0
$\begingroup$

Number of hands:

Two Pair: $\binom{13}2 \binom 42^2 \binom{44}1$

Full House: $\binom{13}1 \binom 43 \binom{12}1 \binom42$

I was wondering why we don't do $\binom{13}1 \binom{12}1 \binom 42^2 \binom{44}1$ for Two Pair and $\binom{13}2 \binom 43 \binom42$ for Full House instead. I made these mistakes and would like to understand why they don't work.

$\endgroup$
1
$\begingroup$

In this answer, I am assuming that a 'hand' is a set of five cards (i.e., there is no associated order to the hand). In some cards games this would be a good assumption, in others (such as 5-card stud [https://en.wikipedia.org/wiki/Five-card_stud], where cards are received and bet on sequentially), this may not be a good assumption.

Let's start with a very much simplified example: a deck with three cards, 1, 2, and 3. How many two-card hands are there? Answer: 3. The 3 hands are $\{1,2\}, \{1,3\},\mbox{and } \{2,3\}.$ If we tried to count this as pick a card ($3$ choices), pick another card ($2$ choices), giving an answer of $6$ possible hands, we would have overcounted by a factor of $2$. For example the hand $\{1,2\}$ is counted twice (once as choose $1$, then $2$; and once as choose $2$, then $1$).

The same idea occurs in your question. A hand with a pair of $7$s and a pair of jacks is the same as a hand with a pair of jacks and a pair of $7$s. But if you count the rank choices as ${{13}\choose{1}}\cdot {{12}\choose{1}}$, you are double counting these two-pair hands. That's why you get the correct number of choices if you choose the two ranks simultaneously as ${13}\choose{2}$.

By contrast in a full-house rank choices are distinguished by one rank getting three-of-a-kind and one rank getting a pair. Thus three $7$s and two jacks is a different hand than three jacks and two sevens.

$\endgroup$
1
$\begingroup$

In your two pair counting, you counted "AAKKJ" cases twice - once when your $\binom{13}{1}$ choice was the ace and your $\binom{12}{1}$ choice was the king, and once when they were in reverse.

On the other hand, the denominations of the full house case are not symmetric - you need to pick one denomination for the three cards, and one for the two. So if you pick one of the $\binom{13}{2}$ pairs of denominations, say, $\{A,K\}$, you still don't know whether to pick three aces or three kings.

So you can think of it as $2\binom{13}{2}\binom{4}{3}\binom{4}{2}$. But $2\binom{13}{2}=\binom{13}{1}\binom{12}{1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.