Why is this not a valid algorithm for counting poker hands?

Question

In my probability class, we had an example asking what is the probability of getting three of a kind when randomly drawing $5$ cards from a typical deck ($13$ denomination and $4$ suits, total of $52$ cards). Three of a kind is defines as getting three cards with the same denomination and two other cards which each have a unique denomination (looks like $AAABC$). The correct answer is as follows:

1) Choose denomination $A$, this can be done in $\binom{13}{1}$ ways

2) Choose the 3 suits for denomination $A$, this can be done in $\binom{4}{3}$ ways

3) Choose the remaining two denominations, $B$ and $C$, this can be done in $\binom{12}{2}$ ways

4) Choose the suits for the remaining two denominations, this can be done in $\binom{4}{1}$ ways for each

So the probability of getting three of a kind is: $$\frac {\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}}{\binom{52}{5}} \approx 0.02112845$$

Now my solution is mostly the same, however when selecting the denominations $B$ and $C$ I say that this can be done by selecting denomination $B$ in $\binom{12}{1}$ ways with $\binom{4}{1}$ possible suits. Then selecting denomination $C$ from the remaining $11$ denominations for any $4$ suits, so my answer is:

$$\frac {\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} \approx 0.0422569$$

which turns out to be exactly double the answer above, this comes from the fact that $\binom{12}{1}\binom{11}{1}=2\binom{12}{2}$. My question is, which solution is correct and why? To me they both seem like valid solutions. Am I double counting certain hands?

The order in which you chose $B$ and $C$ does not matter, and there are $2$ possible orders for those choices, which is exactly why $\binom{12}{1}\binom{11}{1}=2\binom{12}{2}$ — Justin Benfield, Jul 03 '16 at 22:38
Are you not allowed to have 4 cards of the same denomination? — Stack Exchange Broke The Law, Jul 04 '16 at 01:21
@immibis Not for three of a kind. Four of the same denomination would be four of a kind, which is rarer and thus bears three of a kind. — TripeHound, Jul 04 '16 at 08:55

score 6 · Accepted Answer · answered Jul 03 '16 at 22:46

The first answer is correct, in this case you're counting unordered sets, however in your solution when you write selecting $B$ and then $C$ separately you're in a sense counting ordered pairs. What I mean by this is consider choosing $3$ Aces($A$), a king ($K$) and a queen ($Q$)as your hand. Then your algorithm is counting the hands $AAAKQ$ and $AAAQK$ as distinct hands when they should be considered the same since you counted your sample space using unordered sets. This also explains why your answer was exactly twice the correct one, it's because you counted every possibility exactly twice with the order flipped.

score 2 · Answer 2 · answered Jul 03 '16 at 22:35

In your solution, suppose it is for three jacks and two other cards. With your method of counting the other two cards, the pair $(2S,3D)$ is counted again as $(3D,2S).$ So you're double counting the remaining cards, explaining your result being twice the right one.

Why is this not a valid algorithm for counting poker hands?

2 Answers2