Probability of getting full house in poker

Question

I have this problem which I have solved, but using two different methods. I am quite new to combinatorics and want to know how to intuitively understand the difference between the following two methods.

The problem consist of calculating the probability of getting a full house being dealt a 5-card poker hand.

First of, I solve this by simply saying that

$P($get full house$)=\frac{2 {13\choose 2} {4 \choose 3} {4 \choose 2}}{{52 \choose 5}}$. This is the right answer according to my text-book. However, at my first attempt at solving this I forgot the factor 2 in the numerator. My reasoning goes like this: All the possible ways of getting a full house consists of all the ways we can combine two different ranks (i.e. ${13 \choose 2}$) times all the possible ways of choosing 3 cards out of 4 suits, times all the possible ways of choosing 2 cards out of 4 suits. Now, all this seems logical to me. The thing that makes me doubt whether I truly understand what I'm doing is how the factor 2 comes in place. I'm thinking: Because we choose two different ranks without regards to order, we have to compensate for those combinations and therefore multiply with $2!$, because obviously it does matter if I (for example) choose the ranks (ace,knights) and in this sequence choose three ace and two knights.

On the other hand, I can solve the problem using the method described here: https://math.stackexchange.com/a/808328/518320

Which as well seems intuitively clear, thinking the way the user describe the process in that thread.

What is the difference in the two methods? Maybe this is obvious, but I'm new to combinatorics. And is my reasoning above accurate?

Thanks!

Answer 1

They are essentially the same giving $156$ ways of choosing ranks, but if you want to draw a distinction:

• Choose two ranks and then choose one of those two to be the three so the other is the pair $${13 \choose 2}{2 \choose 1}$$

• Choose one rank to be the three then choose another rank to be the pair $${13 \choose 1}{12 \choose 1}$$

and you then multiply this by ${4 \choose 3}{4 \choose 2}$ and divide by ${52 \choose 5}$

Doing the same thing for two pairs to get $858$ ways of choosing the ranks by several methods:

• Choose three ranks and then choose one of those three to be the single so the others are the pairs $${13 \choose 3}{3 \choose 1}$$

• Choose three ranks and then choose two of those three to be the pairs so the other is the single $${13 \choose 3}{3 \choose 2}$$

• Choose one rank to be the single then choose two other ranks to be the pairs $${13 \choose 1}{12 \choose 2}$$

• Choose two ranks to be the pairs then choose another rank to be the single $${13 \choose 2}{11 \choose 1}$$

and you then multiply this by ${4 \choose 1}{4 \choose 2}{4 \choose 2}$ and divide by ${52 \choose 5}$

Answer 2

The two ranks are distinguishable once you get a particular full house. Say you get three kings and two queens. That differs from getting three queens and two kings. So it should be $13 \cdot 12$ ways to pick the two ranks. This is the same as $2 \cdot \binom{13}{2}.$

Answer 3

The expressions $2\binom{13}{2}$ and $\binom{13}{1}\binom{12}{1}$ are both equivalent to "choose $2$ objects from $13$ where the order of choosing matters." In other words, we want permutations of $2$ objects out of $13$ rather than combinations.

The way to get permutations of $k$ objects out of $n$ is to multiply $k$ consecutive integers together, where $n$ is the largest of the integers multiplied: $n(n-1)(n-2) \cdots (n-k+1).$ For example, permutations of $2$ objects out of $n$ give $13\cdot12$ possible permutations. And since $\binom n1 = n,$ $13\cdot12 = \binom{13}{1}\binom{12}{1}.$

But often the number of permutations is written $\dfrac{n!}{(n-k)!},$ because $$ \frac{n!}{(n-k)!} = n(n-1)(n-2) \cdots (n-k+1). $$ To get the number of combinations of $k$ objects out of $n$ (without regard for the order of the objects), we divide by $k!,$ since that's the number of different sequences in which each combination of $k$ objects can occur as a permutation. So we get $$\binom nk = \frac{n!}{(n-k)!k!}.$$ This also means that $k!\binom nk$ is yet another way to write the number of permutations of $k$ objects out of $n.$

In short, it's all really the same thing written in different ways.