This question is from Introduction to Probability, question 32b. I've seen the solution and I understand it clearly. $$\frac{\binom{13}{2}\times \binom{4}{2}\times \binom{4}{2} \times 44}{\binom{52}{5}}$$

The $\binom{13}{2}$ is from the different pairs of values I can have. The two $\binom{4}{2}$ is for the two same values from the possible four cards. The 44 is the fifth card I can have which is different from the previous four.

I came up with my own answer (which is obviously wrong) but I can't understand why it doesn't equal to the above. My own answer is $$\frac{\frac{52}{2!} \times \frac{(52-4)}{2!} \times 44} {3!}$$

$\frac{52}{2!}$ is for the first possible card I can choose. Therefore, the second one must be the same. The 2! is because order doesn't matter.

$\frac{52-4}{2!}$ is for the third possible card I can choose. Therefore, the fourth one must be the same.

44 is the rest of the card that I can choose.

3! is for the order of the two pairs and the last card doesn't matter.

It seems like I undercount it but I can't understand why. Please show me what I am missing. Thank you very much.

  • $\begingroup$ It's hard to know what you are missing, when you don't tell us how you arrived at your answer. You have analyzed nicely where each term in the correct answer comes from, but you have told us nothing about where any of the terms in your answer come from. $\endgroup$ Feb 12 at 8:17
  • $\begingroup$ You should probably write down your reasoning in the post, how exactly you get this answer. What’s your logic behind $52\times(52-4)$? $\endgroup$
    – Aig
    Feb 12 at 8:17
  • $\begingroup$ Sorry, I just added in the explanation. Thank you for viewing my question! $\endgroup$
    – Ron Luu
    Feb 12 at 8:23
  • $\begingroup$ Are you counting the pairs that are subsets of three-of-a-kinds and full houses, or just "pure pairs"? $\endgroup$
    – John
    Feb 12 at 8:25
  • $\begingroup$ The question is about numbers (not probabilities) so I would advice you just to leave out denominator $\binom{52}5$. $\endgroup$
    – drhab
    Feb 12 at 9:02

3 Answers 3


It appears you are not even accounting for the five cards. If I understand correctly, here's what you are doing.

Step 1. The first card can be any of the cards, so $52$ ways.

Take the $3$ other cards with the same rank / denomination out of consideration. (Why? Don't you need another card to make a pair?)

Step 2. Pick one of the $48$ cards remaining - $48$ ways.

Again exclude the $3$ other cards with the same rank. (Again, the same question as above.)

Step 3. Pick any of the $44$ cards remaining.

Hopefully you see the problem. There could be more problems. but I'll wait until you clarify if this is actually what you were thinking. And where you get making the necessary corrections.

Post OP's edit.

You say -

...the second one must be the same. The 2! is because order doesn't matter.

No. The second card cannot be the same. You are not repeating. It has to be of the same rank. So you have $3$ options. Meaning, you must multiply by $3$ - both for the first pair and the second.

The division by $2!$ at this stage is correct.

However, you need to divide by $2!$ at the end. Not by $3!$.

The odd card is already fixed to be the last. It's not leading to double-counting. Only the same two pairs are counted as $2$ instances.


Following your route and avoiding mistakes I arrive at:$$\frac{\frac{52\times3}{2!}\times\frac{48\times3}{2!}\times44}{2!}=2808$$

After choosing the first card of a pair there are $3$ ways to choose its "mate". You overlook that and speak of "the same".

There are $2$ pairs and to avoid double counting we must not divide by $3!$ but by $2!$. The single card that does not belong to any pair plays no part in this

  • $\begingroup$ You are awesome man. I was looking for a way to fix my answer instead of just using the solution answer. However, I'm still confused on the not diving by 3!. I thought that (33) (77) (A) would be the same as (A) (33) (77) or (33) (A) (77). So I divided 3! $\endgroup$
    – Ron Luu
    Feb 12 at 9:18
  • $\begingroup$ First we pick a pair. Then we pick a second pair. Finally we pick a single. This order is fixed. Then there are exactly $2$ ways to arrive at picking $(33)$, $(77)$ and $(A)$ which are: $((33),(77),(A))$ and $((77),(33),(A))$. To repair this double counting we must divide by $2$. $\endgroup$
    – drhab
    Feb 12 at 10:21

