By poker_boy | 1 Comment
Last updated: Monday, August 1, 2011 | 69 Views Tags: card poker, cards, full house, poker hand, poker hands, probability question, two pairs
This week’s question is as follows:
Question: How many ways are there to get a five card poker hand with two pairs? i.e. AAJJ5
This is what I get:
13C1 * 4C2 * 12C1 * 4C2 * 11C1 * 4C1 = 247104
The actual answer is 247104 / 2 = 123552
Why do you need to divide by 2? I think it has something to do with the order you pick the cards for the pairs, but that doesn’t make sense to me. To get a full house, I would just do this:
i.e. KKKQQ
13C1 * 4C3 * 12C1 * 4C2 , So here why do I not need to divide by 2? I could pick a K first or a Q first.
Answer: Dividing by 2 is necessary because, for example, two aces and two eights are the same hand as two eights and two aces. By comparison, a full house with three aces and two eights is different from a full house with three eights and two aces.
When calculating these probabilities, this is a subtle but critical point to understand. I know this from experience!
Comments
1 comment
Mike Smith
August 1, 2011
I asked exactly the same question three years ago. You can look at my profile and the question (it is the first question of the 11 i posed so far) to see the answers.
You just counted double.
You have to see it this way. There are three types of cards in your hand :
the first pair A, the second pair B and the isolated card C
We can arrange these in several ways :
There are 5 possible positions for the isolated card
For the 4 positions of the two pair, there are 3 possibilities instead of 6 :
AABB
ABAB
ABBA
(BBAA, BABA, BAAB is just the same, we don’t know which pair A is
so interchanging A with B doesn’t yield new position-configurations)
So because there are only 3 possible positions instead of 4!/2!2! = 6
for the two pair, we have to divide by 2.