# Thread: Where did I go wrong in 2 pairs poker probability question?

1. ## Where did I go wrong in 2 pairs poker probability question?

The question asks what the probability is of being dealt 2 pairs in a 5 card poker hand (from a shuffled deck of 52 cards).

So you need 1 pair and another numerically different pair (otherwise it will be four of a kind) and a final card which matches neither pair (otherwise it will be a full house).

The answer is approx. 0.0475 which can be found in plenty of places as the answer to this question with a web search and is explained here: https://math.stackexchange.com/a/1770961/484087

I got the answer wrong by exactly a factor of 2, i.e. 0.095. Clearly I have over-counted but I can not see what is wrong with the logic I used. Can someone explain where I went wrong please? Here's what I did:

Num. of combinations of 2 things and 2 things and 1 thing from 5 things: $\displaystyle \frac{5!}{2!·2!·1!}=\frac{120}{4}=30$

1st card can be anything: $\frac{52}{52}$
2nd card must match the 1st: $\frac{3}{51}$
3rd card must not match 1st or 2nd: $\frac{48}{50}$
4th card must match the 3rd: $\frac{3}{49}$
5th card must not match 1st, 2nd, 3rd, or 4th: $\frac{44}{48}$

$\displaystyle 30·\frac{52}{52}·\frac{3}{51}·\frac{48}{50}·\frac{ 3}{49}·\frac{44}{48}=\frac{396}{4165}≈0.095078$

2. ## Re: Where did I go wrong in 2 pairs poker probability question?

Originally Posted by mattstan
The question asks what the probability is of being dealt 2 pairs in a 5 card poker hand (from a shuffled deck of 52 cards). So you need 1 pair and another numerically different pair (otherwise it will be four of a kind) and a final card which matches neither pair (otherwise it will be a full house).
$\dbinom{13}{2}\dbinom{4}{2}\dbinom{4}{2}\dbinom{4 4}{1}=0.047539015606242496998799519807923169267707 082833$

SEE HERE

3. ## Re: Where did I go wrong in 2 pairs poker probability question?

Yes, I am aware of the correct answer, as I wrote and linked in my post.

My question was asking what the flaw was in the logic I used?

4. ## Re: Where did I go wrong in 2 pairs poker probability question?

Originally Posted by mattstan
Yes, I am aware of the correct answer.

My question was asking what the flaw was in the logic I used?
You are double counting the pairs that you will generate. For instance, you choose Aces for the first pair and Kings for the second pair, that is the same as if you choose Kings for the first pair and Aces for the second pair. So, instead of multiplying by 30 permutations, you should be multiplying by 15.

5. ## Re: Where did I go wrong in 2 pairs poker probability question?

Okay, I see. Thanks for pointing that out.

6. ## Re: Where did I go wrong in 2 pairs poker probability question?

Originally Posted by mattstan
Yes, I am aware of the correct answer, as I wrote and linked in my post.

My question was asking what the flaw was in the logic I used?
I really do not see a flaw,
$\dbinom{13}{2}\text{ pick two numbers }{2}\\\dbinom{4}{2}\text{ pick two cards of numbers}\\\dbinom{4}{2}\text{ pick two more cards of the other number}\\\dbinom{44}{1}\text{ pick one card from the numbers not used}$.

7. ## Re: Where did I go wrong in 2 pairs poker probability question?

Originally Posted by Plato
I really do not see a flaw,
$\dbinom{13}{2}\text{ pick two numbers }{2}\\\dbinom{4}{2}\text{ pick two cards of numbers}\\\dbinom{4}{2}\text{ pick two more cards of the other number}\\\dbinom{44}{1}\text{ pick one card from the numbers not used}$.
Look at the OP's logic in the first post, then look at my reply in reply #4 above. It explains the OP's logic and the reason it did not give the same answer.

8. ## Re: Where did I go wrong in 2 pairs poker probability question?

Originally Posted by mattstan
1st card can be anything: $\frac{52}{52}$
2nd card must match the 1st: $\frac{3}{51}$
2nd card can also be anything.....51/51

9. ## Re: Where did I go wrong in 2 pairs poker probability question?

Originally Posted by DenisB
2nd card can also be anything.....51/51
This is a far more complicated way of looking at the problem, as you now need to account for the possibility that the card was the same as the previous card and then for if it is different. While it is possible to approach the problem this way, the OP's logic is more concise.

10. ## Re: Where did I go wrong in 2 pairs poker probability question?

Agree. I was looking at it as in a "looper/simulation" computer program (dumb code).