Starting out with the wrong characterization of the formula you present seems to have set you on the wrong path. Your first formula gives the probability that a random set of five distinct cards from a 52-card Poker deck will meet the criteria for "two pairs" but not for any better hand. That is, that there will be exactly three distinct ranks represented among the five cards, two of them by two cards each, and the other by the remaining card. As I wrote in comments, that's not the same as how many such hands there are, which is what you seem to have tried to compute, much less "How many possible pairs in a random 5-card poker hand?"

@Haris described how to count the number of possible "two pairs" hands, but the question was why your attempt did not match the formula, and if you follow through with their computation, you will find that it does not match the original formula either. That's of course because it is (correctly) answering a different question than the first formula answers.

If we understand the question answered by the first formula as "What is the probability of being dealt a 'two pairs' hand in 5-Card Draw", we can take an approach along lines somewhat similar to what you attempted. The probability is: $$ \big(\frac{52}{52} \times \frac{3}{51} \times \frac{1}{2!} \big) \times \big( \frac{48}{50} \times \frac{3}{49} \times \frac{1}{2!} \big) \times \frac{1}{2!}\times \frac{44}{48} \times 5!$$ That is, 1 for the first card, $\frac{3}{51}$ that the second card forms a pair with the first, but the order of these doesn't matter, $\frac{48}{50}$ that the third differs from the first two, $\frac{3}{49}$ that the fourth pairs with the third, but the order of these two doesn't matter, and the relative order of the two pairs doesn't matter, $\frac{44}{48}$ that the fifth doesn't pair with the others, and 5! orders for each set of five cards.

Now we rearrange to get: $$\frac{52 \times 3 \times 48 \times 3 \times 44}{2 \times 2 \times 2} \times \frac{5!}{52 \times 51 \times 50 \times 49 \times 48}$$ Now the right-hand fraction is $1 / \binom{52}{5}$, and we can refactor the left to get $$\frac{13 \times 12}{2} \times \frac{4 \times 3}{2} \times \frac{4 \times 3}{2} \times \frac{44}{\binom{52}{5}}$$ And behold! $\frac{13 \times 12}{2}$ is $\binom{13}{2}$ and $\frac{4 \times 3}{2}$ is $\binom{4}{2}$, so that gets us to $$\frac{\binom{13}{2}\times\binom{4}{2}\times\binom{4}{2}\times44}{\binom{52}{5}}$$ , which is the same as the original formula.


Note well that the above was written with the known formula in mind. Had that not been the case, I probably would not have arrived by that route at that specific formulation of the answer. It is easier to come to that particular formula more directly, by a line of reasoning similar to your initial explanation, and I would account such a route less prone to error.

Specifically, look up at my characterization of the problem. It is essential to work from a precise understanding of what you are trying to compute, and that characterization serves well. Quoting myself:

there will be exactly three distinct ranks represented among the five cards, two of them by two cards each, and the other by the remaining card.

From that, we can say

  • there are $\binom{13}{2}$ ways to choose two of the 13 ranks for the two distinct pairs;
  • for each of those two ranks, there are $\binom{4}{2}$ ways two choose two distinct cards with that rank;
  • for each set of two pairs, there are 44 ways to choose a fifth card whose rank differs from the ranks chosen for the pairs.

The product of all the above terms, $\binom{13}{2}\times\binom{4}{2}\times\binom{4}{2}\times 44$, is the number of distinct "two pairs" hands. To get to the probability of being dealt one of those hands, we observe that

  • $\binom{52}{5}$ is the total number of distinct 5-card hands.

The probability of being dealt a "two pairs" hand is the number of such hands divided by the number of all distinct 5-card hands, which is then the original formula.

  • $\begingroup$ You are such a legend man!!! Probability is such a hard subject but thanks to you, it's getting clearer for me!!! $\endgroup$
    – Ron Luu
    Feb 26 at 11:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .