then spades, and so on. All in all, our original count counts each distinct three-of-a-kind exactly $3!$ times. So the number of distinct three-of-a-kinds is $$\frac{52\times3\times2}{3!}$$

Next, we just do the pairs in exactly an analogous way to get $$\frac{48\times3}{2!}$$Finally, the number of ways to get a full house is the number of ways to get one three-of-a-kind and one pair, so we multiply the two together.

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